Vedanta Excel in Mathematics Book 9 Final (2080)_compressed

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 151 Vedanta Excel in Mathematics - Book 9 Sequence and Series The first term (t1 ) = 2 × 1 + 3 = 5 The second term (t2 ) = 2 × 2 + 3 = 7 The third term (t3 ) = 2 × 3 + 3 = 9 The fourth term (t4 ) = 2 × 4 + 3 = 11 Thus, the required the first four terms 5, 7, 9, 11 (ii) The given general term (tn) = (–1) n + 1 (n + 7) The first term (t1 ) = (–1) 1 + 1 (1 + 7) = (–1)2 (8) =1 × 8 = 8 The second term (t2 ) = (–1) 2 + 1 (2 + 7) = (–1)3 (9) = –1 × 9 = –9 The third term (t3 ) = (–1) 3 + 1 (3 + 7) = (–1)4 (10) =1 × 10 = 10 The fourth term (t4 ) = (–1) 4 + 1 (4 + 7) = (–1)5 (11) = –1 × 11 = –11 Thus, the required the first four terms are 8, –9, 10 and –11. The general term of linear sequence The sequence in which each pair of successive terms have a constant difference is called a linear sequence. The general term of linear sequence is given by tn = an + b; where a = common difference and b = first term – common difference Example 3: Find the nth term of the sequence 7, 10, 13, 16, … Solution: Here, the sequence 7, 10, 13, 16, … has the common difference 3. So, it is a linear sequence. Let, the nth term (tn) be tn = an + b Since, the common difference = 3. So, a = 3 Now, the first term (t1 ) = 7 or, 3 × 1 + b = 7 [∵ t n = an + b] or, b = 4 Hence, tn = 3n + 4 The general term of quadratic sequence In quadratic sequence the first difference between each pair of consecutive term is not constant. However, the second differences are constant. The general term of quadratic sequence is given by tn = an2 + bn + c. Example 4: Find the general term of the sequence 0, 5, 12, 21, 32, … Solution: Let, the nth term (tn) of the sequence be tn = an2 + bn + c The second constant difference = 2. So, it is quadratic sequence. Now, 2a = 2 ∴a = 1 Also, the first term (t1 ) = 0 Alternatively, In 7, 10, 13, 16, … t 1 = 7 = 3×1 + 4 t 2 = 10 = 3×2 + 4 t3 = 13 = 3×3 + 4 …………………… ∴t n = 3n + 4 7 10 +3 +3 +3 13 16 0 5 5 2 2 2 7 9 11 12 21 32 1st diff→ 2nd diff→

Vedanta Excel in Mathematics - Book 9 152 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur or, 1 × 12 + b × 1 + c = 0 [∵tn = an2 +bn c] or, b = – c – 1 … (i) Again, the second term (t2 ) = 5 or, 1 × 22 + b × 2 + c = 5 [∵ t n= an2 +bn+c] or, 2b + c = 1 … (ii) Putting the value of b from equation (i) in (ii), we get 2 (–c – 1) + c = 1 or, –2c –2 + c = 1 or, c = –3 Putting the value of c in equation (i), we get b = 3 – 1 = 2 Hence, tn = n2 + 2n – 3 Facts to remember 1. The general term can be generated by using formula or making the patterns of numbers. 2. (i) If the second common difference, 2a = 1 then tn = an2 +bn + c contains 1 2 n2 . (ii) If the second common difference, 2a = 2 then tn = an2 +bn + c contains n2 . (iii) If the second common difference, 2a = 3 then tn = an2 +bn + c contains 3 2n2 and so on. Example 5: The nth term of a sequence is given by an = an-1 + an-2, n>2. If a1 = a2 = 1, find the first six terms of the sequence. Solution: Here, an = an-1 + an-2 and a1 = a2 = 1 On putting n = 3, a3 = a3-1 + a3-2 = a2 + a1 = 1 + 1 = 2 On putting n = 4, a4 = a4-1 + a4-2 = a3 + a2 = 2 + 1 = 3 On putting n = 5, a5 = a5-1 + a5-2 = a4 + a3 = 3 + 2 = 5 On putting n = 6, a6 = a6-1 + a6-2 = a5 + a4 = 5 + 3 = 8 Hence, the first six terms are 1, 1, 2, 3, 5 and 8. Example 5: Study the pattern in the given figures and answer the questions. (i) Draw one more figures in the same pattern. (ii) Find the nth term of the sequence of number of dots. (iii) Find the 7th term of the sequence. Solution: (i) One more figures in the same patterns is given alongside Alternatively, 0, 5, 12, 21, 32, ... t 1 = 0 = 12 - 1 = 12 + (2×1 – 3) t 2 = 5 = 22 + 1=22 + (2×2 – 3) t3 = 12 = 32 + 3 = 32 + (2×3 – 3) t 4 = 21 = 42 + 5 = 42 + (2×4 – 3) ………………………………… ∴t n = n2 + 2n – 3 Sequence and Series , , , ...

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 153 Vedanta Excel in Mathematics - Book 9 Sequence and Series (ii) The sequence of number of dots is 1, 5, 9, … Constant difference (d) = 5 – 1 = 9 – 5 = 4 Now, t 1 = 1 = 4 × 1 – 3 t 2 = 5 = 4 × 2 – 3 t 3 = 9 = 4 × 3 – 3 …………………… ∴t n = 4n – 3 (iii) The 7th term (t7 ) = 4 × 7 – 3 = 25 EXERCISE 8.1 General section 1. Define the following terms with an example. a) Sequence b) Finite sequence c) Infinite sequence 2. What are the next two terms of each of the following sequences? a) 4, 9, 14, 19, … b) 20, 17, 14, 11, … c) 1, 2, 4, 8, … d) 96, 48, 24, 12, … e) 1, 4, 9, 16, … f) 1 4, 2 9, 3 16, 4 25 , ... 3. Find the first four terms of the sequences whose general terms are given below. a) tn = 3n + 2 b) tn = 5n – 7 c) an = n2 + n d) an = (-1)n. (n+1) e) tn = n(n + 1) 2 f) tn = n – 1 n + 1 g) tn = 2n – 1 (n + 1)2 h) tn = (–1)n.n n2 + n + 1 Creative section 4. Find the nth terms of the following sequences. Also, find the 6th and 9th terms of each sequence. (a) 4, 8, 12, 16, … (b) 6, 11, 16, 21, … (c) 9, 5, 1, –3, … (d) 1, 4, 9, 16, … (e) 2, –4, 8, –16, …. (f) 1 2 , 2 3, 3 4, 4 5, ... (g) 1 4 , 4 9 , 9 16, 16 25 (h) 3, 6, 11, 18, … 5. a) a1 , a2 , a3 ,… are the terms of a sequence. If a1 = 3, an = 2an–1, find the first five terms of the sequence. b) If t1 = 4, t2 = –3 and tn = tn–1 + tn–2, n≥ 3, n∈N, find the first six terms of the sequence. c) The general term of a sequence is defined as an = 2n + 1 when n ∈ N is odd n2 { + 1 when n ∈ N is even. Find its 7th and 10th terms. 6. Study the following patterns in each of the figures and answer the following questions for each case. (i) Draw one more figures in the same pattern. (ii) Find the general term of the sequence of number of dots. (iii) Find the 10th term of the sequence. a) , , , ... b) 1, 5, 9, ... +4 +4 , , , ...

Vedanta Excel in Mathematics - Book 9 154 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) , , , ... d) 8.3 Series The sum of the terms of a sequence is called the series corresponding to the sequence. For examples a) The series corresponding to the sequence 2, 5, 8, 11, 14 is 2 + 5 + 8 + 11 + 14. b) The series corresponding to the sequence 5, 3, 1, -1, -3 is 5 + 3 + 1 – 1 – 3 Facts to remember (i) If a series has finite number of terms, then it is called a finite series. For example, 2 + 4 + 6+ …+ 20 is a finite series. (ii) If a series has infinite number of terms, then it is called an infinite series. For example, 1 + 3 + 5+ … is an infinite series. 8.4 Sigma notation If a1 , a2 , a3 , …, an is a finite sequence, then the series corresponding to this sequence is Sn = a1 + a2 + a3 + … + an This sum can be represented by using sigma notation as n n = 1 an , where the Greek letter ‘Σ’ is the summation notation and n n = 1 an read as ‘summation of an , when n runs from 1 to n’. 8.5 Partial sum The sum of the finite series a1 , a2 , a3 , …, an is Sn = a1 + a2 + a3 + … + an = n n = 1 an The sum of limited terms of the series is called the partial sum. Here, S1 = a1 , S2 = a1 + a2 = 2 n = 1 an, S3 = a1 + a2 + a3 = 3 n = 1 an, S4 = a1 + a2 + a3 +a4 = 4 n = 1 an and so on. are the partial sum of n n = 1 an. Let’s take 3rd and 4th partial sums of Sn. Then, S3 = a1 + a2 + a3 and S4 = a1 + a2 + a3 +a4 Now, S4 – S3 = (a1 + a2 + a3 +a4 ) – (a1 + a2 + a3 ) = a4 . It shows that the nth term of the series is calculated by using the following relation. an = Sn – Sn – 1, for n>1 Facts to remember 1. In partial sum q n = p an , the number of terms = (q – p) + 1 2. The general term (an) = Sn – Sn-1, for n>1. , , , ... Sequence and Series

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 155 Vedanta Excel in Mathematics - Book 9 Sequence and Series Worked-out Examples Example 1: Find the value of 7 n = 4 (3n + 2) by expanding. Solution: Here, 7 n = 4 (3n + 2) = (3 × 4+2) + (3 × 5+2) + (3 × 6+2) + (3 × 7+2) = 14 + 17 + 20 + 23 = 74 Hence, the value of 7 n = 4 (3n + 2) is 74. Example 2: Find the nth term and then express the following series in sigma notation. (i) 7 + 11 + 15 + 19 +23 (ii) –1+2–4+16 –32+…to 9 terms (iii) 2+5+10+17+…to 10 terms (iv) 1 4 + 3 9 + 5 16 + 7 25 + ... to 15 terms Solution: (i) Here, the constant difference = 11– 7 = 4 The 1st term (t1 ) = 7 = 4 × 1 + 3 The 2nd term (t2 ) = 11 = 4 × 2 + 3 The 3rd term (t3 ) = 15 = 4 × 3 + 3 The 4th term (t4 ) = 19 = 4 × 4 + 3 ………………………………………….. ∴The nth term (tn) = 4n + 3 Hence, 7 + 11 + 15 + 19 + 23 = 5 n = 1 (4n +3) (ii) Here, the terms are powers of 2 having – and + sign in alternate places. The 1st term (t1 ) = –1 = (–1)1 .20 = (–1)1 .21–1 The 2nd term (t2 ) = 2 = (–1)2 .21 = (–1)2 .22–1 The 3rd term (t3 ) = –4 = (–1)3 .22 = (–1)3 .23–1 The 4th term (t4 ) = 8 = (–1)4 .23 = (–1)4 .24–1 ……………………………………………………… ∴The nth term (tn) = (–1)n.2n–1 Hence, –1 + 2 – 4 + 8 – 16 +… to 9 terms = 9 n = 1 (–1)n.2n – 1 (iii) Here, the first differences are, 5 – 2 = 3, 10 – 5 = 5, 17 – 10 = 7, …. The second differences are, 5 – 3 = 2, 7 – 5 = 2. Since, the second differences are constant. So, 2 + 5 + 10 + 17 + … is a quadratic series. The 1st term (t1 ) = 2 = 1+1 = 12 + 1 The 2nd term (t2 ) = 5 = 4 +1= 22 + 1 The 3rd term (t3 ) = 10 = 9 +1= 32 + 1 The 4th term (t4 ) = 17 = 16 + 1=42 + 1 ……………………………………………………… ∴The nth term (tn) = n2 + 1 Hence, 2 + 5 + 10 + 17 + … to 10 terms = 10 n = 1 (n2 + 1) 2 5 3 2 2 5 7 10 17

Vedanta Excel in Mathematics - Book 9 156 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (iv) Here, the sequence of numerators is 1, 3, 5, 7, … which has the constant difference of 2 and the sequence of denominators is 4, 9, 16, 25, … which are perfect square numbers. The 1st term (t1 ) = 1 4 = 2 × 1 – 1 22 = 2 × 1 – 1 (1 + 1)2 The 2nd term (t2 ) = 3 9 = 2 × 2 – 1 32 = 2 × 2 – 1 (2 + 1)2 The 3rd term (t3 ) = 5 16 = 2 × 3 – 1 42 = 2 × 3 – 1 (3 + 1)2 The 4th term (t4 ) = 7 25 = 2 × 4 – 1 52 = 2 × 4 – 1 (4 + 1)2 ……………………………………………………… ∴The nth term (tn) = 2n – 1 (n + 1)2 Hence, 1 4 + 3 9 + 5 16 + 7 25 to 15th terms = 15 n =1 2n – 1 (n + 1)2 EXERCISE 8.2 General section 1. Define series with an example. 2. Identify the following as sequence or series. a) 2, 4, 8, 10, … b) 1+ 3+ 5+ 7+ … c) 3, –1, –5, –9, .. d) 8 + 4+0 – 4 – 8 … 3. Write down the series associated with the following sequences. a) 5, 10, 15, 20, 25 b) 1, –2, 4, –8, 16 c) –7, –3, 1, 5, … d) 1, 1 2 , 1 4, 1 8 , ... 4. Find the number of terms of the following partial sums. a) 5 n = 1 (2n + 1) b) 6 n = 3 5n c) 8 n = 4 (–1)k .k d) 10 i = 3 (i2 – i) 5. Evaluate the following sums by expanding. a) 5 n = 2 (3n + 2) b) 7 n = 3 (5n – 1) c) 4 n = 1 (n2 + 3) d) 5 k = 1 2k e) 5 n = 1 (3n – n3 ) f) 8 i = 4 (–1)i .i g) 5 n = 1 [1 + (–1)n] Creative section 6. a) If 5 n = 1 (2n + k) = 45, find the value of k. b) If 5 n = 2 (2n + m) = 48, find the value of m. 7. t n and Sn are the nth term and sum of the first n terms of a series respectively. a) If Sn = n(n + 1) 2 , find S3 , S4 and t4 . b) If Sn = n(n + 1) (n + 2) 6 , find S4 , S5 and t5 . 8. Find the general term of each of the following series and express them in sigma notation. a) 2 + 5 + 8 + 11 + 14 + 17 b) –10 – 6 – 2 + 2 + 6 + 10 Sequence and Series

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 157 Vedanta Excel in Mathematics - Book 9 Sequence and Series c) 3 + 9 +27 + 81 + 243 d) 1 + 4 + 16 + 64 + …. to 9 terms e) 1 – 2 + 3 – 4 + 5 - ….. to 10 terms f) –2 + 4 – 8 + 16 – 32 + … to 12 terms g) 1 2 + 2 3 + 3 4 + 4 5 + ... to 7 terms h) 1 4 + 4 9 + 9 16 + 16 25 + ... to 9 terms i) 2 3 + 3 5 + 4 7 + 5 9 + ... to 8 terms j) 5 4 – 10 9 + 15 16 – 20 25 + ... to 8 terms k) 2 + 6 + 12 + 20 + 30 + 42 + 56 l) 6 + 11 + 18 + 27 + … to 10 terms 8.6 Arithmetic sequence Let’s make the following figures using match sticks (i) What is the sequence of numbers of match sticks required for each figure? 3, 5, 7, 9, … (ii) Can we find the difference between the successive terms? 5 – 3 = 2, 7 – 5 = 2, 9 – 7 = 2. Here, the difference between successive terms is always 2. This illustration leads us to get idea of an arithmetic sequence. Facts to remember 1. A sequence in which the difference between any two consecutive terms remains constant is known as arithmetic sequence (A.S.) or arithmetic progression (A.P.) 2. The constant difference between the consecutive terms of an A.P. is called common difference. 3. The common difference of an A.P. can be positive, negative or zero. 4. Common difference (d) = t2 – t1 or t3 – t2 or t4 – t3 8.7 Terms and Common Difference of an A.P. Let, ‘a’ be the first term of an A.P. and ‘d’ be the common difference. Then, the terms of the A.P. can be written as follows. t 1 = a = a + (1 – 1) d t 2 = a + d = a + (2 – 1) d t 3 = a + 2d = a + (3 – 1) d t 4 = a + 3d = a + (4 – 1) d ……………………………… t n = a + (n – 1) d Thus, the general term (nth) term, (tn) = a + (n – 1) d Worked-out Examples Example 1: Check whether the following sequences are in A.P. or not. (i) 4, 7, 10, 13, 16, … (ii) 1, 5, 25, 125, … Solution: Here, to check whether the given sequence is an A.P. or not, it is enough to check the differences between the consecutive terms. , , , +d +d +d t 1 = a t 2 = a + d t 3 = a + 2d t 4 = a + 3d

Vedanta Excel in Mathematics - Book 9 158 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur (i) t 2 – t1 = 7 – 4 = 3, t3 – t2 = 10 – 7 = 3, t4 – t3 = 13 –10 = 3, t5 – t4 = 16 – 13 = 3 and so on. Since, the difference between any two consecutive terms is equal, the given sequence is an A.P. (ii) t 2 – t1 = 5 – 1 = 4, t3 – t2 = 25 – 5 = 20, t4 – t3 = 125 –25 = 100 and so on. Since, the difference between any two consecutive terms is not equal, the sequence is not an A.P. Example 2: Find the general term (tn) of the arithmetic sequence -2, 1, 4, 7, … Solution: Here, the first term (a) = -2, common difference (d) =t2 – t1 = 1 – (–2) = 3, nth term (tn) =? Now, tn = a + (n – 1) d = –2 + (n – 1) × 3 = –2 + 3n – 3 = 3n – 5 Hence, the general term of the sequence (tn) = 3n – 5. Example 3: Find the 10th term of an arithmetic sequence 7, 11, 15, … Solution: Here, the first term (a) = 7, common difference (d) = t2 – t1 = 11 – 7 = 4, 10th term (t10) =? Now, tn = a + (n – 1) d ∴t 10 = 7 + (10 – 1) × 4 = 7 + 36 = 43 Hence, the 10th term of the sequence is 43. Example 4: How many terms are there in the arithmetic series 3+15+27+…+123? Solution: Here, the first term (a) = 3, common difference (d) = t2 – t1 = 15 – 3 = 12, nth term (tn) =123, number of terms (n) =? Now, t n = a + (n – 1) d or, 123 = 3 + (n – 1) × 12 or, 123 = 3 + 12n – 12 or, 132 = 12n or, n = 11 Hence, the required number of terms is 11. Example 5: Is 88 a term of the series 11+ 18+ 25 + ……? Solution: Here, the first term (a) = 11, common difference (d) = t2 – t1 = 18 – 11 = 7 Let, nth term (tn) =88 Checking Last term = 11th term = a + 10d = 3 + 10 × 12 = 123 Which is given in the question. Sequence and Series

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 159 Vedanta Excel in Mathematics - Book 9 Sequence and Series Now, t n = a + (n – 1) d or, 88 = 11 + (n – 1) × 7 or, 88 = 11 + 7n – 7 or, 84 = 7n or, n = 12 Yes, 88 is the 12th term of the series. Example 6: The third and the thirteenth terms of an arithmetic series are –40 and 0 respectively. Find the 28th term. Solution: Let ‘a’ and ‘d’ be the first term and the common difference of an AP respectively. Then, 3rd term (t3 ) = –40 ∴a + 2d = –40 … (i) Also, 13th term (t13) = 0 ∴a +12d = 0 … (ii) Now, subtracting (i) from (ii), we get a + 12d – (a + 2d) = 0 – (–40) or, a + 12d – a – 2d = 40 or, 10d = 40 ∴ d = 4 Also, putting the value of ‘d’ in (i), we get a + 2×4 = –40 ∴ a = –40 – 8 = –48 Again, 28th term (t28) = a + 27d = –48 + 27× 4 = –48 +108 = 60 Example 7: If 6 times the 6th term of an A.P. is equal to 9 times its 9th term, find the 15th term. Solution: Let ‘a’ and ‘d’ be the first term and the common difference of an AP respectively. Then, 6th term (t6 ) = a + 5d and 9th term (t9 ) = a + 8d According to question, 6t6 = 9t9 or, 6 (a + 5d) = 9 (a + 8d) or, 6a + 30d = 9a + 72d or, –3a = 42d or, a = –14d … (i) Again, 15th term (t15) = a + 14d = –14d + 14d = 0 [Using (i)] Example 8: A taxi-meter shows Rs 50 at the time of starting and then runs up by Rs 50 for each additional kilometer travelled during 6:00 am to 9:00 pm. If Mr. Gurung travelled a journey of 7 km, how much fare did he pay for the journey? Solution: Here, the sequence of taxi fare for the journey of 1 km, 2 km, 3 km, … is Rs (50 + 50), Rs (50 + 2 × 50), Rs (50 + 3 × 50), … i.e., Rs 100, Rs 150, Rs 200, … The first term (a) = Rs 100, common difference (d) = Rs 50, number of terms (n) = 7 Now, tn = a + (n – 1) d ∴t 7 = 100 + (7 – 50) × 36 = 400 Hence, he paid Rs 400 for travelling a journey of 7 km. Checking Last term = 12th term = a + 11d = 11 + 11 × 7 = 88 Which is given in the question. Checking Third term = a + 2d = –48 + 2 × 4 = –40 Thirteenth term = a + 12d = –48 + 12 × 4 = 0 Which are given in the question.

Vedanta Excel in Mathematics - Book 9 160 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 9: Every year, a tree grows 10 cm less than it does in the preceding year. If it grows by 2 m in the first year, in how many years will it have stop growing? Solution: Here, the sequence of heights of a tree in the 1st year, 2nd year, 3rd year, … is 200 cm, (200 – 10)cm, (200 – 2 × 10) cm, … i.e., 200 cm, 190 cm, 180 cm, ….. The first term (a) = 200 cm, common difference (d) = 190 – 200 = –10, Last term (tn) = 0, of terms (n) =? Now, t n = a + (n – 1) d or, 0 = 200 + (n – 1) × (–10) or, 0 = 200 – 10n + 10 or, 0 = 210 – 10n or, 10n = 210 or, n = 21 Hence, the tree would stop growing after 21 years. EXERCISE 8.3 General section 1. Define: (a) arithmetic sequence. (b) common difference 2. (a) Write down the formula for finding the nth term (tn) when the first term (a), common difference (d) and number of terms (n) of an arithmetic sequence are given. b) What is the 6th term of an A.P. having the first term ‘a’ and the common difference ‘d’? 3. Examine whether the following sequences form an A.P. or not. a) 2, 5, 8, 11, … b) 7, 3, –1, –5, … c) 2, 4, 8, 16, … d) 0.2, 0.22, 0.222, …. 4. The first term (a) and common difference (d) of the arithmetic sequences are given below. Find their general terms. a) a = 3, d = 2 b) a = –5, d = 4 c) a = 10, d = –7 d) a = –2, d = –3 5. a) Find the 7th term of the sequence 3, 6, 9, 12, … b) What is the 10th term of the sequence 7, 13, 19, 25, …? c) Find the 15th term of an arithmetic sequence 125, 120, 115, 110, ... 6. a) If the fifth term of an arithmetic sequence with first term 2 is 14, find the common difference. b) The first and 8th terms of an A.P. are –5 and 23, find the common difference. c) The common difference of an arithmetic series is 5 and its tenth term is 40, find its first term. d) The first term of an AP is 14, common difference is 7 and last term is 98, find the of terms. 7. a) How many terms are there in the sequence 7, 12, 17, …., 82? b) How many terms are there in the sequence 3, 6, 9, ..., 111? c) Find the of terms of the series 25 + 50 + 75 + 1000. 8. a) Which term of the series 2 + 5 + 8 + .. is 56? b) Which term of the sequence 6, 9, 12, 15… is 66? 9. a) Is 49 a term of the sequence 1, 9, 17, 25, …? b) Is 111 a term of the A.P. 3, 6, 9, 12, ….? c) Does 36 belongs to the sequence 8, 11, 14, 17, …? Sequence and Series

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 161 Vedanta Excel in Mathematics - Book 9 Sequence and Series Creative section - A 10. a) If the fifth and the tenth terms of an arithmetic sequence are 14 and 29 respectively. Find the first term and the common difference. Also, find the 17th term. b) If the third and the ninth terms of arithmetic sequence are 15 and 39 respectively. Find the 15th term. c) If the 5th and the 12th terms of an arithmetic series are –16 and 5 respectively, find its 20th term. 11. a) If the fifth and the tenth terms of an arithmetic sequence are 23 and 48 respectively, find the series. b) The fourth and the sixth terms of an arithmetic sequence are 41 and 35 respectively, find the series. 12. a) If the sixth and the ninth terms of an arithmetic sequence are 52 and 76 respectively, which term is 100? Find it. b) If the 5th and the 10th terms of an arithmetic sequence are 30 and 55 respectively, which term is 80? Find it. 13. a) If 6 times the 6th term of an arithmetic sequence is equal to 9 times its 9th term, show that its 15th term is zero. b) If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, find the 18th term. Creative section - B 14. a) The starting salary of a lady working as receptionist in an office is Rs 15,000 per month. If she gets an increment of Rs. 2,000 every year, what will be her salary in the 5th year? b) A taxi-meter shows Rs 50 at the time of stating and then runs up by Rs 50 for each additional kilometer travelled during 6:00 am to 9:00 pm. If Mrs. Rai travelled a journey of 5 km by the taxi, how much fare did she pay for journey? c) A cinema hall has managed 20 rows of seats and there are 12 seats in the front row. If each successive row contains two additional seats than its front row, how many seats are there in the last row? 15. a) Suppose that you are attending an interview for a job and the company gives you the following two offers for 5 years. Offer A: Rs 20,000 salary to start with followed by an annual increment of Rs 2,500. Offer B: Rs 25,000 salary to start with followed by an annual increment of Rs 1,500. Which offer would you choose and why? b) A dealer sold 50 computers in B.S. 2077, 70 computers in B.S. 2078, 90 computers in B.S. 2079 and so on forming an arithmetic sequence. (i) How many computers will it sell in B.S. 2083? (ii) In which year, will the dealer sell 250 computers? 16. a) Four people can sit in each table in a restaurant. When two tables are placed together, 6 people can be seated. When three tables are placed together to form a long table, 8 people can be seated. How many such tables should be placed together to form a long table such that 16 people could have the dinner together? b) On the first day strike of physicians in a hospital, the attendance of the OPD was 450 patients. As the strike continued, the attendance declined by 50 patients every day, find from which day of strike, the OPD would have no patient?

Vedanta Excel in Mathematics - Book 9 162 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) Every year, a tree grows 5 cm less than it did in the preceding year. If it grows by 1 m in the first year, in how many years would it stop growing? d) A construction company will be penalized each day of delay in construction for bridge. The penalty will be Rs 25,000 for the first day and increases by Rs 5,000 for each following day. Based on this budget, the company can afford to pay a maximum of Rs 1,00,000 toward penalty. Find the maximum number of days by which the completion of work can be delayed. Project work and activity section 17. a) Think the values of the first term (a) and common difference (d) of your choice for the arithmetic sequence. Then, find the general term and the first five terms of the sequence. b) Measure the heights of the first four steps of the stair/ladder from its base which is available in your house/school. Find the general term formula for the heights of the steps from the base and the height of the last step. 8.8 Geometric sequence Let’s consider that a viral disease is spreading in a way such that at any stage two new persons get affected from an infected person. At first stage, one person is infected, at second stage two persons are infected and is spreading to four persons and so on. Then, number of persons infected at each stage are 1, 2, 4, 8, ... where except the first term, each term is twice the previous term. This idea helps us to get the concept of geometric sequence. Facts to remember 1. A sequence in which each term, except first term, is obtained on multiplying the preceding term by a fixed non-zero number is called a geometric sequence. The fixed number is called common ratio and denoted by r. 2. A sequence in which the ratios between any two consecutive terms are constant, is known as geometric sequence (G.S.) or geometric progression (G.P.) 3. Common ratio (r) = t 2 t 1 or t 3 t 2 or t 4 t 3 and so on. 8.9 Terms and Common Ratio of a G.P. Let, ‘a’ be the first term of a G.P. and ‘r’ be the common ratio. Then, the terms of the G.P. can be written as follows. t 1 = a = ar0 = ar1-1 t 2 = ar= ar2-1 t 3 = ar2 = ar3-1 t 4 = ar3 = ar4-1 ……………… t n = arn-1 Thus, the general term or (nth) term, (tn) = arn-1, r ≠ 0 ×r ×r ×r t 1 = a t 2 = ar t 3 = ar2 t 4 = ar3 Sequence and Series

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 163 Vedanta Excel in Mathematics - Book 9 Sequence and Series Worked-out Examples Example 1: Examine whether the following sequences are in G.P. or not. (i) 2, 6, 18, 54, … (ii) 3, 33, 333, 3333,… (iii) 8, 12, 18, 27, … Solution: Here, to check that the given sequence is in G.P., it is sufficient to check if the ratios between the consecutive terms are equal or not. (i) t 2 t 1 = 6 2 = 3, t 3 t 2 = 18 6 = 3, t 4 t 3 = 54 18 = 3, … Since, the ratios between the two consecutive terms are equal, the given sequence is a G.P. (ii) t 2 t 1 = 33 3 = 11, t 3 t 2 = 333 33 = 111 11 , t 4 t 3 = 3333 333 = 1111 111 , … Since, the ratios between the two consecutive terms are not equal, the given sequence is not a G.P. (iii) t 2 t 1 = 12 8 = 3 2 , t 3 t 2 = 18 12 = 3 2, t 4 t 3 = 27 18 = 3 2 , … Since, the ratios between the two consecutive terms are equal, the given sequence is a G.P. Example 2: Find the 7th term (t7 ) of the geometric sequence 120, 60, 30, 15,… Solution: Here, the first term (a) = 120, common ratio (r) = t 2 t 1 = 60 120 = 1 2 , number of terms (n) = 7, 7th term (t7 ) =? Now, tn = arn-1 ∴ t 7 = 120 × ( 1 2)7 – 1 = 120 × ( 1 2) 6 = 120 × 1 64 = 15 8 Hence, the 7th term is 15 8 . Example 3: The fourth term of a GP whose first term is 2 is 54, find the common ratio. Solution: Here, the first term (a) = 2, 4th term (t4 ) = 54, common ratio (r) =?, Now, tn = arn-1 or, t 4 = 2 × r4-1 or, 54 = 2 × r3 or, 27 = r3 or, r3 = 33 or, r = 3 Hence, the common ratio is 3.

Vedanta Excel in Mathematics - Book 9 164 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 4: How many terms are there in the geometric series 16 + 24 + 36 + …+ 81? Solution: Here, the first term (a) = 16, common ratio (r) = 24 16 = 3 2, nth term (tn) =81, number of terms (n) =? Now, t n = arn-1 or, 81 = 16 × ( 3 2) n – 1 or, 81 16 = (3 2) n – 1 or ( 3 2) 4 = (3 2) n – 1 or, n – 1 = 4 or, n = 5 Hence, there are 5 terms in the given series. Example 5: Which term of GP 1 + 1 2 + 1 4 +….. is 1 128? Solution: Here, the first term (a) = 1, common ratio (r) = 1 2 1 = 1 2 Let, nth term (tn) = 1 128 Now, t n = arn-1 or, 1 128 = 1× ( 1 2 ) n – 1 or, ( 1 2) 7 = (1 2) n – 1 or, n – 1 = 7 or, n = 8 Hence, 1 128 is the 8th term of the series. Example 6: The fifth and the eighth terms of a G.P. are 8 and 64 respectively. Find the 10th term. Solution: Let ‘a’ and ‘r’ be the first term and the common ratio of a G.P. respectively. Then, 5th term (t5 ) = 8 ∴ar4 = 8 … (i) Also, 8th term (t8 ) = 64 ∴ ar7 = 64 … (ii) Now, dividing equation (ii) by equation (i), we get ar7 ar4 = 64 8 or, r3 = 8 or, r = 2 Also, putting the value of ‘r’ in (i), we get a × 24 = 8 or, 16a = 8 ∴ a = 8 16 = 1 2 Checking Last term = 5th term = ar4 = 16 × (3 2) 4 = 81 Which is given in the question. Checking Last term = 8th term = ar7 = 1 × (1 2) 7 = 1 128 Which is given in the question. Checking Fifth term = ar4 = 1 2 × (2)4 = 8 Eighth term = ar7 = 1 2 × (2)7 = 64 Which are given in the question. Sequence and Series

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 165 Vedanta Excel in Mathematics - Book 9 Sequence and Series Again, 10th term (t10) = ar9 = 1 2 × 29 = 1 2 × 512 = 256 Hence, the 10th term of the G.P. is 256. Example 7: A man joined a company as the post of a manager. The company gave him a starting salary of Rs 80,000 and agreed to increase his salary 10% annually. What will be his salary after 5 years? Solution: Here, the sequence of monthly salary in the 1st, 2nd, 3rd years is Rs 80,000, Rs (110% of Rs 80,000), Rs (110% of 110% of Rs 80,000), …. i.e., Rs 80,000, Rs 88,000, Rs 96,800,… The first term (a) = Rs 80,000, common ratio (r) = 88000 80000 = 11 10, number of terms (n) = 5 Now, tn = arn-1 ∴t 5 = 80000 × (11 10) 5 – 1 = 80000 × (11 10) 4 = 80000 × 14641 10000 = 1,17,128 Hence, his salary will be Rs 1,17,128 after 5 years. EXERCISE 8.4 General section 1. Define: (a) geometric sequence. (b) common ratio 2. a) Write down the formula for finding the nth term (tn) when first term (a), common ratio (r) and number of terms (n) of a geometric sequence are given. b) A G.P. has the first term x and common ratio y, what will be its 4th term. 3. Examine whether the following sequences form a G.P. or not. a) 2, 4, 8, 16, … b) 1 3, 1, 3, 9, 27, … c) 5, –20, 80, –320, … d) 6, 12, 18, 24, … 4. Find the first three terms of the G.P. whose first term (a) and common ratio (r) are given below. a) a = 2, r = 3 b) a = –9, r =3 c) a = 1 8, r = –2 d) a =27 , r = 1 3 5. a) Find the 8th term of the sequence 3, 6, 12, 24, … b) What is the value t7 in the sequence 729, 243, 81,…? c) The first and the second terms of a geometric sequence are 9 and 18 respectively. What is the fifth term? 6. a) In a G.P., the common ratio is 2 and the 8th term is 256, find its first term. b) If the sixth term of a geometric sequence with common ratio 3 is 972, find the first term. c) If the fourth term of a GP with first term 2 is 54, find the common ratio. d) A geometric sequence has the first term 8. If its fourth term is 1, what is its common ratio? 7. a) How many terms are there in the geometric sequence 3, 6, 12, … , 192? b) How many terms are there in the geometric series 16 + 24 + 36 + …+ 81?

Vedanta Excel in Mathematics - Book 9 166 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 8. a) Which term of GP 1, 2, 4, 8, 18, .. is 256? b) Which term of the series 27 + 9 + 3 + ….is 1 81 ? 9. a) Is 486 a term of a geometric sequence 2, 6, 18, 54, …? b) Is 1 128 a term of a geometric progression 1 + 1 2 + 1 4 + ...? Creative section - A 10. a) The 3rd and the 8th terms of a geometric sequence are 12 and 384 respectively, find the 10th term. b) In a geometric sequence, the 2nd term is 6 and the 5th term is 162, what is the 7th term? c) If the 6th and the 9th terms of a G.P. are 1 and 8 respectively, find its 10th term. 11. a) If the third and the sixth terms of a geometric series are 45 and 1215 respectively, find the series. b) If the fifth term and the ninth term of a G.P. are 144 and 2304 respectively, find the geometric series. Creative section - B 12. a) A person has two parents (father and mother), 4 grandparents and so on. Find the of his ancestors of 6th generations preceding his own. b) A bookshelf has 5 shelves. The top shelf has 4 books, and each shelf below has 3 times as many books as the shelf directly above it. If all of the books are different in the bookshelf and you want to read a book from the bottom shelf, how many choices would you have? c) The number of bacteria in a certain culture doubles every hour. There were 20 bacteria present in the culture originally. (i) How many bacteria will be present at the end of 2nd hour? (ii) How many bacteria will be present at the end of of 4th hour? 13. a) The man joined a company as an assistant manager. The company gave him a starting salary of Rs 60,000 and agreed to increase his salary 10% annually. What will be his salary after 5 years? b) The present value of a scanner machine is Rs 50,000 and its value depreciates each year by 20%. What will be the value of the machine after 5 year? c) The population of a town grows by 15% per year. If the population of the town was 2,00,000 at the end of 2021 A.D., find the projected population of the town at the end of 2031 A.D. d) Shaswat purchased a motorcycle for Rs 4,00,000 this year. The value of a motor cycle depreciates at the rate of 15% per year. What will be the value of the motor cycle 4 year hence? 14. If the numbers a, b and c are in an A.P., show that 5a , 5b and 5c form a G.P. Project work and activity section 15. a) Think the values of the first term (a) and common ratio (r) of your choice for the geometric sequence. Then, find the general term and the first five terms of the sequence. Sequence and Series

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 167 Vedanta Excel in Mathematics - Book 9 Sequence and Series b) Draw a square having each side of length 20 cm on a chart paper. Mark the mid-points of each side and join them to form a new square. Also, mark the mid-points of each side of this new square and join them to form another square. Continue this process till to get the fourth square. Find the area of each square and explain the sequence formed by these areas. c) Assume that you were symptomatic and positive patient of COVID-19. You forgot to cover your mouth when two of your friends came to visit you at home. They left and on the next day, they also became positive with virus. Each infected friend in turn spread the virus to two other friends on the following day and continued this pattern. In this scenario, how can we determine the number of infected persons? What are the possible ways to detect the infected persons? Form a sequence and justify your answer. d) Visit the available website and search the population of Nepal in B.S. 2078. Also, find the annual growth rate of population. Estimate the projected population of Nepal in B.S. 2088. OBJECTIVE QUESTIONS 1. The fifth term of the sequence 2, 5, 8, 11, … is (A) 11 (B) 14 (C) 15 (D) 19 2. Which term comes next in the sequence 1, 2, 4, 8,..? (A) 8 (B) 12 (C) 16 (D) 20 3. The series associated to the sequence 1, 3, 5, 7 is (A) 1 + 3 + 5 + 7 (B) –1 – 3 –5 –7 (C) 1 – 3 + 5 – 7 (D) –1+3–5+7 4. What will the 6th term of a sequence whose nth term is given by tn = {1 + (-1)n}? (A) -1 (B) 0 (C) 1 (D) 2 5. The 8th term of the sequence 1, 1, 2, 3, 5, 8, 13, … is (A)21 (B) 18 (C) 16 (D) 34 6. Given a1 = 1 , a2 = 3 and an = 2an-1 + an-2 then a3 is (A)1 (B) 4 (C) 5 (D) 7 7. The general term of the sequence 1, 4, 9, 16, 25, … is (A) n2 (B) (n + 1)2 (C) (n2 – 1) (D) 2n 8. The first term of an arithmetic progression is unity and the common difference is 3. Which of the following will be a term of this A.P.? (A) 8 (B) 14 (C) 19 (D) 24 9. If 5 times of 5th term of an A.P. is equal to 8 times the 8th term, then the 13th term of the A.P. is (A) 0 (B) 5 (C) 8 (D) 13 10. How many terms are there in the series 1 + 3 + 9 + … + 243? (A) 5 (B) 6 (C) 7 (D) 8

Vedanta Excel in Mathematics - Book 9 168 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 9.1 Factorisation –Looking back Classwork-Exercise Let’s say and write the factors of the expressions as quickly as possible. 1. a) ax + ay = ……………. b) 3x – 12y = ……………. c) y2 – y = ……………. d) x2 y + xy2 = ……………. 2. a) a (x + y) + b (x + y) = ……………. b) x (a – b) – y (a – b) = ……………. c) 2p (p – x) – 3 (p – x) = ……………. d) y (y – 3) – (y – 3) = ……………. 3. a) x2 – a2 = ……………. b) p2 – 4 = ……………. c) 9a3 – 16a = ……………. d) 100m3 – m = ……………. 9.2 Factors and Factorisation - Review In the given figure, x unit is the length of a square. So, the area of the square = x2 sq. unit. Let, the length of the square is increased by 2 units. Then, the length of the rectangle = (x + 2) units. Now, the area of the rectangle = x × (x + 2) sq. units = (x2 + 2x) sq. units Here, x2 + 2x is the product of x and (x + 2). Therefore, x and (x + 2) are the factors of the expression x2 + 2x. Thus, factorisation is the process of expressing a polynomial as the product of two or more polynomials. When we factorise an expression, we write it as the product of its factors. The process of getting the factors becomes easier if we apply the selected method of factorisation for the particular type of expression. So, it is important and very much useful to know about the types of expressions which are to be factorised. a) Expression having a common factor in each of its term Let's take an expression ax + ay. Here, both terms contain a common term, a. In such an expression, the factor which is present in all terms of the expression is taken out as common and each term of the expression should be divided by the common factor to get another factor. x2 x x x2 2x x x + 2 x 2 Unit 9 Factorisation of Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 169 Vedanta Excel in Mathematics - Book 9 Worked-out Examples Example 1: Factorise (i) 4ab + 6ac (ii) 2x2 y – 4xy2 + 6xy (iii) 2a (x – y) + 7b (y – x) (iv) 3p2 q (a – b) – 6pq2 (b – a) Solution: (i) 4ab + 6ac = 2a (2b + 3c) (ii) 2x2 y – 4xy2 + 6xy = 2xy (x – 2y + 3) (iii) 2a (x – y) + 7b (y – x) = 2a (x – y) – 7b (x – y) = (x – y) (2a – 7b) (iii) 3p2 q (a – b) – 6pq2 (b – a) = 3p2 q(a – b) + 6pq2 (a – b) = 3pq (a – b) (p + 2q) b) Expression having common factors in the groups of terms Let’s take an expression ax + by – ay – bx. It can be regrouped as ax – ay – bx + by (or ax – bx – ay + by). In ax – ay – bx + by, the group ax – ay has the common factor a and the group – bx + by has the common factor b. Then, ax – ay – bx + by = a (x – y) –b (x – y) = (x – y) (a – b) Similarly, ax – bx – ay + by = x (a – b) –y (a – b) = (a – b) (x – y) In this way, in such expressions, the terms are arranged in suitable groups such that each group has a common factor. Example 2: Factorise a) x2 – ax + ab – bx b) ac (b2 + 1) + b (a2 + c2 ) Solution: a) x2 – ax + ab – bx = x2 – ax – bx + ab = x (x – a) – b (x – a) = (x – a) (x – b) b) ac (b2 + 1) + b (a2 + c2 ) = ab2 c + ac + a2 b + bc2 = ab2 c + a2 b + bc2 + ac = ab (bc + a) + c (bc + a) = (bc + a) (ab + c) c) Expression of the form ax2 + bx + c, where a ≠ 0. x2 + 5x + 6, 2x2 + x – 28, etc. are the trinomial expressions of the form ax2 + bx + c. To factorise such expressions, we need to find the numbers p and q such that p + q = b and pq = ac. Then, the trinomial expression is expanded to four terms and factorisation is performed by grouping. Example 3: Factorise: a) x2 + 7x + 12 b) 2x2 – 5x + 2 c) 2x2 – x – 6 Solution: a) x2 + 7x + 12 = x2 + (4 + 3) x + 12 = x2 + 4x + 3x + 12 = x (x + 4) + 3 (x + 4) = (x + 4) (x + 3) 12 × 1 = 12 4 × 3 = 12 x x2 x x + 3 x x 1 5 9 2 6 10 3 7 11 4 8 12 x x x x x x2 x + 4 Factorisation of Algebraic Expressions

Vedanta Excel in Mathematics - Book 9 170 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) 2x2 – 5x + 2 = 2x2 – (4 + 1) x + 2 = 2x2 – 4x – x + 2 = 2x (x – 2) – 1 (x – 2) = (x – 2) (2x – 1) x x 2x x2 x x 2 2x – 1 x2 2 x x x x 1 2 x2 x x–2 2x – 1 x–2 2 × 2 = 4 4 × 1 = 4 c) 2x2 – x – 6 = 2x2 – (4 – 3) x – 6 = 2x2 – 4x + 3x – 6 = 2x (x – 2) +3 (x – 2) = (x – 2) (2x + 3) 2x + 3 x(x–2) x(x–2) x(x–2) x(x–2) x–2 3x–6 x x 3 2x + 3 x2 x x 2 x x x 2x + 3 x–2 x–2 x–2 2 x x 1 4 2 5 3 x x 6 2 × 6 = 12 4 × 3 = 12 Example 4: Resolve into factors: a) a2 b2 – 3 + 2b2 a2 b) 9 (x + y)2 + x + y – 8 Solution: https://www.geogebra.org/m/kthvykmx Vedanta ICT Corner Please! Scan this QR code or browse the link given below: a) a2 b2 – 3 + 2b2 a2 = a2 b2 – 3. a b. b a + 2b2 a2 = a2 b2 – a b. b a – 2 a b. b a + 2b2 a2 = a b (a b – b a) – 2b a (a b – b a) = ( – b a) (a b – 2b a ) b) Let, x + y = a Then, 9 (x + y) 2 + x + y – 8 = 9a2 + a – 8 = 9a2 + 9a – 8a – 8 = 9a (a + 1) – 8 (a + 1) = (a + 1) (9a – 8) = (x + y + 1) [9 (x + y) – 8] = (x + y + 1) (9x + 9y – 8) EXERCISE 9.1 General section 1. In each of the following figures, write the polynomial as the product of its factors. a) b) c) d) x x 1 2 x x x 2 1 x a 3 3 a Factorisation of Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 171 Vedanta Excel in Mathematics - Book 9 Factorisation of Algebraic Expressions 2. Factorise: a) 2ax + 4bx b) 4p2 – 6p c) 6a2 b + 9ab2 d) 2px2 – 4px + 6p2 x e) 6x3 y2 + 9x2 y3 – 3x2 y2 f) 2x (a + b) + 3y (a + b) g) 3a (x – y) – (x – y) h) x (x + 2) + 3x + 6 i) 2t (t – 1) – t + 1 3. In each of the following figures write the polynomial as the product of its factors. a) b) c) d) e) f) Creative section 4. Resolve into factors: a) ax + by + ay + bx b) pm – qn + pn – qm c) a2 + ab + ca + bc d) mx2 + my2 – nx2 – ny2 e) xy – 2y + 3x – 6 f) x2 + 4x + 3x + 12 g) p2 – 8p – p + 8 h) 16x2 – 4x – 4x + 1 i) a2 – a (b + c) + bc j) x2 – (y – 3) x – 3y k) pq (r2 + 1) – r (p2 + q2 ) l) y (x + z) + z (x + y) + y2 + z2 5. Factorise: a) x2 + 4x + 3 b) x2 – 7x – 8 c) a2 – 27a + 180 d) 2x2 + 7x + 6 e) 3p2 – 7p – 6 f) 2x2 + 3xy – 5y2 g) 3a2 – 16ab + 13b2 h) 9a3 bx + 12a2 b2 x – 5ab3 x i) 12 a2 b2 + a b – 20 j) x2 y2 – 2 – 3y2 x2 k) 2 (x + y) 2 + 9 (x + y) + 7 l) 3 (x – y) 2 – 10 (x – y) + 8 6. a) The area of a rectangular field is (x2 + 8x + 15) sq. m. (i) Find the length and breadth of the field. (ii) Find the perimeter of the field. b) The area of a rectangular plot of land is (x2 + 13x + 40) sq. m. (i) Find the length and breadth of the land. (ii) If the length and breadth of the land are reduced by 2/2 m respectively, find the new area of the land. c) A rectangular ground has area (2x2 + 11x + 12) sq. m. If the length of the ground is decreased by 2 m and the breadth is increased by 2 m, find the new area of the ground. (Take a longer side of the rectangles as length) x x 3 2 x x 3 2 x x 3 4 x x 3 x 1 x2 x2 x x 2 x 2 3 x x x 1 3

Vedanta Excel in Mathematics - Book 9 172 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Project work and activity section 7. a) Write three different expressions of your own each in the form of a3 + b3 and a3 – b3 . Then, factorise your expressions. b) Write an expression in each of the following forms, then, factorise your expressions (i) x2 + ax + b (ii) x2 – ax – b (iii) x2 + ax – b (iv) x2 – ax + b c) Write an expression in each of the following forms, factorise your expressions. (i) ax2 + (ii) ax2 – bx – c (iii) ax2 + bx – c (iv) ax2 – bx + c d) Expression of the form (a + b)2 and (a – b)2 (i) Expression of the form (a + b) 2 Geometrical interpretation: When the length a of the smaller square is increased by b, each side of the bigger square ABCD is (a + b) Now, area of ABCD = a2 + ab + ab + b2 = (a + b) 2 = a2 + 2ab + b2 By usual multiplication: (a + b) 2 = (a + b) (a + b) = a (a + b) + b (a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 Facts to remember 1. (a + b) 2 = (a + b) (a + b) = (a2 + 2ab + b2 ) 2. a2 + b2 = (a + b) 2 – 2ab (ii)Expression of the form (a – b) 2 Geometrical interpretation: Let each side of bigger square ABCD be a. When the length a of the smaller square is decreased by b, each side of the bigger square PQRD is (a – b) Now, area of ABCD = (a – b) 2 + b (a – b) + b (a – b) + b2 or, a2 = (a – b) 2 + ab – b2 + ab – b2 + b2 or, a2 = (a – b) 2 + 2ab – b2 or, (a – b2 ) = a2 – 2ab + b2 By usual multiplication: (a – b) 2 = (a – b) (a – b) = a (a – b) – b (a – b) = a2 – ab – ab + b2 = a2 – 2ab + b2 Facts to remember 1. (a – b) 2 = (a – b) (a – b) = (a2 – 2ab + b2 ) 2. a2 + b2 = (a – b) 2 + 2ab a + b a + b a a2 b2 ab b ab b b a b D C A B a a – b (a – b) 2 a – b b b(a – b) b(a – b) b2 a – b D A P C B b b Q R Factorisation of Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 173 Vedanta Excel in Mathematics - Book 9 Factorisation of Algebraic Expressions e) Expression having the difference of two squared terms (Expression of the form a2 – b2 ) Let’s take a square sheet of paper of length a units. From a corner of the sheet, another square of length b units is cut out. a2 a a a a a – b a – b b b a a b b b2 b a b A Area = (a + b) (a – b) B D C a a + b a – b Area = a2 Area of shaded region is = a2 – b2 Here, area of the rectangle ABCD is a2 – b2 , which is the product of length (a + b) and breadth (a – b) ∴length × breadth = area of rectangle i.e., (a + b) (a – b) = a2 – b2 Here, the expression a2 – b2 is the difference of two squared terms a2 and b2 and it is the product of (a + b) and (a – b). So, (a + b) and (a – b) are the factors of a2 – b2 . Thus, to factorise an expression of the form a2 – b2 , we should use the formula, a2 – b2 = (a + b) (a – b) Worked-out Examples Example 1: Factorise a) 8x3 y – 18xy3 b) 81ax5 – 16ax Solution: a) 8x3 y – 18xy3 = 2xy (4x2 – 9y2 ) = 2xy[(2x) 2 – (3y) 2 ] = 2xy (2x + 3y) (2x – 3y) b) 81ax5 – 16ax = ax (81x4 – 16) = ax [(9x2 ) 2 – (4)2 ] = ax (9x2 + 4) (9x2 – 4) = ax (9x2 + 4) [(3x2 ) – 22 ] = ax (9x2 + 4) (3x + 2) (3x – 2) Example 2: Resolve into factors. a) 1 – 9 (a – b)2 b) a2 – b2 + 2b – 1 Solution: a) 1 – 9 (a – b) 2 = 12 – [3 (a – b)]2 = [1 + 3 (a – b)] [1 – 3 (a – b)] = (1 + 3a – 3b) (1 – 3a + 3b) b) a2 – b2 + 2b – 1 = a2 – (b2 – 2b + 1) = a2 – (b – 1)2 = (a + b – 1) [a – (b – 1)] = (a + b –1) (a – b + 1) Using (a – b) 2 = a2 – 2ab + b2

Vedanta Excel in Mathematics - Book 9 174 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 3: Factorise x2 – 6x – 40 + 14b – b2 Solution: x2 – 6x – 40 + 14b – b2 = (x2 – 6x) – 40 + 14b – b2 = (x2 – 2.x.3 + 32 – 32 ) – 40 + 14b – b2 = (x – 3)2 – 9 – 40 + 14b – b2 = (x – 3)2 – (49 – 14b + b2 ) = (x – 3)2 – (72 – 2.7.b + b2 ) = (x – 3)2 – (7 – b) 2 = (x – 3 + 7 – b) (x – 3 – 7 + b) = (x – b + 4) (x + b – 10) Example 4: Factorise (w2 – x2 ) (y2 – z2 ) – 4wxyz Solution: (w2 – x2 ) (y2 – z2 ) – 4wxyz = w2 y2 – w2 z2 – x2 y2 + x2 z2 – 4wxyz = (wy) 2 – 2.wxyz + (xz)2 – (wz) 2 – 2wxyz – (xy) 2 = (wy) 2 – 2.wy.xz + (xz) 2 – [(wz) 2 + 2.wz.xy + (xy) 2 ] = (wy – xz) 2 – (wz + xy) 2 = (wy – xz + wz + xy) (wy – xz – wz – xy) = (wy + wz + xy – xz) (wy – wz – xy – xz) EXERCISE 9.2 General section 1. Factorise: a) 9x2 – 4 b) 25a2 b2 – 1 c) 48ax2 – 75ay2 d) x4 – y4 e) 16x5 y – 81xy5 f) 625a4 – 256b4 g) 4 – (m – n) 2 h) 1 – (a – b) 2 i) 16 – 25(p – q) 2 j) (2a – b) 2 – (a – 2b) 2 k) a2 + 2ab + b2 – c2 l) p2 – q2 – r2 – 2qr m) a2 – b2 + 4b – 4 n) 16a4 – 4a2 – 4a – 1 o) ax2 – ay2 – x – y p) a2 – (a – b)x – b2 10 cm 2 cm 2 cm 10 cm 18 m 5 m 5 m 18 m 15 cm 3 cm 15 cm 3 cm 21 m 7 m 21 m 7m 25 ft 9 ft 25 ft 9 ft 30 m 12 m 12 m 30 m d) e) f) a) b) c) 2. Let’s find the area of the shaded region using a2 – b2 = (a + b) (a – b). Using a2 – 2ab + b2 = (a – b) 2 https://www.geogebra.org/m/cr273mr4 Vedanta ICT Corner Please! Scan this QR code or browse the link given below: Factorisation of Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 175 Vedanta Excel in Mathematics - Book 9 Factorisation of Algebraic Expressions Creative section 3. Factorise: a) x2 + 6x + 5 – 4y – y2 b) a2 – 10a + 16 - 6b – b2 c) p2 – 12p – 28 + 16q – q2 d) x4 + 8x2 – 65 + 18y – y2 e) 9a2 – 30a + 24 – 8x – 16x2 f) 625y2 + 400y – 36 + 20z – z2 g) 16p2 – 72pq + 80q2 – 6qr – 9r2 h) 25x2 – 20xy – 21y2 + 10yz – z2 4. Resolve into factors. a) (a2 – b2 ) (c2 – d2 ) + 4abcd b) (x2 – 1) (y2 – 1) – 4xy c) (p2 – 4) (9 – q2 ) + 24pq d) (9 – x2 ) (100 – y2 ) – 120xy 5. a) A square sheet of paper is 25 cm long. A small square portion of length 9 cm is cut out from it. Find the area of the remaining portion of the paper. b) A farmer has a square field of length 150 m. He separates a small square portion of length 60 m from it to cultivate vegetables and he cultivates crops in the remaining portion. Find the area of the crops cultivated portion of the field. Project work and activity section 6. a) Take a squared chart paper of length ‘a + b’. Then, show (a + b) 2 in chart paper and present in class room. b) Show (a – b) 2 geometrically in chart paper and present in class. c) Write any three expressions of your own in the form of a2 – b2 . Then, factorize your expressions. f) Expression of the form a4 + a2 b2 + b4 The expressions of the form a4 + a2 b2 + b4 are also factorised by using the similar method of factorisation of the expression of a2 – b2 form. Following formulae are useful while factorising these types of expressions. a2 + 2ab + b2 = (a + b) 2 a2 + b2 = (a + b) 2 – 2ab a2 – 2ab + b2 = (a – b) 2 a2 + b2 = (a – b) 2 + 2ab a2 – b2 = (a + b) (a – b) Example 1: Factorise a) a4 + a2 b2 + b4 b) x4 – 3x2 y2 + y4 Solution: a) a4 + a2 b2 + b4 = (a2 ) 2 + (b2 ) 2 + a2 b2 = (a2 + b2 ) 2 – 2a2 b2 + a2 b2 = (a2 + b2 ) 2 – (ab) 2 = (a2 + b2 + ab) (a2 + b2 – ab) = (a2 + ab +b2 ) (a2 – ab + b2 ) b) x4 – 3x2 y2 + y4 = (x2 ) 2 + (y2 ) 2 – 3x2 y2 = (x2 – y2 ) 2 + 2x2 y2 – 3x2 y2 = (x2 – y2 ) 2 – (xy) 2 = (x2 – y2 + xy) (x2 – y2 – xy) = (x2 + xy – y2 ) (x2 – xy – y2 ) Using a2 + b2 = (a + b) 2 – 2ab Using a2 – b2 = (a + b) (a – b) Using a2 + b2 = (a – b) 2 + 2ab

Vedanta Excel in Mathematics - Book 9 176 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Example 2: Resolve into factors a) 9x4 + 14x2 + 25 b) x4 y4 + x2 y2 +1 c) a4 + 4b4 Solution: a) 9x4 + 14x2 + 25 = (3x2 ) 2 + (5)2 + 14x2 = (3x2 + 5)2 – 2.3x2 .5 + 14x2 = (3x2 + 5)2 – 16x2 = (3x2 + 5)2 – (4x) 2 = (3x2 + 4x + 5) (3x2 – 4x + 5) b) x4 y4 + x2 y2 + 1 = x2 y2 2 + (1)2 + x2 y2 = x2 y2 + 1 2 – 2. x2 y2.1 + x2 y2 = x2 y2 + 1 2 – x y 2 = x2 y2 + x y + 1 x2 y2 – x y + 1 EXERCISE 9.3 General section 1. Factorise: a) x4 + 4 b) 4a4 + 1 c) p4 + 64q4 d) 81m4 + 4n4 e) 64a4 + 81x4 f) 625b4 + 4d4 g) m4 + 1 4 h) 100c4 + t 4 25 Creative section 2. Resolve into factors a) x4 + x2 y2 + y4 b) a4 + 4a2 + 16 c) 36p4 + 3p2 + 1 d) 9m4 + 14m2 n2 + 25n4 e) 4x4 – 13x2 y2 + 9y4 f) 225x4 – x2 y2 + 16y4 g) 32x4 +14x2 y2 + 2y4 h) 20b4 + 55b2 y2 + 45y4 i) 48m5 n –3m3 n3 + 27mn5 3. Find the factors of the following expressions. a) 4x4 – 16x2 + 9 b) 9p4 – 28p2 + 16 c) 25a4 – 19a2 + 1 d) y4 – 13y2 z2 + 4z4 e) 81a4 – 99a2 x2 + 25x4 f) 121p4 – 133p2 q2 + 36q4 g) 4x5 – 56x3 y2 + 25xy4 h) 27b4 p – 21b2 p3 + 3p5 i) 60a5 b – 240a3 b3 + 135ab5 4. Factorise the following expressions. a) 9x4 – 25x2 + 16 b) 4y4 – 5y2 + 1 c) 25m4 – 29m2 + 4 d) 9a4 – 34a2 b2 + 25b4 e) 100c4 – 325c2 d2 + 225d4 f) 169p4 – 365p2 q2 + 196q4 5. Factorise. a) x4 y4 + x2 y2 + 1 b) a4 b4 + b4 a4 + 1 c) x2 + 1 + 1 x2 d) x4 y4 – 7x2 y2 + 1 e) m4 n4 – 29m2 n2 + 4 f) b4 d4 – 21 b2 d2 + 100 c) a4 + 4b4 = (a2 ) 2 + (2b2 ) 2 = (a2 + 2b2 ) 2 – 2a2 .2b2 = (a2 + 2b2 ) 2 – (2ab) 2 =(a2 + 2ab + 2b2 ) (a2 – 2ab + 2b2 ) Factorisation of Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 177 Vedanta Excel in Mathematics - Book 9 Factorisation of Algebraic Expressions Project work and activity section 6. a) Write any three expressions of your own in the form of a4 + b4 . Then, factorize your expressions. b) Write any three expressions of your own in the form of a4 + a2 b2 + b4 . Then, factorize your expressions. c) Write any three expressions of your own in the form of a4 – a2 b2 + b4 . Then, factorize your expressions. g) Expression of the form (a + b) 3 and (a – b) 3 i.e., cubes of binomials (i) Expression of the form (a + b) 3 Geometrical interpretation: Let’s take a cube such that its each side is (a + b). Then, the volume of the cube is (a + b) 3 which is shown below. a a a (a + b) 3 = a3 + 3a2 b + 3ab2 + b3 a a b a+b a+b a+b b b b a b b Thus, (a + b) 3 = a3 + 3a2 b + 3ab2 + b3 = a3 + 3ab (a + b) + b3 By usual multiplication: Let’s find the product of (a + b) 3 . (a + b) 3 = (a + b) (a + b) (a + b) = (a + b) ( ) 2 = (a + b) (a2 + 2ab + b2 ) = a (a2 + 2ab + b2 ) + b (a2 + 2ab + b2 ) = a3 + 2a2 b + ab2 + a2 b + 2ab2 + b3 = a3 + 3a2 b + 3ab2 + b3 ∴(a + b) 3 = a3 + 3a2 b + 3ab2 + b3 = a3 + 3ab (a + b) + b3 Facts to remember 1. The expanded form of (a + b) 3 = a3 + 3a2 b + 3ab2 + b3 2. The factors of (a + b) 3 are (a + b), (a + b) and (a + b). (ii) Expression of the form (a – b) 3 Let’s take a cube such that its each side is (a – b). Then, the volume of the cube is (a – b) 3 which is shown below. a–b a–b a–b a–b a–b a–b a–b b b a b a–b b b b b b From the above model, the volume of cube = volume of parts of the cube

Vedanta Excel in Mathematics - Book 9 178 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur or, a3 = (a – b) 3 + 3 (a – b) 2 b + 3 (a – b) b2 + b3 or, a3 – 3 (a – b) 2 b – 3 (a – b) b2 – b3 = (a – b) 3 or, a3 – 3 (a2 –2ab +b2 ) b – 3 (a – b) b2 – b3 = (a – b) 3 or, a3 – 3a2 b + 6ab2 – 3b3 – 3ab2 + 3b3 – b3 = (a – b) 3 or, a3 – 3a2 b + 3ab2 – b3 = (a – b) 3 Thus, (a – b) 3 = a3 – 3a2 b + 3ab2 – b3 = a3 – 3ab (a – b) – b3 Facts to remember 1. The expanded form of (a – b) 3 = a3 – 3a2 b + 3ab2 – b3 2. The factors of (a + b) 3 are (a – b), (a – b) and (a – b). Worked-out Examples Example 1: Express 8x3 + 36x2 + 54x + 27 as a cube of a binomial. Solution: Here, 8x3 + 36x2 + 54x + 27 = (2x) 3 + 3. (2x) 2 . 3 + 3. 2x. 32 + 33 = (2x + 3)3 Example 2: Factorise: 64a3 – 240a2 b + 300ab2 – 125b3 Solution: Here, 64a3 – 240a2 b + 300ab2 – 125b3 = (4a) 3 – 3. (4a) 2 . 5b + 3. 4a. (5b) 2 – (5b) 3 = (4a – 5b) 3 = (4a – 5b) (4a – 5b) (4a – 5b) h) Expression of the form (a3 + b3 ) and (a3 – b3 ) Let’s find the product of the expressions (a + b) and (a2 – ab + b2 ). (a + b) (a2 – ab + b2 ) = a (a2 – ab + b2 ) + b (a2 – ab + b2 ) = a3 – a2 b + ab2 + a2 b – 2 + b3 = a3 + b3 Also, we have (a + b) 3 = a3 + 3ab (a + b) + b3 or, (a + b) 3 – 3ab (a + b) = a3 + b3 or, (a + b) {(a + b) 2 – 3ab} = a3 + b3 or, (a + b) (a2 + 2ab + b2 – 3ab) = a3 + b3 or, (a + b) (a2 - ab + b2 ) = a3 + b3 So, (a + b) and (a2 – ab + b2 ) are the factors of a3 + b3 Similarly, (a – b) (a2 + ab + b2 ) = a3 – b3 and (a – b) and (a2 + ab + b2 ) are the factors of a3 – b3 . Facts to remember 1. a3 + b3 = (a + b) (a2 – ab + b2 ) 2. a3 – b3 = (a – b) (a2 + ab + b2 ) Example 3: Factorise: a) 8a4 + 125a b) x6 – y6 c) a7 + 1 a5 Factorisation of Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 179 Vedanta Excel in Mathematics - Book 9 Factorisation of Algebraic Expressions Solution: a) 8a4 + 125a = a (8a3 + 125) = a [(2a) 3 + 53 ] = a (2a + 5) [(2a) 2 – 2a.5 + 52 ] = a (2a + 5) (4a2 – 10a + 25) b) x6 – y6 = (x3 ) 2 – (y3 ) 2 = (x3 + y3 ) (x3 – y3 ) = (x + y) (x2 – xy + y2 ) (x – y) (x2 + xy + y2 ) = (x + y) (x – y) (x2 – xy + y2 ) (x2 + xy + y2 ) c) a7 – 1 a5 = a7 – a × 1 a6 = a(a6 – 1 a6) = a[(a3 ) 2 – 1 a3 2 ] = a a3 + 1 a3 a3 – 1 a3 = a a + 1 a (a2 – a × 1 a + 1 a2) a – 1 a (a2 + a × 1 a + 1 a2) = a a + 1 a a – 1 a a2 – 1+ 1 a2 a2 + 1+ 1 a2 =a a + 1 a a – 1 a a2 – 1+ 1 a2 a + 1 a 2 – 2 × a × 1 a + 1 = a a + 1 a a – 1 a a2 – 1+ 1 a a + 1 a 2 – (1)2 = a a + 1 a a – 1 a a2 – 1+ 1 a a + 1+ 1 a a – 1+ 1 a Example 4: Resolve into factors: 8x3 – 20x2 y + 30xy2 – 27y3 . Solution: 8x3 – 20x2 y + 30xy2 – 27y3 = (2x) 3 – (3y) 3 – 20x2 y + 30xy2 = (2x – 3y) [(2x) 2 + 2x.3y + (3y) 2 ] – 10xy (2x – 3y) = (2x – 3y) (4x2 + 6xy + 9y2 – 10xy) = (2x – 3y) (4x2 – 4xy + 9y2 ) Example 5: Factorise: 2a6 – 19a3 + 24 Solution: 2a6 – 19a3 + 24 = 2a6 – 16a3 – 3a3 + 24 = 2a3 (a3 – 8) – 3 (a3 – 8) = (a3 – 8) (2a3 – 3) = (a3 – 23 ) (2a3 – 3) = (a – 2) (a2 + 2a + 4) (2a3 – 3) = (a – 2)(2a3 – 3) (a2 + 2a + 4) EXERCISE 9.4 General Section 1. Let’s say and write the expanded forms of the following cubes. a) (a + x) 3 = ............................................ b) (a – x) 3 = ............................................ c) (p + q) 3 = ............................................ d) (p – q) 3 = ............................................ e) (x+1)3 = .............................................. f) (x – 1)3 = ............................................. Using a3 + b3 = (a + b) (a2 – ab + b2 )

Vedanta Excel in Mathematics - Book 9 180 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 2. Let’s say and write the factors of the expressions. a) m3 + n3 = ............................................ b) m3 – n3 = .............................................. c) x3 + y3 = ............................................. d) x3 – y3 = ............................................... e) a3 + 1 = .............................................. f) a3 – 1 = ............................................. Creative section-A 3. Express the following expressions as cubes of binomials a) 8x3 + 60x2 + 150x + 125 b) 27a3 + 108a2 + 144a + 64 c) 64y3 – 240y2 + 300y – 125 d) 343m3 – 882m2 + 756m – 216 e) 1000a3 – 900a2 b + 270ab2 – 27b3 f) 8x3 +132x2 y + 726xy2 + 1331y3 4. Find the factors of the following expressions. a) 8a3 + 36a2 + 54a + 27 b) 125x3 + 75x2 + 15x + 1 c) 27p3 – 108p2 + 144p – 64 d) 729w3 – 1944w2 + 1728w – 512 e) 216m3 – 540m2 n+ 450mn2 – 125n3 f) 2197e3 +3549e2 g+1911eg2 + 343g3 5. Resolve into factors. a) 8x3 + y3 b) 1 + 27a3 c) 128t 4 – 2t d) x3 y – 64y4 e) a6 + b3 f) 64x6 y3 – 125 g) a6 – 64 h) 729x6 – y6 Creative section-B 6. Factorise. a) (a + b) 3 + 1 b) (x + 2)3 – 27 c) (x – y) 3 – 8(x + y) 3 d) p3 + 1 p3 e) a3 b3 – b3 a3 f) x4 + 1 x2 g) y7 – 1 y5 h) p4 q – q2 p2 7. Resolve into factors. a) 27x3 + 30x2 y + 40xy2 + 64y3 b) 8 – 6a – 9a2 + 27a3 c) 64m3 – 28m2 n + 35mn2 – 125n3 d) 2p6 – 19p3 + 24 e) 40a6 + 11a3 – 2 f) 8x6 – 9x3 + 1 8. a) The volume of a rectangular tank is (2a3 + a2 – 2a – 1) cu. ft. where a>4 ft., the length is longer than its breadth and the breadth is longer than its height. (i) Find its length, breadth and height. (ii) Find the area of the its floor. b) From a cubical log of each side x cm, a small cube of each side 3 cm is removed. (i) What is the volume of removed part of the log? (iii) What is the volume of remaining part of the log? Project work and activity section 9. a) Write any three expressions of your own in the form of a3 + b3 . Then, factorize your expressions. b) Write any three expressions of your own in the form of a3 – b3 . Then, factorize your expressions. Factorisation of Algebraic Expressions

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 181 Vedanta Excel in Mathematics - Book 9 Factorisation of Algebraic Expressions OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. The factorisation of 16 – p4 is (A)(4 + p2 ) (2 + p) (2 – p) (B) (4 + p2 ) (2 – p) (2 – z) (C) (4 – p2 ) (2 – p) (2 – p) (D) (4 + p2 ) (2 + p) (2 + p) 2. The factors of x2 – x – 12 are (A)(x+4), (x – 3) (B) (x – 4), (x – 3) (C) (x – 4), (x + 3) (D) (x+4), (x + 3) 3. a2 + b2 = (a + b) 2 + x, what is the value of x? (A) 2ab (B) –2ab (C) 4ab (D) –4ab 4. a + b is a factor of …. (A)a2 – b2 (B) a2 + b2 (C) (a – b) 2 (D) a3 – b3 5. (x + y) (x2 – xy + y2 ) is equal to (A)(x + y) 3 (B) (x – y) 3 (C) x3 + y3 (D) x3 – y3 6. p3 – q3 is equal to (A)(p + q) (p2 – pq + q2 ) (B) (p – q) (p2 – pq + q2 ) (C) (p – q) (p2 + pq + q2 ) (D) (p + q) (p2 + pq + q2 ) 7. The factors of x3 – 64 are (A)(x – 4), (x2 + 4x + 16) (B) (x + 4), (x2 + 4x + 16) (C) (x + 4), (x2 – 4x + 16) (D) (x – 4), (x2 – 4x + 16) 8. The factors of 125x3 + 1 are (A)(5x – 1), (25x2 + 5x + 1) (B) (5x + 1), (25x2 – 5x + 1) (C) (5x – 1), (25x2 – 5x + 1) (D) (25x + 1), (25x2 – 5x + 1) 9. The factorisation of x4 + x2 y2 + y4 is (A)(x2 + xy + y2 ) (x2 – xy + y2 ) (B) (x2 – xy + y2 ) (x2 – xy + y2 ) (C) (x2 + xy + y2 ) (x2 + xy + y2 ) (D) (x2 – xy + y2 ) (x2 + xy – y2 ) 10. The factors of a4 + 4 are (A)(a2 + 2a + 2), (a2 + 2a – 2) (B) (a2 + 2a + 2), (a2 + 2a + 2) (C) (a2 – 2a + 2), (a2 + 2a + 2) (D) (a2 – 2a + 2), (a2 + 2a – 2) https://www.geogebra.org/m/hgfn6vtt Vedanta ICT Corner Please! Scan this QR code or browse the link given below:

Vedanta Excel in Mathematics - Book 9 182 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 10.1 H.C.F. and L.C.M. –Looking back Classwork-Exercise Let’s answer the following questions as quickly as possible. 1. a) For any two expressions 2a – 4 and a2 – 4, (i) What are the factors of 2a– 4? (ii) What are the factors of a2 – 4? (iii) what is the common factor among the factors of these expressions? (iv) what is the common factor of these expressions called? b) ax + ay and bx + by are two given expressions. (i) What are the factors of ax + ay? (ii) What are the factors of bx + by. (iii) What is the common factor among the factors of these expressions? (iv) What are the remaining factors of these expressions? (v) What is the product of common factor and remaining factors of these expressions called? 10.2 H.C.F. of algebraic expressions Let's take any two monomial expressions 4a2 and 6a3 . Here, all the possible factors of 4a2 are a, a2 , 2a, 2a2 , 4a, 4a2 . Also, all the possible factors of 6a3 are a, a2 , a3 , 2a, 2a2 , 2a3 , 3a, 3a2 , 3a3 . Now, the common factors of 4a2 and 6a3 are a, a2 2a and 2a2 . Among these common factors, 2a2 is the highest one. So, the Highest Common Factor (H.C.F.) of 4a2 and 6a3 is 2a2 . Thus, to find the H.C.F. of the monomial expressions, at first we should find the H.C.F. of the numerical coefficients. Then, the common variable with the least power is taken as the H.C.F. of the expressions. For example, In 12x4 y3 and 8x3 y2 4 is the H.C.F. of 12 and 8. x3 is the H.C.F. of x4 and x3. y2 is the H.C.F. of y3 and y2 . So, the H.C.F. of 12x4 y3 and 8x3 y2 is 4x3 y2 . Unit 10 Highest Common Factor (H.C.F.) and Lowest Common Multiples (L.C.M.)

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 183 Vedanta Excel in Mathematics - Book 9 We can find the H.C.F. of polynomial expressions either by the process of factorisation or by division process. In case of factorisation process, the given polynomials are to be factorised, then a common factor or the product of common factors is obtained as their H.C.F. Facts to remember 1. The H.C.F. of two or more algebraic expressions is an expression of the highest degree which exactly divides the given expressions. 2. If there is no common factor other than one, then H.C.F. is 1. Worked-out Examples Example 1: Find the highest common factors (H.C.F.) of each pair of expressions. Also, represent the H.C.F. in Venn-diagram. a) x3 – 4x, x2 + 7x + 10 b) a2 – 6a + 6b – b2 , b2 + ab – 6b c) m3 + 8, m3 – 8 d) (p – q)2 + 4pq, p2 – pq – 2q2 Solution: Here, a) The 1st expression = x3 – 4x = x (x2 – 4) = x (x + 2) (x – 2) The 2nd expression = x2 + 7x + 10 = x2 + (5 + 2) x + 10 = x2 + 5x + 2x + 10 = x (x + 5) + 2 (x + 5) = (x + 5) (x + 2) Hence, the H.C.F. = common factor = (x + 2) b) The 1st expression = a2 – 6a + 6b – b2 = a2 – b2 – 6a + 6b = (a + b) (a – b) – 6 (a – b) = (a – b) (a + b – 6) The 2nd expression = b2 + ab – 6b = b (b + a – 6) Hence, the H.C.F. = common factor = (a + b – 6) c) The 1st expression = m3 + 8 = m3 + 23 = (m + 2) (m2 – m.2 + 22 ) = (m + 2) (m2 – 2m + 4) The 2nd expression = m3 – 8 = m3 – 23 = (m – 2) (m2 + m.2 + 22 ) = (m – 2) (m2 + 2m + 4) Hence, the H.C.F. = common factor = 1 1st expression 2nd expression x(x – 2) (x + 5) (x+2) 1st expression 2nd expression 1 m3 + 8 m3 – 8 1st expression 2nd expression (a+b–6) (a – b) b H.C.F. and L.C.M.

Vedanta Excel in Mathematics - Book 9 184 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur d) The 1st expression = (p – q) 2 + 4pq = p2 – 2pq + q2 + 4pq = p2 + 2pq + q2 = (p + q) 2 = (p + q) (p + q) The 2nd expression = p2 – pq – 2q2 = p2 – (2 – 1) pq – 2q2 = p2 – 2pq + pq – 2q2 = p(p – 2q) + q (p – 2q) = (p – 2q) (p + q) Hence, the H.C.F. = common factor = (p + q) Example 2: Find the highest common factors (H.C.F.) of each pair of expressions. a) 18(2x3 – x2 – x), 20 (24x4 + 3x) b) 16a4 + 4a2 b2 + b4 , 8a3 + b3 c) x4 + 3x2 + 196, x3 + x (x + 14) + 4x2 d) (1 – m2 ) (1 – n2 ) + 4mn, 1 – 2m + n – m2 n + m2 Solution: Here, a) The 1st expression = 18(2x3 – x2 – x) = 18x (2x2 – x – 1) = 2 × 3 × 3x (2x2 – 2x + x – 1) = 2 × 3 × 3x {2x (x – 1) + 1 (x – 1)} = 2 × 3 × 3x (x – 1) (2x + 1) The 2nd expression = 20 (24x4 + 3x) = 20 × 3x (8x3 + 1) = 2 × 2 × 5 × 3x {(2x) 3 + (1)3 } = 2 × 2 × 3 × 5x (2x + 1) (4x2 – 2x + 1) Hence, the H.C.F. = common factor = 2 × 3x (2x +1) = 6x (2x + 1) b) The 1st expression = 16a4 + 4a2 b2 + b4 = (4a2 ) 2 + (b2 ) 2 + 4a2 b2 = (4a2 + b2 ) 2 – 2. 4a2 .b2 + 4a2 b2 = (4a2 + b2 ) 2 – 4a2 b2 = (4a2 + b2 ) 2 – (2ab) 2 = (4a2 + b2 +2ab) (4a2 + b2 – 2ab) = (4a2 +2ab + b2 ) (4a2 – 2ab + b2 ) The 2nd expression = 8a3 + b3 = (2a) 3 + b3 = (2a + b) [(2a) 2 – 2a. b + b2 ] = (2a + b) (4a2 – 2ab + b2 ) Hence, the H.C.F. = common factor = (4a2 – 2ab + b2 ) 1st expression 2nd expression (p + q) (p + q) (p – 2q) H.C.F. and L.C.M.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 185 Vedanta Excel in Mathematics - Book 9 c) The 1st expression = x4 + 3x2 + 196 = (x2 ) 2 + (14)2 + 3x2 = (x2 + 14)2 – 2. x2 .14+ 3x2 = (x2 + 14)2 – 25x2 = (x2 + 14)2 – (5x) 2 = (x2 +5x +14) (x2 – 5x +14) The 2nd expression = x3 + x (x + 14) + 4x2 = x3 + x2 + 14x + 4x2 = x3 + 5x2 + 14x = x (x2 + 5x + 14) Hence, the H.C.F. = common factor = (x2 + 5x + 14) d) The 1st expression = (1 – m2 ) (1 – n2 ) + 4mn = 1 – n2 – m2 + m2 n2 + 4mn = m2 n2 + 2mn + 1 – m2 + 2mn – n2 = (mn) 2 + 2.mn. 1 + (1)2 – (m2 – 2mn + n2 ) = (mn + 1)2 – (m – n) 2 = {(mn + 1) + (m – n)} {(mn + 1) – (m – n)} = (mn + 1 + m – n) (mn + 1 – m + n) = (mn + m – n + 1) (mn – m + n + 1) The 2nd expression = 1 – 2m + n – m2 n + m2 = m2 – 2m + 1 – m2 n + n = (m – 1)2 – n (m2 – 1) = (m –1)2 – n (m + 1) (m – 1) = (m – 1) {m – 1 – n (m + 1)} = (m – 1) (m – 1 – mn – n) = (1 – m) (mn – m + n + 1) Hence, the H.C.F. = common factor = (mn – m + n + 1) Example 3: Find the highest common factors (H.C.F.) of the following expressions. a) 9p2 – 4q2 – 4qr – r2 , r2 – 4q2 – 9p2 – 12pq and 9p2 + 6pr + r2 – 4q2 b) 27x3 + 18x2 + 6x +1, 81x4 – 9x2 – 6x – 1 and 81x4 + 9x2 + 1 c) x2 – 12x – 28 + 16y – y2 , x2 + 2x – y2 + 2y and x2 – y2 + 4y – 4 Solution: Here, a) The 1st expression = 9p2 – 4q2 – 4qr – r2 = (3p) 2 – (4q2 + 4qr + r2 ) = (3p) 2 – {(2q) 2 + 2. 2q. r + (r) 2 } = (3p) 2 – (2q + r) 2 = (3p + 2q + r) (3p – 2q – r) The 2nd expression = r2 – 4q2 – 9p2 – 12pq = (r) 2 – (4q2 + 12pq + 9p2 ) = (r) 2 – {(2q) 2 + 2. 2q. 3p + (3p) 2 } = (r) 2 – (2q + 3p) 2 = (r + 2q + 3p) (r – 2q – 3p) The 3rd expression = 9p2 + 6pr + r2 – 4q2 = (3p) 2 + 2. 3p.r + (r) 2 – (2q) 2 = (3p + r) 2 – (2q) 2 = (3p + r + 2q) (3p + r – 2q) Hence, the H.C.F. = common factor = (3p + 2q + r) H.C.F. and L.C.M.

Vedanta Excel in Mathematics - Book 9 186 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur b) The 1st expression = 27x3 + 18x2 + 6x +1 = (3x) 3 +(1)3 + 6x (3x +1) = (3x +1) (9x2 – 3x + 1) + 6x (3x +1) = (3x + 1) (9x2 – 3x + 1 + 6x) = (3x + 1) (9x2 + 3x + 1) The 2nd expression = 81x4 – 9x2 – 6x – 1 = (9x2 ) 2 – (9x2 + 6x + 1) = (9x2 ) 2 – {(3x) 2 + 2. 3x .1 + (1)2 } = (9x2 ) 2 – (3x +1)2 = (9x2 + 3x + 1) (9x2 – 3x – 1) The 3rd expression = 81x4 + 9x2 + 1 = (9x2 ) 2 + (1)2 + 9x2 = (9x2 + 1)2 – 2. 9x2 .1+ 9x2 = (9x2 + 1)2 – 9x2 = (9x2 + 1)2 – (3x) 2 = (9x2 +3x +1) (9x2 – 3x +1) Hence, the H.C.F. = common factor = (9x2 +3x +1) c) The 1st expression = x2 – 12x – 28 + 16y – y2 = x2 – 2.x.6 + 62 – 62 – 28 + 16y – y2 = (x – 6)2 – (64 – 16y + y2 ) = (x – 6)2 – (82 – 2.8.y + y2 ) = (x – 6)2 – (8 – y) 2 = (x – 6 + 8 – y) (x – 6 – 8 + y) = (x – y + 2) (x + y – 14) The 2nd expression = x2 + 2x – y2 + 2y = x2 – y2 + 2(x + y) = (x + y) (x – y) + 2(x + y) = (x + y) (x – y + 2) The 3rd expression = x2 – y2 + 4y – 4 = x2 – (y2 – 2. y. 2 + 22 ) = x2 – (y – 2)2 = (x + y – 2) (x – y + 2) Hence, the H.C.F. = common factor = (x – y + 2) EXERCISE 10.1 General section 1. a) Define the H.C.F. of algebraic expressions. b) What is the H.C.F. of 4ax and 6xy? c) What will be the H.C.F. of x and y if x is a factor of y? d) Write the condition under which the H.C.F. of the given algebraic expressions is 1. 2. Find the H.C.F. of the following expressions: a) 2x2 (x + 2) (x – 2) and 4x (x + 2) (x + 3) b) 4xy2 (x – 1) (x + 2) and 6x2 y (x – 1) (x – 4) c) 6a2 b2 (a – b) (2a + 3b) and 9a3 b3 (2a + 3b) (a + b) d) 12a2 b3 (a – 3b) (a + b – 2) and 16a3 b2 (a + 3b) (a + b – 2) e) (p + q) (p – q), (p – q) (p2 + pq + q2 ) and (p – q) (p + q) (p2 + q2 ) H.C.F. and L.C.M.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 187 Vedanta Excel in Mathematics - Book 9 H.C.F. and L.C.M. 3. Find the H.C.F. of the following expressions: a) a2 – b2 and a3 + b3 b) a2 – 2a and a4 – 8a c) x2 – 9 and 3x + 9 d) x2 – y2 and x2 + 2xy + y2 e) 4x2 – 100 and 4x + 20 f) 4x3 + 8x2 and 5x3 – 20x g) a2 – b2 – 2a + 1 and a2 – ab – a h) x3 – 8y3 and x2 + 2xy + 4y2 i) p4 – 16 and p2 – p – 6 j) x4 + 4y4 and 2x3 y + 4xy3 + 4x2 y2 Creative section 4. Find the H.C.F. of the following expressions: a) a3 – 1 and a4 + a2 + 1 b) p3 + 8 and p4 + 4p2 + 16 c) 16x4 – 4x2 – 4x – 1, 8x3 – 1 d) 81a4 – 9a2 + 6a – 1, 27a3 + 1 e) 36(x3 +x2 y –12xy2 ), 54(x3 y+4x2 y2 –21xy3 ) f) 6(5x3 y–500xy3 ), 20(x3 – 1000y3 ) g) x4 – x2 + 144, x3 + x (x + 12) + 4x2 h) a4 + 4a2 +100, a3 + a (a+10) + 3a2 i) 2a3 – a2 – a + 2, a3 + a2 – a – 1 j) 3x3 – x2 + x – 3, x3 – x2 + x –1 5. Find the highest common factors (H.C.F.) of the following expressions: a) a2 – 4, a3 + 8, a2 + 5a + 6 b) x2 – 9, x3 – 27, x2 + x – 12 c) 4x2 – 9, 2x2 + x – 3, 8x3 + 27 d) 3x2 – 8x + 4, 2x2 – 5x + 2, x4 – 8x e) 5a3 – 20a, a3 – 3a2 – 10a, a3 – a2 – 2a + 8 f) m3 – m2 – m + 1, 2m4 – 2m, 3m2 + 3m – 6 g) x3 – 64y3 , x2 – 6xy + 8y2 , x2 – 16y2 h) 4x4 + 16x3 – 20x2 , 3x3 + 14x2 – 5x, x4 + 125x i) 8x3 + 27y3 , 16x4 + 36x2 y2 + 81y4 , 4x3 – 6x2 y + 9xy2 j) x3 y + y4 , x4 + x2 y2 + y4 , 2ax3 – 2ax2 y + 2axy2 6. Find the highest common factors (H.C.F.) of the following expressions: a) x3 – 1, x4 + x2 + 1, x3 + 2x2 + 2x + 1 b) 5p3 – 20p, p3 – p2 – 6p, p3 – p2 – 2p + 8 c) (a + b) 2 – 4ab, a3 – b3 , a2 + ab – 2b2 d) (a – b) 3 + 3ab (a – b), a4 + a2 b2 + b4 , a3 + a2 b + ab2 e) a2 + 2ab + b2 – c2 , b2 + 2bc + c2 – a2 , c2 + 2ca + a2 – b2 f) 9x2 – 4y2 – 8yz – 4z2 , 4z2 – 4y2 – 9x2 – 12xy, 9x2 + 12xz + 4z2 – 4y2 g) x2 – 18x – 19 + 20y – y2 , x2 – x – y2 – y, x2 – y2 + 2y – 1 h) a2 – 6a – 7 + 8b – b2 , a2 + a – b2 + b, a2 - b2 + 2b – 1 i) 1+4x+4x2 – 16x4 , 1+2x – 8x3 – 16x4 , 1 + 4x2 + 16x4 j) 81x4 – 9x2 – 6x – 1, 81x4 + 27x3 – 3x – 1, 81x4 + 9x2 + 1 7. a) Sunayana and Bishwant have rectangular plots of land of same width. The area of Sunayana’s and Bishwant’s plots are (x2 + 35x + 300) sq. m. and (x2 + 27x + 180) sq. m. respectively. Find the width of their plots. b) The area of four walls of a rectangular room is (4x2 + 20x – 56) sq. m and the volume of the room is (x3 – 8) cu. m, find the height of the room.

Vedanta Excel in Mathematics - Book 9 188 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Project work and activity section 8. a) Write a pair of expressions of your own in the forms of a2 – b2 and a2 + bx + c = 0, a ≠ 0. Then, factorize your expressions and find their highest common factor. b) Write a pair of expressions of your own in the forms of a3 + b3 and a4 + a2 b2 +b4 . Then, factorize your expressions and find their H.C.F. 3.3 L.C.M. of algebraic expressions Let's take any two monomial expressions x2 and x3 . A few multiplies of x2 are x2 , x3 , x4 , x5 , x6 ,..... A few multiplies of x3 are x3 , x4 , x5 , x6 , x7 , .... Here, the common multiplies are x3 , x4 , x5 , x6 , ... Among these common multiplies, x3 is the lowest one. So, the Lowest Common Multiple (L.C.M.) of x2 and x3 is x3 . Thus, to find the L.C.M. of the monomial expressions, at first we should find the L.C.M. of the numerical coefficients. Then, the common variable with the highest power is taken as the L.C.M. of the expressions. For Example, In 6x4 y2 and 8x3 y3 24 is the L.C.M. of 6 and 8. x4 is the L.C.M. of x4 and x3 . y3 is the L.C.M. of y2 and y3 . So, the L.C.M. of 6x4 y2 and 8x3 y3 is 24x4 y3 . In case of polynomial expressions, their L.C.M. is obtained by the process of factorisation. By this process the product of common factors and the factors which or not common is taken as the L.C.M. of the polynomials Facts to remember 1. The L.C.M. of two or more algebraic expressions is an expression of the lowest degree which is exactly divisible by the given expressions. 2. L.C.M. of three expressions is the product of factors common to at least two and the remaining factors of the given expressions. 3. For any two algebraic expressions, the product of the expressions is equal to the product of the H.C.F. and L.C.M. of the expressions i.e., if H and L be the H.C.F. and L.C.M. of two algebraic expressions A and B, then A×B = H×L Example 1: Find the lowest common multiple (L.C.M.) of each pair of expressions. Also, represent the L.C.M. in Venn-diagram. a) a3 – a and a4 – a b) 6x6 + 6x4 + 6x2 and 4x6 – 4x3 c) (m + 5)3 – 15m (m + 5) and (m – 5)2 + 20m Solution: Here, a) The 1st expression = a3 – a = a (a2 – 1) = a (a + 1) (a – 1) 1st expression 2nd expression (a + 1) a (a – 1) a2 + a + 1 H.C.F. and L.C.M.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 189 Vedanta Excel in Mathematics - Book 9 H.C.F. and L.C.M. The 2nd expression = a4 – a = a (a3 – 1) = a (a – 1) (a2 + a + 1) Hence, the L.C.M. = common factors × remaining factors = a (a – 1) (a + 1) (a2 + a + 1) = a (a +1) (a3 – 1) b) The 1st expression = 6x6 + 6x4 + 6x2 = 6x2 (x4 + x2 + 1) = 2×3x2 [(x2 ) 2 + 12 + x2 ] = 2×3x2 [(x2 + 1)2 – 2x2 + x2 ] = 2×3x2 [(x2 + 1)2 – x2 ] = 2×3x2 (x2 + x + 1) (x2 – x + 1) The 2nd expression = 4x6 – 4x3 = 4x3 (x3 – 1) = 2×2x3 (x – 1) (x2 + x + 1) ∴ The L.C.M. = common factors × remaining factors = 2×3×2x3 (x2 + x + 1) (x – 1) (x2 – x + 1) = 12x3 (x3 – 1) (x2 – x + 1) c) The 1st expression = (m + 5)3 – 15m (m + 5) = (m + 5) {(m + 5)2 – 15m} = (m + 5) (m2 + 10m+ 25 – 15m) = (m + 5) (m2 – 5m + 25) The 2nd expression = (m – 5)2 + 20m = m2 – 10m + 25 + 20m = m2 + 10m + 25 = m2 + 2×m× 5 + 52 = (m + 5)2 = (m + 5) (m + 5) ∴ The L.C.M. = common factors × remaining factors = (m + 5) (m2 – 5m + 25) (m + 5) = (m +5) (m3 + 125) Example 2: Find the L.C.M. of following expressions. a) x2 + 2xy + y2 – z2 , y2 + 2yz + z2 – x2 and z2 + 2xz + x2 – y2 b) 2a4 – 54a, 2a5 + 18a3 + 162a and 4a3 + 12a2 + 36a c) 1+4p+4p2 – 16p4 , 1+2p – 8p3 – 16p4 and 1 + 4p2 + 16p4 Solution: Here, a) The 1st expression = x2 + 2xy + y2 – z2 = (x + y) 2 – z2 = (x + y + z) (x + y – z) The 2nd expression = y2 + 2yz + z2 – x2 = (y + z) 2 – x2 = (y + z + x) (y + z – x) The 3rd expression = z2 + 2zx + x2 – y2 = (z + x) 2 – y2 = (z + x + y) (z + x – y) ∴ The L.C.M. = common factors × remaining factors = (x + y + z) (x + y – z) (y + z – x) (x – y + z) b) The 1st expression = 2a4 – 54a = 2a (a3 – 27) = 2a (a3 – 33 ) = 2a (a – 3) (a2 + 3a + 9) 1st expression 2nd expression 3(x2 – x+1) 2x2(x2+x+1) 2x(x – 1) 1st expression 2nd expression (m2–5m+25) (m + 5) (m + 5)

Vedanta Excel in Mathematics - Book 9 190 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur The 2nd expression = 2a5 + 18a3 + 162a = 2a (a4 + 9a2 + 81) = 2a {(a2 ) 2 + 92 + 9a2 } = 2a {(a2 + 9)2 – 2.a2 .9 + 9a2 } = 2a {(a2 + 9)2 – 9a2 } = 2a {(a2 + 9)2 – (3a) 2 } = 2a (a2 + 3a + 9) (a2 – 3a + 9) The 3rd expression = 4a3 + 12a2 + 36a = 4a (a2 + 3a + 9) ∴ The L.C.M. = common factors × remaining factors = 4a (a2 + 3a +9) (a – 3) (a2 – 3a + 9) = 4a (a – 3) (a4 + 9a2 + 81) c) The 1st expression = 1 + 4p + 4p2 – 16p4 = (1)2 + 2×1×2p + (2p) 2 – 16p4 = (1 + 2p) 2 – (4p2 ) 2 = (1 + 2p + 4p2 ) (1 + 2p – 4p2 ) The 2nd expression = 1+2p – 8p3 – 16p4 = 1 (1 + 2p) – 8p3 (1 + 2p) = (1 + 2p) (1 – 8p3 ) = (1 + 2p) {1 – (2p) 3 } = (1 + 2p) (1 – 2p) (1 + 2p + 4p2 ) The 3rd expression = 1 + 4p2 + 16p4 = 1 + (4p2 ) 2 + 4p2 = (1 + 4p2 ) 2 – 2 × 1 × 4p2 + 4p2 = (1 + 4p2 ) 2 – 4p2 = (1 + 4p2 ) 2 – (2p) 2 = (1 + 2p + 4p2 ) (1 – 2p + 4p2 ) ∴ The L.C.M. = common factors × remaining factors = (1 + 2p + 4p2 ) (1 + 2p – 4p2 ) (1 + 2p) (1 – 2p) (1 – 2p + 4p2 ) = (1 + 8p3 ) (1 – 8p3 ) (1 + 2p – 4p2 ) = (1 – 64p6 ) (1 + 2p – 4 2 ) Example 3: The H.C.F. and L.C.M. of any two expressions are (a – 9) and (a + 9) (a – 9)2 respectively. If the first expression is (a2 – 81), find the second expression. Solution: Here, the 1st expression = a2 – 81 = (a + 9) (a – 9) H.C.F. = (a – 9) and L.C.M. = (a + 9) (a – 9)2 We have, the first expression × the second expression = H.C.F. × L.C.M. or, The second expression = H.C.F. × L.C.M. First expression or, The second expression = (a – 9) (a + 9) (a – 9)2 (a + 9) (a – 9) =(a – 9)2 Hence, the second expression is (a – 9)2 . H.C.F. and L.C.M.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 191 Vedanta Excel in Mathematics - Book 9 H.C.F. and L.C.M. EXERCISE 10.2 General section 1. a) Define the L.C.M. of algebraic expressions. b) What is the L.C.M. of a2 b3 and a3 b2 ? c) If p is the multiple of q, what is the L.C.M. of p and q? d) Write down the relation of any two algebraic expression and their H.C.F. and L.C.M. 2. Find the L.C.M. of the following expressions: a) 3x (x +1) (x – 1) and 2x2 (x – 1) (x + 3) b) 4x3 (x – 3) (x + 2) and 6x2 (x + 2) (x + 3) c) 8a2 b (a – b) (a2 + ab + b2 ) and 12ab2 (a – b) (a + b) d) (x + 2) (x + 3), (x + 3) (x – 2), (x – 2) (x – 3) e) (a – 3) (a – 4), (a – 4) (a – 5), (a – 5) (a – 3) 3. Find the L.C.M. of the following expressions: a) 3x2 + 6x, 2x3 + 4x2 b) ax2 + ax, ax2 – a c) 4x5 y4 + 2x4 y5 , 10x4 y3 + 5x3 y4 d) x2 – xy, x3 y – xy3 e) 4x2 – 2x, 8x3 – 2x f) a3 – b3 , a2 + ab + b2 g) x2 + 5x + 6, x2 – 4 h) x2 – 9, 3x3 + 81 i) a4 b – ab4 , a4 b2 – a2 b4 j) x4 + x2 y2 + y4 , x3 – y3 k) a4 + a2 b2 + b4 , a3 + b3 l) 6x2 – x – 1, 54x4 + 2x Creative section 4. Find the L.C.M. of the following expressions: a) 4x3 + 500y3 , 2x4 + 50x2 y2 + 1250y4 b) 24a3 – 81b3 , 32a5 + 72a3 b2 + 162ab4 c) 1 + 4x + 4x2 – 16x4 , 1 + 8x3 d) 16a4 – 4a2 – 4a – 1 , 16a4 + 8a3 – 2a – 1 e) x4 – 7x2 + 81, x3 + 5x (x + 2) – x f) 100y4 – 36y2 +1, 10y3 – y (3y + 1) – y2 5. Find the lowest common multiple (L.C.M.) of the following expressions: a) a2 – 4, a3 – 8, (a + 2)2 b) (a – 3)2 , a2 – 9, a3 + 27 c) a3 – 2a2 + a, a3 + a2 – 2a, a3 – 4a d) x4 – y4 , x2 – y2 , x3 – y3 e) 4x3 – 10x2 + 4x, 3x4 – 8x3 + 4x2 , x4 – 8x f) x3 – 9x, x4 – 2x3 – 3x2 , x3 – 27 g) a3 – 4a, a4 – a3 – 2a2 , a3 – 8 h) a4 + a2 + 1, a3 – 1, a3 – a2 + a i) x3 – 2x2 y + 2xy2 – y3 , x4 – y4 , x3 + y3 j) x2 + 3x + 2, x2 + 5x + 6, x2 + 4x + 3 6. Find the L.C.M. of the following expressions: a) a2 + 2ab + b2 – c2 , b2 + 2bc + c2 – a2 and c2 + 2ca + a2 – b2 b) x2 – y2 – 2yz – z2 , y2 – z2 – 2zx – x2 and z2 – x2 – 2xy – y2 c) 9x2 - 4y2 - 8yz – 4z2 , 4z2 - 4y2 - 9x2 – 12xy, 9x2 + 12xz + 4z2 - 4y2 d) x2 +(a + b) x + ab, x2 + (a – b) x – ab and x2 – (a + b) x + ab e) x3 + 2x2 – x – 2, x3 + x2 – 4x – 4 and x4 – 5x2 + 4 f) a2 – 4a – 5 + 6b – b2 , a2 + 2ab + b2 – 25 and a2 + ab – 5a g) x2 + 8x – 33 – 42y – 9y2 , x2 – 6xy + 9y2 – 121 and x2 y + 11xy – 3xy2 7. a) The H.C.F. and L.C.M. of any two expressions are 2ab (a – b) and 12a2 b2 (a + b) (a2 – b2 ) respectively. If the first expression is 6a4 b + 12a3 b2 + 6a2 b3 , find the second expression. b) The H.C.F. and L.C.M. of any two expressions are x2 – x + 1 and (x3 +1) (x2 + x + 1) respectively. If the second expression is x4 + x2 +1, find the first expression.

Vedanta Excel in Mathematics - Book 9 192 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Project work and activity section 8. a) Write a pair of expressions of your own in the forms of a3 + b3 and a4 + a2 b2 +b4 . Then, factorize your expressions and find their lowest common multiple. b) Write a pair of expressions of your own, then show that the product of the expressions is equal to the product of their H.C.F. and L.C.M. OBJECTIVE QUESTIONS Let’s tick (√) the correct alternative. 1. The H.C.F. of a3 bx and abx3 is (A) x (B) ab (C) abx (D) a3 bx3 2. The expression of highest degree which exactly divides the given expression is called (A) H.C.F. (B) L.C.M. (C) Factor (D) Multiple 3. What is the H.C.F. of (x – y) 3 and x3 – y3 ? (A) x – y (B) x3 – y3 (C) (x – y) 3 (D) x2 – y2 4. The L.C.M. of a2 + ay and a3 y – ay3 is (A) a (a + y) (B) ay (a + y) (C) a (a2 – y2 ) (D) ay (a2 – y2 ) 5. If p is a factor of q, what is the H.C.F. of p and q? (A) p (B) q (C) 1 (D) pq 6. If x and y are H.C.F. and L.C.M. of two expressions respectively, what is the relation between x and y? (A) x is always a multiple of y (B) x is always a factor of y (C) y is always a factor of x (D) H.C.F. of x and y is always 1. 7. The highest common factor of a3 + 1 and a4 + a2 + 1 is (A) a+1 (B) a2 + a + 1 (C) a2 – a + 1 (D) (a + 1)3 8. If H and K be the H.C.F. and L.C.M. of the expressions P and Q respectively, what is the relation among H, K, P and Q? (A) PK = HQ (B) HP = KQ (C) HK = PQ (D) H + K = P + Q 9. If the product of two expressions is m (m2 – 9), what is the product of their H.C.F. and L.C.M.? (A) m (m – 3) (B) m (m + 3) (C) m (m2 – 9) (D) m (m – 9) 10. The greatest common divisor of x4 + x2 y2 + y4 and x3 + y3 is (A) (x2 + xy + y2 ) (B) (x2 – xy + y2 ) (C) x3 + y3 (D) 1 https://www.geogebra.org/m/jbytnwfv Vedanta ICT Corner Please! Scan this QR code or browse the link given below: H.C.F. and L.C.M.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 193 Vedanta Excel in Mathematics - Book 9 11.1 Indices – review An index is a number that shows how many times a base is multiplied by itself. An index is also called an exponent. Let's take a few examples of repetitive multiplication of the same base. a = a1 , a × a = a2 , a × a × a = a3 , a × a × a × a = a4 , etc. Here, 1 is the index of a1 , 2 is the index of a2 , 3 is the index of a3 , 4 is the index of a4 , and so on. Thus, the index refers to the power to which a number is raised. For example, in 23 the base 2 is raised to the power 3. Indices is the plural of index. 11.2 Laws of Indices While performing the various operations of indices, we apply different proven rules like product rule, quotient rule, power rule, etc. These rules are well known as laws of indices. The table given below shows the laws of indices at a glance. Name of laws Rules Examples Product law am × an = am + n 23 × 25 = 23+5 = 28 Quotient law am ÷ an = am – n when m > n am ÷ an = 1 an – m when m < n 37 ÷ 33 = 37 – 3 = 34 and 33 ÷ 37 = 1 37 – 3 = 1 34 Power law (am) n = am × n, (ab) m = ambm, a m b = am bm (24 ) 2 = 24 × 2 = 28 , (2x)5 = 25 × x5 , … Negative index law a– m = 1 am or am = 1 a–m 5–3 = 1 53 or 53 = 1 5–3 Zero index law a0 = 1, (ab) 0 = 1, (a + b)0 = 1 20 = 1, 30 = 1, 990 = 1 Root laws of indices m n a = n am or n am = m n a 1 2 3 = 2 31 or 3 , 2 3 3 = 3 32 (i) Product law of indices If am and an are the two algebraic terms, where m and n are the positive integers, then am × an = am + n Unit 11 Indices

Vedanta Excel in Mathematics - Book 9 194 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur Proof We know that, a2 = a × a (two factors) a3 = a × a × a (three factors) am = a × a × a × ... ('m' factors) an = a × a × a × ... ('n' factors) \ am × an = (a × a × a × ... 'm' factors) × (a ×a × a × ... 'n' factors) = a × a × a × ... ('m + n' factors) = am + n Thus, am × an = am + n Similarly, if m, n, p, q, r, ... are the positive integers, then am × an × ap × aq × ar × ... = am + n + p + q + r + ... (ii) Quotient law of indices If am and an are the two algebraic terms, where m and n are the positive integers, then am ÷ an = am – n when m > n am ÷ an = 1 an – m when n > m Proof am ÷ an = am an = a × a × a × ... 'm' factors a × a × a × ... 'n' factors = a × a × a × ... ('m – n' factors), where m > n = am – n Thus, am ÷ an = am – n, when m > n But, in the case of n > m, am ÷ an = am an = a × a × a × ... 'm' factors a × a × a × ... 'n' factors = 1 a × a × a × ... 'n – m' factors = 1 an – m Thus, am ÷ an = 1 an – m, when n > m (iii) Power law of indices If am be an algebraic term, then (am) n = am × n = amn, where m and n are the positive integers. Proof (am) n = am × am × am × ... (n) factors = am + m + m + ... 'n' times 'm' = amn Thus, (am) n = amn cor 1. amn = (am) n cor 2. amn = (an) m cor 3. am bm = m a b cor 4. m a b = am bm cor 5. (ab) m = ambm cor 6. ambm = (ab) m Indices

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 195 Vedanta Excel in Mathematics - Book 9 Indices (vi) Root law of indices If a m n is an algebraic term, where m and n are the positive integers, then a m n = n am Proof nth order of root in represented as n In this way the 2nd order of root is represented as 2 or only The 3rd order of root is represented as 3 and so on. The square root of 32 = 3 = 3 2 2 = 2 32 or 32 The cube root of 53 = 5 = 5 3 3 = 3 53 Thus, 3 2 2 = 2 32 5 3 3 = 3 53 In general, a m n = n am Example 1: Find the value of: a) (16)3 4 b) (9–3) 1 6 c) 243 32 – 2 5 d) (160.5) 3 2 e) 3 729 f) 3 64–1 Solution: a) (16)3 4 = (24 ) 3 4 = (2)4 × 3 4 = 23 = 8 b) (9–3) 1 6 = (32 ) –3 × 1 6 = 3–6 × 1 6 = 3–1= 1 3 c) 243 32 – 2 5 = 35 25 – 2 5 = 25 35 2 5 = 2 3 5 × 2 5 = 2 3 2 = 4 9 d) (160.5) 3 2 = (24 ) 0.5 × 3 2 = 24 × 0.5 × 3 2 = 23 = 8 (iv) Law of negative index If a–m is an algebraic term, where m is a negative integer, a–m = 1 am or, 1 am = a–m or, am = 1 a–m Proof Here, a–m = am – 2m = am ÷ a2m = am a2m = am am × am = 1 am Thus, a–m = 1 am Similarly, 1 am = a–m and am = 1 a–m (v) Law of zero index If a0 is an algebraic term, where a ≠ 0, then a0 = 1. Proof Here, a0 = am – m = am × a–m = am am = 1 Thus, a0 = 1, where a ≠ 0 Worked-out Examples

Vedanta Excel in Mathematics - Book 9 196 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur e) 3 729 = 3 93 = 9 3 3 = 9 = 32 = 3 f) 3 64–1 = 3 1 64 = 1 8 3 2 = 1 8 3 = 1 2 3 3 = 1 2 Example 2: Evaluate: a) 25 16 – 1 2 125 64 1 3 ÷ 16 81 – 1 4 b) 127 × 286 217 × 166 c) (1 – 6–9) –1 + (1 – 69 ) –1 d) 1 (a – b)–1 + 1 (b – c)–1 + 1 (c – a)–1 Solution: a) 25 16 – 1 2 125 64 1 3 ÷ 16 81 – 1 4 = 16 25 1 2 125 64 1 3 ÷ 81 16 1 4 = 4 5 2 × 1 2 5 4 3× 1 3 ÷ 3 2 4× 1 4 = 4 5 5 4 ÷ 3 2 = 4 5 5 4 × 2 3 = 4 5 × 5 6 = 2 3 b) 127 × 286 217 × 166 = (22 × 3)7 × (22 × 7)6 (3 × 7)7 × (24 ) 6 = 214 × 37 × 212 × 76 37 × 77 × 224 = 214 + 12 × 76 224 × 77 = 226 – 24 27 – 6 = 4 7 c) (1 – 6–9) –1 + (1 – 69 ) –1 = 1 – 1 69 –1 + (1 – 69 ) –1 = 69 – 1 69 –1 + (1 – 69 ) –1 = 69 69 – 1 + 1 1 – 69 = 69 69 – 1 – 1 69 – 1 = 69 – 1 69 – 1 = 1 d) 1 (a – b) –1 + 1 (b – c) –1 + 1 (c – a) –1 = (a – b) + (b – c) + (c – a) = 0 Example 3: Prove that a) 7m + 2 + 4 × 7m 7m + 1 × 8 – 3 × 7m = 1 b) 5x – 5x – 1 4 × 5x – 1 = 1 c) 273n + 1 × (243)–4n 5 9n + 5 × 33n – 7 = 1 d) 3–p × 92p – 2 33p – 2 × (3 × 2)–1 = 2 3 Solution: a) LHS = 7m + 2 + 4 × 7m 7m + 1 × 8 – 3 × 7m = 7m × 72 + 4 × 7m 7m × 71 × 8 – 3 × 7m = 7m (49 + 4) 7m (7 × 8 – 3) = 53 53 = 1 = RHS b) LHS = 5x – 5x – 1 4 × 5x – 1 = 5x – 5 5 x 4 × 5 5 x = 5x 1 – 1 5 4 × 5 5 x = 4 5 4 5 = 1 = RHS c) LHS = 273n + 1 × (243)–4n 5 9n + 5 × 33n – 7 = (33 ) 3n + 1 × (35 ) –4n 5 (32 ) n + 5 × 33n – 7 = 39n + 3 × 3–4n 32n + 10 × 33n – 7 = 39n + 3 – 4n 32n + 10 + 3n – 7 = 35n + 3 35n + 3 = 35n + 3 – (5n + 3) = 30 = 1 = RHS d) LHS = 3–p × 92p – 2 33p – 2 × (3 × 2)–1 = 3–p × (32 ) 2p – 2 33p – 2 × 3–1 × 2–1 = 3–p + 4p – 4 33p – 2 –1 × 2–1 = 33p – 4 33p – 3 × 2–1 = 33p – 4 – 3p + 3 2–1 = 3–1 2–1 = 2 3 = RHS Indices

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 197 Vedanta Excel in Mathematics - Book 9 Indices Example 4: Simplify: a) 50ba5 c–2 3 × 20ab5 c8 3 b) 27x3 y6 3 ÷ 81x4 y8 4 c) (a + b)–8 3 × (a + b)2 3 Solution: a) 50ba5 c–2 3 × 20ab5 c8 3 = 1000a5+1 b1+ 5 c–2 + 8 3 = 1000a6 b6 c6 3 = (103 a6 b6 c6 ) 1 3 = 10a2 b2 c2 b) 27x3 y6 3 ÷ 81x4 y8 4 = (33 x3 y6 ) 1 3 ÷ (34 x4 y8 ) 1 4 = 3xy2 ÷ 3xy2 = 1. c) (a + b) –8 3 × (a + b) 2 3 = (a + b) – 8 3 × (a + b) 2 3 = (a + b) – 8 3 + 2 3 = (a + b) – 8 + 2 3 = (a + b) – 6 3 = (a + b) –2 = 1 (a + b) 2 Example 5: Simplify: xa xb a2 + ab + b2 × xb xc b2 + bc + c2 × xc xa c2 + ca + a2 Solution: xa xb a2 + ab + b2 × xb xc b2 + bc + c2 × xc xa c2 + ca + a2 = (xa – b) a2 + ab + b2 × (xb – c) b2 + bc + c2 × (xc – a) c2 + ca + a2 = xa3 – b3 × xb3 – c3 × xc3 – a3 = xa3 – b3 + b3 – c3 + c3 – a3 = x0 = 1 Example 6: Simplify: a+b xa2 xb2 × b+c xb2 xc2 × c+a xc2 xa2 Solution: a+b xa2 xb2 × b+c xb2 xc2 × c+a xc2 xa2 = x a2 – b2 a + b × x b2 – c2 b + c × x c2 – a2 c + a = xa – b × xb – c × xc – a = xa – b + b – c + c – a = x0 = 1 Example 7 : Simplify: 1 1 + xa – b + xc – b + 1 1 + xb – c + xa – c + 1 1 + xc – a + xb – a Solution: 1 1 + xa – b + xc – b + 1 1 + xb – c + xa – c + 1 1 + xc – a + xb – a = xa xb xc x 1 + b 1 + + xb xc xa x 1 + c 1 + + xc xa xb x 1 + a 1 + = xb + xa + xc xb 1 + xc + xb + xa xc 1 + xa + xc + xb xa 1 = xb xa + xb + xc + xc xa + xb + xc + xa xa + xb + xc = xb + xc + xa xa + xb + xc = 1 Example 8 : Simplify m + (mn2 ) 1 3 + (m2 n)1 3 m – n × 1 – n 1 3 m1 3 Solution: m + (mn2 ) 1 3 + (m2 n)1 3 m – n × 1 – n 1 3 m1 3 = m +m1 3 n 2 3 + m2 3 n 1 3 m – n × m1 3 – n 1 3 m1 3 = m1 3 (m2 3 + n2 3 + m1 3 n1 3 ) m – n × m1 3 – n 1 3 m1 3

Vedanta Excel in Mathematics - Book 9 198 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur = (m1 3 – n1 3) [(m1 3) 2 + m1 3.n 1 3+ (n 1 3) 2 ] m – n = (m1 3) 3 – (n1 3) 3 m – n = m – n m – n = 1 Example 9 : Show that x2 – 1 y2 x × x – 1 y y – x y2 – 1 x2 y × y + 1 x x – y = x y x + y Solution: Here, LHS = x2 – 1 y2 x × x – 1 y y – x y2 – 1 x2 y × y + 1 x x – y = x + 1 y x x – 1 y x x – 1 y y – x y + 1 x y y – 1 x y y + 1 x x – y = x + 1 y x x – 1 y x + y – x y + 1 x y + x – y y – 1 x y = x + 1 y x x – 1 y y y + 1 x x y – 1 x y = x + 1 y x y + 1 x × x – 1 y y y – 1 x = xy + 1 y x xy + 1 x × xy – 1 y y xy – 1 x = xy + 1 y × x xy + 1 x × xy – 1 y × x xy – 1 y = x y x × x y y = x y x + y = RHS Example 10 : If pqr = 1, prove that 1 1 + p + q–1 + 1 1 + q + r–1 + 1 1 + r + p–1 = 1. Solution: Here, pqr = 1, then qr = 1 p = p–1 Now, LHS = 1 1 + p + q–1 + 1 1 + q + r–1 + 1 1 + r + p–1 = qr qr(1 + p + q–1) + r r(1 + q + r–1) + 1 1 + r + p–1 = qr qr + pqr + r + r r + qr + 1 + 1 1 + r + qr = qr qr + 1 + r + r qr + r + 1 + 1 qr + r + 1 = qr + 1 + r qr + 1 + r = 1 = RHS Example 11: If x – 1 = 2 2 3 + 2 1 3 , show that x3 – 3x2 – 3x = 1. Solution: Here, x – 1 = 2 2 3 + 2 1 3 or, (x – 1)3 = (2 2 3 + 2 1 3 ) 3 [Cubing on both sides] or, x3 – 3x2 + 3x – 1 = 2 2 3 3 + 2 1 3 3 + 3. 2 2 3 . 2 1 3 2 2 3 + 2 1 3 or, x3 – 3x2 + 3x – 1 = 4 + 2 + 6 (x – 1) or, x3 – 3x2 + 3x – 1 = 6 + 6x – 6 or, x3 – 3x2 + 3x – 6x = 1 or, x3 – 3x2 – 3x = 1 Hence, x3 – 3x2 – 3x = 1 proved. Indices

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 199 Vedanta Excel in Mathematics - Book 9 Indices General section 1. a) Express am × an as a single base. b) What is the value of (5a)0 , a ≠ 0? c) Find the value of am – n × an – m. d) What is the value of (p + q) 0 + 1p + q? 2. Evaluate: a) 29 × 2-6 b) 5–7 × 58 c) 11–5 ÷ 11–3 d) (25) 3 2 e) (64)– 2 3 f) 1 64 1 6 g) 8 27 – 4 3 h) 169 196 – 1 2 i) (32–1) 5–1 j) a0 125 – 2 3 k) (70.5) 2 l) 9 25 0.5 × 32 243 0.2 m) 3 64–1 n) 729 64 3 –1 o) 1003 4 × 1 100 4 p) 3 9 3 9 9 Creative section - A 3. Find the value of: a) 8 27 – 1 3 ÷ 4 9 – 1 2 b) 125 64 – 2 3 ÷ 625 256 – 1 2 c) 27 8 – 1 3 81 16 1 4 ÷ 4 25 – 1 2 d) 25 16 – 1 2 125 64 1 3 ÷ 8 27 – 1 3 4. Simplify: a) (8a3 ÷ 27x–3) – 2 3 b) (125p3 ÷ 64q–3) – 2 3 c) 146 × 155 356 × 65 d) 409 × 498 569 × 358 5. Simplify: a) (xa ) b – c × (xb ) c – a × (xc ) a – b b) (ax + y) x – y × (ay+z) y – z × (az – x) z + x c) x2a + 3b × x3a – 4b xa + 2b × x4a – 3b d) 1 1 + ax – y + 1 1 + ay – x e) 1 1 – xm – n + 1 1 – xn – m f) (1 – 3–5) –1 + (1 – 35 ) –1 g) (a + b) –1.(a–1 + b–1) h) y–1 x–1 + x–1 y–1 –1 6. Prove that: a) 3x + 1 + 3x 4 × 3x = 1 b) 5n + 2 – 5n 24 × 5n = 1 c) 72p + 1 – 3 × 49p 4 × 49p = 1 d) 6m + 2 – 6m 6m+1 – 6m = 7 e) 7n + 2 + 4 × 7n 7n + 1 × 8 – 3 × 7n = 1 f) 5a + 3 – 55 × 5a – 1 5a + 2 + 89 × 5a = 1 7. Simplify: a) 3p – 3p – 1 3p + 1 + 3p b) 5x – 5x – 1 4 × 5x – 1 c) 5 × 2m – 4 × 2m – 2 3 × 2m + 2 – 5 × 2m + 1 d) 2n +2 × (2n – 1) n + 1 2n(n – 1) ÷ 4n e) 5–n × 625n – 1 53n – 2 × (5 × 2)–1 f) 9x × 3x – 1 – 3x 32x + 1 × 3x – 2 – 3x 8. Simplify: a) 25a2 b2 × 27a3 3 b) a6 b–2c4 ÷ a4 b–4c8 4 EXERCISE 11.1

Vedanta Excel in Mathematics - Book 9 200 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur c) 16x8 y4 4 ÷ 8x6 y3 3 d) 56p7 q4 3 7p4 q7 3 e) 216m7 n5 4 ÷ 6–1m–1n 4 f) (a + b) –7 3 × (a + b) 1 3 g) (2x – y) –8 3 ÷ (2x – y) – 2 3 h) (a + b) –1 × (a – b) (a2 – b2 ) 9. Simplify: a) xa xb a + b × xb xc b + c × xc xa c + a b) ax a–y x – y × ay a–z y – z × az a–x z – x c) xa xb a2 + ab + b2 × xb xc b2 + bc + c2 × xc xa c2 + ca + a2 d) xl 2+m2 x–lm l – m × xm2+n2 x–mn m–n × xn2+l 2 x–nl n – l e) xa + b xa – b c – a × xb + c xb – c a – b × xc + a xc – a b – c f) xm + n xp m – n × xn + p xm n–p × xp + m xn p – m g) p + (pq2 ) 1 3 + (p2 q) 1 3 p – q × 1 – q 1 3 p 1 3 h) x b b – c 1 b – a × x c c – a 1 c – b × x a a – b 1 a – c i) x + 1 y a × 1 y – x a y + 1 x a × 1 x – y a j) a2 – 1 b2 a × a – 1 b b – a b2 – 1 a2 b × b + 1 a a – b 10. Simplify: a) ay az yz × az ax zx × ax ay xy b) xr – p xr – q pq × xq – r xq – p pr × xp – q xp – r qr c) x+y ax2 ay2 × y+z ay2 az2 × z+x az2 ax2 d) xb xc 1 bc × xc xa 1 ca × xa xb 1 ab e) 1 1 + xa – b + xc – b + 1 1 + xb – c + xa – c + 1 1 + xc – a + xb – a f) 1 1 + ax – y + ax – z + 1 1 + ay – z + ay – x + 1 1 + az – x + az – y Creative section - B 11. a) If a3 + b3 + c3 = 0, prove that (xa + b) a2 – ab + b2 × (xb + c) b2 – bc + c2 × (xc + a) c2 – ca + a2 = 1 b) If a = xq + r.yp , b = xr + p.yq and c = xp + q.yr , prove that aq – r × br – p × cp – q = 1 c) If xyz = 1, prove that 1 1 + x + y–1 + 1 1 + y + z–1 + 1 1 + z + x–1 = 1. d) If a + b + c = 0, prove that 1 1 + xa + x–b + 1 1 + xb + x–c + 1 1 + xc + x–a = 1. 12. a) If x = 2 1 3 + 2 – 1 3 , prove that 2x3 – 6x = 5. b) If a = p 1 3 – p – 1 3 , prove that a3 + 3a = p – 1 p . c) If x – 2 = 3 1 3 + 3 2 3 , show that x(x2 – 6x + 3) = 2. Indices