The Leading Maths - 8

ALGEBRA 145 (d) Multiply: (i) x3 × x2x (ii) 82a × 23a (e) Divide: (i) 23x 2x (ii) 42x ÷ 43x (f) Write the product of am and an in index form. (g) What is the value of (ax)0 ? (h) Why a0 = 1? Justify with suitable example. 2. Write whether the following statements are true or false. (a) The product of 4x and 4x is 44x2 . (b) The quotient of 9x ÷ 3x is 3. (c) The value of any base with zero power is 1. (d) The relation am – n = 1 am – n is correct. 3. Circle ( ) the correct answers. (a) What is the value of 32a × 33a in index form? i. 35a ii. 3a iii. 3 iv. 36a (b) What is the value of 2x2 y 0 ? i. 0 ii. 1 iii. x y iv. 2x y (c) What is the value of 2x0 + 3? i. 3 ii. 4 iii. 5 iv. 6 (d) Which is the correct value of (3x0 – a0 )2 ? i. 1 ii. 2 iii. 3 iv. 4 (e) Which is the correct value of 9x2 ? i. 18x ii. 9x iii. 2x iv. 3x (f) Which is the exponent form of (– 2a) × (– 2a) × (–2a)? i. 8a ii. 8a3 iii. – (3a)3 iv. (– 2a)3 (g) What is the value of 32x × 33x × 1 3x? i. 34x ii. 33x iii. 32x iv. 35x (h) What is the value of (a + b)2 when a = 2 and b = 1? i. 2 ii. 3 iii. 4 iv. 9

146 The Leading Maths - 8 4. Rewrite the following without using radical sign ( ). (a) 9 (b) 27 3 (c) 32 5 5. Rewrite the following by using the radical sign. (a) 3 1 2 (b) a– 1 3 (c) b 2 3 6. Simplify: (a) 8– 2 3 (b) 27– 2 3 (c) 8 27 – 2 3 7. Write simplified equivalent expressions by using positive indices only. (a) x3 x– 4 (b) a2 x3 y–2 b–2xy–4 (c) 2– 1 × 3– 2 × 40 8. Write in index notation: (a) i. (x + y)2 (x + y)3 ii. (2x + 3y)2m (2x + 3y)3n iii. 23 .23 .24 (b) i. (3a)2 ii. (xy)3 iii. (ab)2x (c) i. 3 2 2 ii. a b 3 iii. x + y x – y 3 (d) i. (52 )3 ii. (a2 ) 3 iii. [(a + b)2 ]3 9. Simplify each of the following by removing negative index and radical signs. (a) (a–2)3 , (x–3)–4, (x–2y3 z0 )–1, x–2y5 x–3y–2. (b) 16 – 3 2 , 2 –3 × 2 –3 , 4 –5 × 2 10 , 27 3 × 81 4 . (c) x4 y–2, x4 × xy 3 2 × x2 y 3 , x– 3 2 ÷ y– 3 2 6 . 10. Prove that: (a) x–1y y–1z z–1x = 1 (b) ab2 3 a2 b4 3 a–3b–6 3 = 1 (c) xa xb 2 xb xc 2 xc xa 2 =1 (d) a x – y × ay –z × az – x = 1 (e) xa +b ÷ x c + b × xc + d ÷ x d + a = 1

ALGEBRA 147 (f) xm xn m + n × xn xp m + p × xp xm p + m = 1 (g) ax a– y x – y × ay a– z y – z × az a– x z – x = 1 (h) ap(x – y) × ap(y – z) × ap(z – x) = 1 (i) xa xb a2 + ab + b2 × xb xc b2 + bc + c2 × xc xa c2 + ca + a2 = 1 (j) ax2 + y2 a– xy x – y × ay2 + yz a– z2 y – z × ax2 + zx a– z2 z – x = 0 ANSWERS 4. (a) 3 (b) 3 (c) 2 5. (a) 3 (b) 1 a 3 (c) b2 3 6. (a) 1 4 (b) 1 9 (c) 9 4 7. (a) 1 x (b) (abxy)2 (c) 1 18 8. (a) (x + y)5 , (2x + 3y)2m + 3n , 210 (b) 9a2 , x3 y3 , a2x b2x (c) 9 4 , a3 b3 , (x + y)3 (x – y)3 (d) 56 , a6 , (a + b)6 9. (a) 1 a6 ; x+2 ; x2 y3 ; xy7 (b) 1 8 ; 1 8 ; 1 ; 9 (c) x2 y ; x2 y ; y9 x9

148 The Leading Maths - 8 CHAPTER 9 ALGEBRAIC EXPRESSION Lesson Topics Pages 9.1 Factorization By Grouping 149 9.2 Factorization of Difference of Two Squares 153 9.3 Factorization of Trinomial 157 9.4 HCF 166 9.5 LCM 170 9.6 Rational Algebraic Expression 175 ” What are constant and variable? Give some examples. ” What is the value of 'a' in a = 23? Is it constant or variable? ” If x2 = 4, what is the value of x? Is x constant or variable? ” What is the total number of the students in this class 8? ” What are the heights of the students in this class 8? ” What are the factors of (a + b)2 , (a – b)2 and a2 – b2 ? WARM-UP

ALGEBRA 149 9.1 Factorization By Grouping Terms At the end of this topic, the student will be able to: ¾ factorise the polynomials by grouping the terms. Learning Objectives I. Polynomials A term is a product of members denoted by digits and letters. e.g. 1, a, 2a, 3x2 y, 1 2a, etc. Usually in a term counting of a number and a letter, the number is called a coefficient. An algebraic expression is a combination of terms separated by addition or subtraction. e.g., 1 + a, 1 2a + a, a – 1 2a , 3 x2 + 2a – 1, etc. A polynomial is an algebraic expression that has only terms of the form axn where a is a real number, n is a whole number or non-negative integer and x is any variable. A polynomial is said to be a polynomial with integer if each term has integer coefficient. eg, 2a + 1, 3x2 + 2xy + y are polynomials with integer coefficient. Note: In this chapter we are considering polynomials with integer coefficient integer only. II. Factor of a polynomial If a polynomial is expressed as a product of other polynomial, each of the other polynomials are called factors of the polynomial. eg, (i) 2a = 2. a. Then 2 and a are factors of the polynomial 2a. (ii) 2ab + 4b2 = 2b(a + b). The polynomials a – b and a + b are factors of the polynomial a2 – b2 . III. Prime polynomial A polynomial is said to be prime if it cannot be exposed as a product of other polynomials lower degree. e.g., x + 1 is a prime polynomial. a2 + 1 is a prime polynomial. a 2 2ab 4b2 2b a b

150 The Leading Maths - 8 But x2 – 1 is not a prime polynomial because x2 – 1 = (x – 1) (x + 1) which is a product of lower degree polynomial. A polynomial is said to be factored completely if it is written as product of factors where each factor is a monomial, a prime polynomial or a power of a prime polynomial. e.g., (a) 2x + 2 = 2 (x + 1) Here, the factor 2 is also called common factor of the terms 2x and 2. (b) a2 + 2a + 1 = (a + 1)2 , etc. are completely factorized. CLASSWORK EXAMPLES Example : 1 Factorize : (i) 2a + 16 (ii) 2a – a2 Solution: (i) Here, each term of the expression 2a + 16 has 2 as a factor. Then, 2a + 16 = 2.a + 2.2.2.2 = 2(a + 8) (ii) Here, each term of the expression 2a – a2 has a as a factor. Then, 2a – a2 = 2.a – a.a = a(2 – a ) Example : 2 Resolve into factors: a2 bc + ab2 c + abc2 . Solution: Here, a2 bc = a × a × b × c ab2 c = a × b × b × c = a × b × b × c abc2 = a × b × c × c = a × b × c × c Thus, the common factor is abc. Hence, a2 bc + ab2 c + abc2 = abc(a + b + c) a 8 ← → 2a 1616 2← → a a 2 2 – a a 2 a b c a2 bc abc2 ← →abc ab2 c

ALGEBRA 151 Example : 3 Factorize: x2 + ax + bx + ab Solution: Here, We first group the terms and then factorize. x2 + ax + bx + ab = (x2 + ax) + (bx + ab) = x(x + a) + b(x + a) = (x + a) (x + b) Example : 4 Resolve into factors: x3 + x2 + x + 1 Solution: Here, x3 + x2 + x + 1 = (x3 + x) + (x2 + 1) = x(x2 + 1) + 1(x2 + 1)} = (x2 + 1) (x + 1). EXERCISE 9.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is term? Write its parts with an example. (b) Write the difference between constant and variable. Which symbols are generally used for constants and variables? (c) Define polynomial with examples. (d) Why is polynomial different from algebraic expression? (e) What are the factors of the polynomials 2x + x2 ? 2. Circle ( ) the correct answers. (a) The algebraic expression axn + b is a polynomial when n is ............. i. a natural number ii. a whole number iii. a negative number iv. a rational number (b) What are the factors of the polynomial 3x2 – 2x? i. x ii. 3x iii. 3x, (x – 2) iv. x, (3x – 2) ← x→ ← a → x2 xb ab ax ← →← → x b x2 1 x3 x2 1 x x 1

152 The Leading Maths - 8 (c) What is the area of the given rectangle? i. 3x – 2x ii. 6x2 – 1 iii. 6x2 – 2x iv. 6x – 2x2 (d) What is the product of (a + b) and (a – b)? i. (a + b)2 ii. (a – b)2 iii. a2 + b2 iv. a2 – b2 (e) What is the expended form of (a + b)2 ? i. a2 + b2 ii. a2 + 2ab + b2 iii. a2 – b2 iv. a2 – 2ab + b2 3. Factorize: (a) a2 + ab2 (b) 3xy – 6y2 (c) x3 – x2 (d) p3 – 3p2 q (e) a(a – b) + b(a – b) (f) a(a – b) + b(a – b) + (a – b)2 4. Group the terms and factorize: (a) abc + bcd + cda + cd2 (b) a2 bc + ab2 c + abc2 + abcd (c) x2 + 3x + 4x + 12 (d) x2 – 3x + x – 3 (e) ab + bc + ac + b2 (f) 2x2 + xy – 2x – y (h) 2x2 y + x2 z – 2xy2 – xyz (h) 2p2 qr – 2p2 r + 2pq2 r – 2pqr (i) a3 + a2 + ax + x + a + 1 (j) x5 + x4 + x3 + 2x2 + 2x + 2 ANSWERS 3. (a) a (a + b2 ) (b) 3y (x – 2y) (c) x2 (x – 1) (d) p2 (p – 3q) (e) (a – b) (a + b) (f) 2a (a – b) 4. (a) c (a + d) (b + d) (b) bc (a + b) (a + c) (c) (x + 3) (x + 4) (d) ( x – 3) (x + 1) (e) (a + b) (a + c) (f) (x – 1) (2x + y) (g) x(x – y) (2y + z) (h) pr(2p + q) (q – 2) (i) (a + 1) (a2 + x + 1) (j) (x3 + 2) (x2 + x + 1) 3x – 1 2x

ALGEBRA 153 9.2 Factorization of Difference of Two Squares At the end of this topic, the student will be able to: ¾ factorize the algebraic expressions in the terms of a2 – b2 . Learning Objectives ACTIVITY - 1 The length and height of a rectangular room are (a + b) and (a – b) respectively. What is its area? We know, Area of rectangular room (A) = l × b (a + b)(a – b) = a (a – b)+ b(a – b) = a2 – ab + ba – b2 = a2 – ab + ab – b2 (ab = ba. Why?) = a2 – b2 . Hence, a2 – b2 = (a + b)(a – b). Geometric interpretation of a2 – b2 a + b a – b Draw a square with length 'a' units whose area is a2 sq. units, small square with length 'b' units whose area is b2 sq. units at one corner from the big previous square as shown in the adjoining figure. Also, cut the remaining part with area a2 – b2 and arrange as shown in the figure such that it is a rectangle. Introduction This section begins with a short review of the fundamental methods of factoring certain polynomials with rational coefficients, i.e., algebraic expressions with positive integral indices and rational coefficients. To factorize a polynomial is to express the polynomial as a product of other polynomials called factors of the given polynomial. In other words, the process of finding the factors whose product will be a given expression is called factorization or resolution into factors. Note that we shall not deal with factorization processes that deal with numbers or quantities such as 2, x, 3x, 3, x–1, etc. It is not difficult to show by direct multiplication that x2 – 4 = (x + 2) (x – 2) and x – 4 = (x + 2) (x – 2). We however do not deal with such cases. A. Standard Forms of Factorization We shall factorize the following cases only by using algebraic formulae: (i) Terms with common factor Consider an expression of the form ab + ac. Then ab + ac = a(b + c). (ii) Difference of two squares Let us recall the standard identity (a + b)(a – b) = a2 – b2 . Obviously, a2 – b2 = (a + b)(a – b). (iii) Trinomial of the form x2 + px + q We consider the following product of (x + 3) and (x + 2): Multiplication Factorization (x + 3) (x + 2) = x2 + (3 + 2) x + 3× 2 Considering the reverse process we see that the trinomials of the form x2 + (3 + 2) x + 3 × 2 can be resolved into two factors of the type x + 3 and x + 2 by finding two factors 3 and 2 of the constant term 3 × 2 such that their algebraic sum is the numerical coefficient of the middle term x in the given expression. Thus to factorize an expression of the type x2 + px + q (i.e., with unity as the coefficient of x2 ), we shall find two numbers a and b such that ab = q and a + b = p. Then, x2 + px + q = x2 + (a + b)x + ab = x2 + ax + bx + ab = x(x + a) + b(x + a) = (x + a)(x + b). ∴ x2 + px + q = (x + a)(x + b), where ab = q and a + b = p. b a b a a – b a – b b a b a a – b a – b a b b a A = a2 – b2 A = (a + b)(a – b) x2 2x 3× 2 x 3x x 2 3 x + 3 x + 2 x2 + (3 + 2) x + 3 × 2 = x2 + 5x + 6 where, 6 = 3 × 2 and 5 = 3 + 2 4.1 FACTORIZATION At the end of this topic, the students will be able to: � factorize the given polynomial expressions. Learning Objectives 112 Active Mathematics-9 The rectangle has length (a + b) and breadth (a – b). So, the are of rectangle is l × b = (a + b) (a – b). Hence, a2 – b2 = (a + b) (a – b). Thus, the difference of the squares of any two numbers taken in a given order is the product of the sum of the two numbers and their difference in the given order.

154 The Leading Maths - 8 (a – b) a b (a – b) (a + b) (a + b) (a – b) (a – b) b b a (a2 – b2 ) Using this identity, we can factorize any expression which is the difference of two squares. CLASSWORK EXAMPLES Example : 1 Factorize each of the following: (a) x2 – 4 (b) 4x2 y2 – y4 (c) x4 – y4 (d) (3a2 + 2b2 )2 – (a2 – 3b2 )2 Solution: (a) x2 – 4 = (x2 ) – (2)2 = (x + 2)(x – 2) [ a2 – b2 = (a + b) (a – b)] (b) 4x2 y2 –y4 = y2 (4x2 – y2 ) = y2 {(2x)2 – y2 } = y2 (2x + y)(2x – y). (c) x4 – y4 = (x)4 – (y)4 = (x2 )2 – (y2 )2 = (x2 + y2 ) (x2 – y2 ) [ a2 – b2 = (a + b) (a – b)] = (x2 + y2 )(x + y)(x – y). (d) (3a2 + 2b2 )2 – (a2 – 3b2 )2 [ a2 – b2 = (a + b) (a – b)] = [(3a2 + 2b2 ) + (a2 – 3b2 )][(3a2 + 2b2 ) – (a2 – 3b2 )] = [3a2 + 2b2 + a2 – 3b2 ][3a2 + 2b2 – a2 + 3b2 ] = [4a2 – b2 ][2a2 + 5b2 ] = [(2a)2 – b2 ][2a2 + 5b2 ] = (2a + b)(2a – b)(2a2 + 5b2 )

ALGEBRA 155 EXERCISE 9.2 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is the meaning of factorization? (b) What are the factors of a2 – b2 ? (c) If one factor of x2 – 4 is x – 2, write its another factor. (d) Write the area of the shaded region in the given figure. 2. Circle ( ) the correct answers. (a) Which are the correct factors of x2 – y2 ? i. (x – y)(x – y) ii. (x + y)(x – y) iii. (x + y)(x + y) iv. x2 + y2 (b) Which are the factors of x4 – 9? i. (x + 3)(x – 3) ii. (x2 – 3)(x + 3) iii. (x2 + 3)(x2 – 3) iv. (x2 + 3)(x2 + 3) (c) What is the area of the shaded portion in the figure? i. a2 – b2 – c2 ii. ab – c2 iii. abc iv. a2 – bc (d) Which are the factors of p4 – q4 ? i. (p2 + q2 ), (p – q), (p + q) ii. (p2 – q2 ), (p2 – q2 ) iii. p3 , (p – q) iv. p2 , q2 , (p2 – q2 ) (e) 4a2 – 9b2 = ............ i. (2a + 3b) (a – b) ii. (2a + 3b) (2a – 3b) iii. (3a + 2b) (3a – 2b) iv. (4a – 9b) (a + b) 3. Factorize the following algebraic expressions. (a) 4x2 – y2 (b) 9x2 – 25y2 (c) 16 – 81a2 b2 (d) a2 x – 49b2 x (e) 25b2 x – b2 xy2 (f) 1 – 9x2 25y2 (g) 81 – x4 (h) 625a4 – b4 (i) (3 + b2 )2 – (1 – 2b2 )2 (j) (3a2 + b2 )2 – (a2 – 2b2 )2 p cm q cm p cm c a b

156 The Leading Maths - 8 4. By using the formula of a2 – b2 , find the value of: (a) 42 – 22 (b) 122 – 82 (c) (4.2)2 – (2.4)2 5. Find the area of the shaded region in the figures below: (a) (b) (c) 6. Find the area of the shaded fraction by using the formula of a2 – b2 . (a) (b) (c) ANSWERS 3. (a) (2x + y) (2x – y) (b) (3x + 5y) (3x – 5y) (c) (4 – 9ab) (4 + 9ab) (d) x (a + 7b) (a – 7b) (e) b2 x (5 + y) (5 – y) (f) 1 + 3x 5y 1 – 3x 5y (g) (9 + x2 ) (3 + x) (3 – x) (h) (25a2 + b2 ) (5a + b) (5a – b) (i) (2 + b) (2 – b) (2 + 3b2 ) (j) (2a + b) (2a – b) (2a2 + 3b2 ) 4. (a) 12 (b) 80 (c) 11.88 5. (a) (ab – cd)cm2 (b) a2 – b2 (c) p2 – q2 cm2 6. (a) 13 cm2 (b) 153 cm2 (c) 189 cm2 d cm c cm a cm b cm A B D C S P Q R S P Q R b cm b cm a cm a cm C D F E M A Q N q cm q cm q cm q cm O G P 6 cm 6 cm 7 cm 7 cm 4 cm 4 cm 13 cm 13 cm 6 cm 6 cm 15 cm 15 cm

ALGEBRA 157 9.3 Factorization of Trinomial At the end of this topic, the student will be able to: ¾ factorise the algebraic expression in the term of x2 + px + q and ar2 + bx + c Learning Objectives I. Introduction By direct multiplication, we have (x + 4) (x + 2) = x2 + 4x + 2x + 8 = x2 + 6x + 8 ......... (i) (x + 4)(x – 2) = x2 + 4x – 2x – 8 = x2 + 2x – 8 … (ii) (x – 4)(x + 2) = x2 – 4x + 2x – 8 = x2 – 2x – 8 … (iii) (x – 4)(x – 2) = x2 – 4x – 2x + 8 = x2 – 6x + 8 … (iv) Note:In each case, the product contains three terms such that; 1. The first term is the product of the first terms of the two factors, 2. The coefficient in the second term is the algebraic sum of the product of the first terms and the constant terms and 3. The third term is the product of the two last constant terms. The product of (x + a) and (x + b) can be expressed as follows: Multiplication (x + a)(x + b) = x2 + (a + b)x + ab Factorization Consider the reverse process. We notice that the trinomials are of the form x2 + (a + b) x + ab. It can be written as or resolved into two factors of the type x + a and x + b. For this, we have to find two factors a and b of the constant term ab. The factors are such that their algebraic sum (a + b) is the numerical coefficient of the middle term in the given expression. II. Factorization of trigonomial of the form x2 + px + q Thus, to factorize an expression of the type x2 + px + q (i.e., with unity as the coefficient of x2 ), we shall find two numbers a and b such that ab = q and a + b = p. x2 4x 2x 8 x x 2 4 ∴ (x + 4) (x + 2) = x2 + 6x + 8

158 The Leading Maths - 8 Then, x2 + px + q = x2 + (a + b) x + ab = x2 + ax + bx + ab = x(x + a) + b(x + a) = (x + a)(x + b). III. Factorization of trinomial of the form ax2 + bx + c Consider the following products: (3x + 2)(2x + 5) = 6x2 + 15x + 4x + 10 = 6x2 + 19x + 10 (3x – 2)(2x – 5) = 6x2 – 15x – 4x + 10 = 6x2 – 19x + 10 (3x – 2)(2x + 5) = 6x2 + 15x – 4x – 10 = 6x2 + 11x – 10 (3x + 2)(2x – 5) = 6x2 – 15x + 4x – 10 = 6x2 – 11 x – 10 Note: In each case, the product contains three terms such that; 1. The first term is the product of the two first terms of the factors. 2. The coefficient in the second term is the algebraic sum of the coefficients of the products of the ‘outer’ terms and ‘inner’ terms and 3. The third term is the product of the two last terms of the factors. To cover all four cases considered above, we consider the following product of (ax + b) and (cx + d): Multiplication (ax + b)(cx + d) = acx2 + (ad + bc)x + bd Factorization Considering the reverse process we see that the trinomials of the form ac x2 + (ad + bc)x + bd can be resolved into two factors of the type ax + b and cx + d by finding two factors ad and bc of the constant term (ac)(bd) = abcd such that their algebraic sum is the numerical coefficient of the middle term x in the given expression, Thus, to factorize an expression of the type px2 + qx + r (i.e., with not-unity as the coefficient of x2 ), we shall find two numbers ac = p and bd = r such that adbc = pr and ad + bc = q. (x + a) (x + b) x + b ⇒ = x2 + (a + b) x + ab = x2 + px + q b x x x2 bx ax ax a x + a 5x 5x 5x 10 x2 x2 x2 2x x2 x2 x2 2x x 5 x x x x 2 2x + 5 → → 3x + 2 (3x + 2) (2x 5) = 6x2 + 19x + 10

ALGEBRA 159 Then, px2 + qx + r = acx2 + adx + bcx + bd = ax(cx + d) + b(cx + d) = (ax + b)(cx + d). CLASSWORK EXAMPLES Example : 1 Resolve the expression x2 + 5x + 6 into factors. Solution: We have to find two numbers whose product is 6 and sum is 5. Obviously, the two factors of 6 are 2 and 3 (i.e. 6 = 2 × 3 and 5 = 3 + 2). Thus, x2 + 5x + 6 = x2 + (x + 2) x + 6 = x2 + 3x + 2x + 6 = x(x + 3) + 2(x + 3) = (x + 3)(x + 2) Example : 2 Resolve x2 – 5x + 6 into factors. Solution: Here, 2 × 3 = 6 and 2 + 3 = 5 (– 2) × (– 3) = 6 and (– 2) + (– 3) = – 5 So, the possible factors of + 6 must be both negative. Thus, x2 – 5x + 6 = x2 – (2 + 3) x + 6 = x2 – 2x – 3x + 6 = x(x – 2) – 3(x – 2) = (x – 2)(x – 3). Example : 3 Factorize x2 + 2x – 15. Solution: Here, we need two numbers whose sum is 2 and product –15. The factors of 15 are 1, 5 and 15. Since the product must be negative, one of the numbers must be positive and the other negative such that their sum is positive. The two possible choices are (i) 5 and –3 and (ii) –5 and 3 x2 x2 x x x x x x x + 3 Area = l × b = (x + 3) (x + 2 x + 2 x x x x x x 1 1 1 1 1 1 1 1 1 1 1 1

160 The Leading Maths - 8 Since 5 + (–3) = +2 and (–5) + 3 = – 2, we must take 5 and –3. Thus, x2 + 2x – 15 = x2 + (5 – 3) x – 15 = x2 + 5x – 3x – 15 = x(x + 5) – 3(x + 5) = (x + 5)(x – 3). Example : 4 Factorize x2 – 2x – 15. Solution: Here, we need two numbers whose sum is – 2 and product is –15. Since the product must be negative, one of the numbers must be positive and the other negative such that their sum is negative. Of the two possible choices: (i) 5 and – 3 and (ii) – 5 and 3, we must take –5 and 3. Thus, x2 – 2x – 15 = x2 – (5 – 3) x – 15 = x2 (5 + 3) x – 15 = x2 – 5x + 3x – 15 = x(x – 5) + 3(x – 5) = (x – 5)(x + 3). Note: In x2 ± (a + b) x + ab = b, the trinomial takes the form x2 ± 2ax + a2 = (x ± a) (x ± a) = (x ± a)2 . This is thus a perfect square trinomial. (a + b)2 = a2 + 2ab + b2 (a – b)2 = a2 – 2ab + b2 ab ab a a a – b b a – b b b b ab ab ab ab a2 (a – b)2 = + + a2 b2 b2 b2 (a + b) a (a + b) a

ALGEBRA 161 Remarks: (i) a2 + b2 = a2 + 2ab + b2 – 2ab = (a + b)2 – 2ab (ii) a2 – b2 = a2 + 2ab + b2 – 2ab = (a – b)2 + 2abc (iii) (a + b)2 = a2 + 2ab + b2 = a2 – 4ab + 2ab = (a – b)2 + 4ab (iv) (a – b)2 = a2 – 2ab + b2 = a2 + 2ab + b2 – 4ab = (a + b)2 – 4ab Example : 5 Express in the form of square of binomial. (a) 4x2 – 4x + 1 (b) 9x2 + 12xy + 4y2 Solution: (a) 4x2 – 4x + 1 = (2x)2 – 2.2x.1 + (1)2 = (2x – 1)2 [(a2 – 2ab + b2 = (a – b)] (b) 9x2 + 12xy + 4y2 = (3x)2 + 2.3x.2y + (2y)2 = (3x + 2y)2 [ (a2 + 2ab + b2 = (a + b)2 ] Example : 6 Find the square root of 9x2 – 24xy + 16y2 Solution: Here, the square root of 9x2 – 24xy + 16y2 = ax2 – 36xy + 16y2 = (3x)2 – 2.3x.4y + (4y)2 = (3x – 4y)2 = 3x – 4y Example : 7 Factorize: x4 + 324 Solution: Here, (x2 )2 and 324 = 182 . So, we use the fraction a2 + b2 = (a + b)2 – 2ab. Thus, x4 + 324 = (x2 )2 + (18)2 = (x2 + 18)2 – 2x2 . 18 = (x2 + 18)2 – (6x)2 = (x2 + 18 + 6x)(x2 + 18 – 6x)2 = (x2 + 6x + 18) (x2 – 6x + 18) Furthermore, (a + b + c)2 = [(a + b) + c]2 = (a + b)2 + 2(a + b) c + c2 = a2 + 2ab + b2 + 2ac + 2ab + c2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

162 The Leading Maths - 8 Example : 8 Factorize: 6x2 + 19x + 10. Solution: Here, the coefficient of x2 is not-unity. It is 6. Consider the product of 6 and (+10). It is 60. Now, we find two numbers whose product is + 60. The two numbers must therefore be both positive or both negative; and their sum must be + 19, the coefficient of the middle term. So, both numbers must be positive. Note also that; 1 × 60 = 60; 60 × 1 = 60; 1 + 60 = 61 = 60 + 1;2 × 30 = 60; 30 × 2 = 60; 2 +30 = 32 = 30 + 2;3 × 20 = 60; 20 × 3 = 60; 3 + 20 = 23 = 20 + 3;4 × 15 = 60;15 × 4 = 60; 4 + 15 = 19 = 15 + 4;5 × 12 = 60;12 × 5 = 60; 5 + 12 = 17 = 12 + 5;6 × 10 = 60;10 × 6 = 60; 6 + 10 = 10 =10 + 6. Of the various possible choices, the suitable numbers are 4 and 15 or 15 and 4. 60 30 15 2 2 2 3 6x2 + 17x + 10 = 6x2 + (4 + 15) x + 10 = 6x2 + 4x + 15x + 10 = 2x(3x + 2) + 5(3x + 2) = (3x + 2)(2x + 5). Example : 9 Resolve into factors: 6x2 – 19x + 10. Solution: Here, we proceed exactly in the same way as in problem 1. The only difference being the fact the sum of the two numbers chosen is negative. So, we take – 4 and – 15 as the required numbers. 6x2 – 19x + 10 = 6x2 – (4 + 15) x + 10 = 6x2 – 4x – 15x + 10 = 2x(3x – 2) – 5(3x – 5) = (3x – 2)(2x – 5).

ALGEBRA 163 Example : 10 Resolve into factors: 6x2 – 11x – 10. Solution: 6x2 – 11x – 10 = 6x2 + (4 – 15) x – 10 = 6x2 + 4x – 15x – 10 = 2x(3x + 2) – 5(3x + 2) = (3x + 2)(2x – 5). Example : 11 Resolve into factors: (a) 6x2 + 11x – 10 (b) 4a2 + 5ab – 6b2 Solution: 6x2 + 11x – 10 = 6x2 + ( – 4 + 15) – 10 = 6x2 – 4x + 15x – 10 = 2x(3x – 2) + 5(3x – 2) = (3x – 2)(2x + 5). (b) 6a2 + 5ab – 4b2 Here, the product of 6 and – 4 is – 24 and the coefficient of the middle term is + 5. The two numbers whose product is – 24 and sum is 5 are 8 and – 3. 6a2 + 5ab – 4b2 = 6a2 + (8 – 3) ab – 4b2 = 6a2 + 8ab – 3ab – 4b2 = 2a(3a + 4b) – b(3a + 4b) = (3a + 4b)(2a – b) In the present case, the product – 60 is negative. So, one of the factors should be negative while the other one should be positive. Since their algebraic sum must be –11, the larger one must be negative. So, the required numbers must be + 4 and – 15. In other words, to factorize the expression 6x2 – 11x – 10, the required numbers whose sum is –11 and the product is – 60 are 4 and – 15.4 + (– 15) = –11, (the coefficient of the middle term) 4 × (–15) = –60, (the product of the coefficient of the first term and the last term).being the fact the sum of the two numbers chosen is negative. So, we take – 4 and – 15 as the required numbers. Here, every step is the same as that in problem no. 3. The only difference is that the algebraic sum of the two numbers is +11, a positive number. Hence, the larger number 15 is taken with positive sign and the smaller number 4 with negative sign. Thus, the required numbers are 15 and – 4.

164 The Leading Maths - 8 EXERCISE 9.3 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do 1. Factorize: (a) x2 + 5x + 6 (b) x2 + 7x + 12 (c) x2 – 5x + 6 (d) x2 – 8x + 12 (e) x2 + 5x – 6 (f) x2 – 5x – 6 (g) x2 + x – 20 (h) x2 – x – 12 (i) x2 – 8x – 9 (j) a2 – 2a – 15 (k) a2 – 5ab – 6b2 (l) x2 – 2ax – 35a2 (m) x2 + 6x + 9 (n) x2 – 8x + 16 (o) 4x2 + 12x + 9 (p) q4 + 4 (q) x4 + 4 (r) 4a4 + b4 2. Find the product of : (a) (a + 2b + c)2 (b) (a + b + 2c)2 (c) (a – b + 2c)2 (d) (2x + y – z)2 (e) (3x – 2y + z)2 (f) (3x – 2y – 1)2 3. Factorize the following expressions. (a) 2a2 + 5a + 2 (b) 2x2 – 5x + 2 (c) 2x2 – 7x + 3 (d) 2x2 – 3x – 20 (e) 6y2 + 8y + 2 (f) a3 – 11a2 + 24a (g) 16x2 y2 – 4xy2 – 30y2 (h) 9a2 – 12ab + 4b2 (i) 4x2 – 5xy – 6y2 (j) 10x2 – 41x – 45 (k) 6x2 + xy – 12y2 (l) 12x2 + 17xy – 5y2 4. Factorize the following expressions. (a) 2x – 15 + x2 (b) 7 – 8x + x2 (c) 11x + x2 + 24 (d) – 2 – 3x + 2x2 (e) – 7x + 3x2 – 6 (f) 7x – 3 + 6x2

ALGEBRA 165 (g) 6 + 17x + 12x2 (h) 6 + 12x – 5x2 (i) 7x – 6x2 – 2 (j) 3 – 10x – 8x2 (i) 7x – 4 – 3x2 (j) 6 + 7x – 9x2 ANSWERS 1. (a) (x + 2) (x – 3) (b) (x + 3) (x + 4) (c) (x – 2) (x – 3) (d) (x – 2) (x – 6) (e) (x + 6) (x – 1) (f) (x – 6) (x + 1) (g) (x + 5) (x – 4) (h) (x – 4) (x + 3) (i) (x – 9) (x + 1) (j) (a – 5) (a + 3) (k) (a + b) (a – 6b) (l) (x + 5a) (x – 7a) (m) (x + 3) (x + 3) (n) (x – 4) (x – 4) (o) (2x + 3) (2x + 3) (p) (q2 + 2q + 2) (q2 – 2q + 2) (q) (x2 + 2x + 2) (x2 – 2x + 2) (r) (2a2 + 2ab + b2 ) (2a2 – 2ab + b2 ) 2. (a) a2 + 4b2 + c2 + 4ab + 4bc + 2ca (b) a2 + b2 + 4c2 + 2ab + 4bc + 4ca (c) a2 + b2 + 4c2 – 2ab – 4bc + 4ca (d) 4x2 + y2 + z2 + 4xy – 2yz (e) 9x2 + 4y2 + z2 – 12xy – 4yz + 6xz (f) 9x2 + 4y2 + 1 – 12xy + 4y – 6x 3. (a) (2a + 1) (a + 4) (b) (2x – 1) (x – 2) (c) (x – 3) (2x – 1) (d) (x – 4) (2x + 5) (e) (y + 1) (6y +2 ) (f) a (a – 8) (a – 3) (g) 2y2 (2x – 3) (4x + 5) (h) (3a – 2b) (3a – 2b) (i) (x – 2y) (4x + 3y) (j) (x – 5) (10x + 9) (k) (2x + 3y) (3x – 4y) (l) 12x2 + 17xy – 5y2 4. (a) (x + 5) (x – 3) (b) (x – 7) (x – 1) (c) (x + 8) (x + 3) (d) (2x + 1) (x – 2) (e) (3x + 2) (x – 3) (f) (2x + 3) (3x – 1) (g) (3x + 2) (4x + 3) (h) (5x + 2) (3 – x) (i) (2 – 3x) (2x – 1) (j) (1 – 4x) (3 + 2x) (k) (3x – 4) (1 –x) (l) (3x + 2) (3 – x)

166 The Leading Maths - 8 9.4 Highest Common Factor (HCF) At the end of this topic, the student will be able to: ¾ find the HCF of the given algebraic expressions. Learning Objectives ” Which numbers divide both numbers 9 and 12? ” Which number other than 1 divides the numbers 10, 15 and 25? ” Which exactly divides the expressions 2x2 y and 3xy2 except 1? ” What exactly divides the polynomials (a + b)2 and a2 – b2 ? ” Which numbers exactly divides the numbers 6, 18 and 30? ” Which is the greatest number that exactly divides the numbers 6, 18 and 30? WARM-UP I. Introduction Now, we consider algebraic expressions having integral coefficients. Suppose they are factorizable. If a factor is contained in two or more algebraic expressions, it is said to be a common factor of the expressions. A common factor is sometimes called a common divisor. From elementary arithmetic, we know that 3 is a common factor of 9, 12, and 21. In other words, 3 divides each of the numbers 9, 12 and 21 without any remainder. Given algebraic monomials of the form ab2 and a2 b, we know that a is a common factor of ab2 and a2 b; b is also a common factor of ab2 and a2 b; and ab is also a common factor of ab2 and a2 b. Thus, each of a, b, and ab is a common factor of ab2 and a2 b. That is, each of the factors a, b, and ab divides each of the two expressions ab2 and a2 b without remainder and ab is the highest or the greatest common factor. In Venn-diagram, the greatest common factor is the intersection part of both sets. In the above example, the HCF is ab. A B U b a a b A ∩ B = HCF

ALGEBRA 167 The highest common factor of two or more algebraic expressions is the factor of the highest degree (or dimensions) that divides each of them without any remainder. It is abbreviated as HCF. Note: If there is no common factor in the given expression, their HCF is 1. Why? C D U 2 3 5 7 1 HCF = 1 In the case of simple monomials, we first find factors that are common to each of the given monomials and then find their product. The result is the required HCF of the given monomials. Hence, the product of the common factors of the greatest degree of the given algebraic expression is called the hight common factor (HCF) of the given expressions. To find the HCF of polynomials with integral coefficients, we proceed as follows: 1. Resolv each of the given expressions into factors, 2. Determine the factors common to each of them, and 3. Multiply all the common factors to get the required HCF. CLASSWORK EXAMPLES Example : 1 Find the highest common factor of a2 b3 and a3 b4 c. Solution: Breaking up the given expressions into prime factors, we have a2 b3 = a × a × b × b × b a3 b4 c = a × a × a × b × b × b × b × c The factors common to the two expressions are a, a, b, b. The product of the common factors is a2 b3 . ∴ The required HCF of a2 b3 and a3 b4 c is a2 b3 . Example : 2 Find the HCF of 35a2 bx3 y and 49ab2 x2 y3 . Solution: Here, 35a2 bx3 y = 5 × 7 × a × a × b × x × x × x × y 49ab2 x2 y3 = 7 × 7 × a × b × b × x × x × y × y × y . Thus, the common factors are 7, a, b, x, x, y. Hence, the HCF = 7abx2 y.

168 The Leading Maths - 8 Example : 3 Find the HCF of ab(x3 + 27) and ab2 (x2 – 9). Solution: Here, the first expression = ab(x3 + 27) = ab (x3 + 33 ) = ab(x + 3)(x2 – 3x + 9) The second expression = ab2 (x2 – 9) = ab2 (x2 – 32 ) = ab2 (x + 3)(x – 3). Thus, the common factors are a, b, and x + 3. The product of the common factors = ab(x + 3) ∴ HCF = ab(x + 3). EXERCISE 9.4 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What are common factors of 18 and 24? (b) What is the greatest common factor of 24 and 36? (c) Write the full form of HCF and define it. (d) What is the HCF of 6x2 yz3 and 9x3 yz? (e) What is the HCF of x2 – 4 and 3x2 – 6x? 2. Circle ( ) the correct answers. (a) Which is the correct full form of the HCF? i. High Common Factor ii. Higher Common Factor iii. Highest Common Factor iv. High Command Factor (b) HCF is the product of ........................ i. Factors ii. Highest Factors iii. All factors iv. All common factors (c) What is the HCF of a2 bc and de2 f? i. abc ii. def iii. 1 iv. 0

ALGEBRA 169 (d) What is the HCF of 2a2 bc3 , 4ab2 c2 and 6abc8 ? i. abc ii. 2abc iii. 48a4 b4 c8 iv. 2abc2 (e) What is the HCF when there is no common factor in the given expressions? i. 0 ii. 1 iii. 2 iv. 3 (f) What is the HCF of 4x2 – y2 and (x – y)2 ? i. (2x + y) ii. 2x – y2 iii. 2x + y iv. 2x – y 3. Find the Highest Common Factor of the following expressions and represent them in the Venn diagram. (a) ab and a2 b2 (b) ab2 and ab3 (c) a2 b3 , a3 b4 and a3 b3 (d) 3a2 x7 and 3a3 x6 (e) 17ax3 y4 z, 34bxz2 , 51abxyz (f) a2 + ab and ab + b2 (g) a2 – b2 and a2 – ab (h) a3 – 3a2 b and a2 – 9b2 (i) (x – 2)3 and x3 – 8 (j) 3ax(a – x)3 and 2a2 x(a – x)2 (k) x4 – x and x3 + x2 + x (l) x2 – x – 2 and x2 + x – 6 4. Find the HCF of the following expressions and show them in the Venn diagram. (a) x2 – 8x + 7 and 2x2 – 3x – 2 (b) 2x2 – 3x – 2 and 6x2 + 7x – 3 (c) – 6x2 + 7x – 2 and – 5x2 + 6 + 12x (d) 6x2 + x – 2 and – 8x2 – 2x + 1 5. Find the HCF of the following expressions and show them in the Venn diagram. (a) a2 – a – 2, a2 + a – 6 and 3a2 – 13a + 14 (b) 2x2 + 5x + 2, 3x2 + 8x + 4 and 2x2 + 3x – 2 (c) x2 + x + 1, x6 – 1 and x3 – 1. ANSWERS 3. (a) ab (b) ab2 (c) a2 b3 (d) 3a2 x6 (e) 17xz (f) a + b (g) a – b (h) a – 3b (i) x – 2 (j) ax (a – x)2 (k) x (x2 + x + 1) (l) x – 2 4. (a) 1 (b) 1 (c) 1 (d) 1 5. (a) a – 2 (b) x + 2 (c) x2 + x + 1

170 The Leading Maths - 8 9.5 Lowest Common Factor (LCM) At the end of this topic, the student will be able to: ¾ find the LCM of the given algebraic expressions. Learning Objectives ” What are the common multiples? ” What are the common multiples of 5, 10, and 15? Which is the lowest? ” What are the common multiples of x2 , x3 and x4 ? Which is the lowest? An expression, which is exactly divisible by two, or more, other expressions, is said to be a common multiple of those expressions. WARM-UP I. Introduction The expression of the lowest degree or of the lowest dimension that is exactly divisible by two or more expressions is called their Lowest Common Multiple. It is shortly abbreviated as LCM. Note: LCM is exactly divided by the given expressions. Thus, the LCM of x2 , x3 and x4 is x4 ; the LCM of ab and bc is abc. Why? LCM = Common factors × Remaining factors = HCF × Remaining factors. U A ∪ B = LCM A B c b a X Y x x x2 Z U X ∪ Y ∪ Z = LCM A general procedure of finding the LCM of two expressions is to write each of them in the factorized form. Then the product of the factors in the highest degree gives the LCM.

ALGEBRA 171 ACTIVITY - 1 Ram and Ali are friends. Ram ans John are also friends. Are Ali and John friends? Not sure, Ali and John may or may not be friends. If they are not friend to each other, Ram is a mediator for both of them. In this case, Ram is the common person for both. Here, Ram is the HCF for their friendship. If Ali and John are friends. They form a friendship group. In this case, they are a group of friends. This group is the LCM. Remarks: ” If there is no common factor in the given expressions, the required LCM is the product of the given expressions. ” The relation between LCM and HCF of two expressions is, LCM × HCF = 1st expression × 2nd expression. CLASSWORK EXAMPLES Example : 1 Find the LCM of 5ab2 and a3 b2 . Solution: Here, 5ab2 = 5 × a × b2 a3 b2 = a3 × b2 ∴ LCM = 5 × a3 × b2 = 5a3 b2 . Example : 2 Find the LCM of 2(a2 – b2 ) and 3a2 – 3ab. Solution: Here, 2(a2 – b2 ) = 2(a + b)(a – b) 3a2 – 3ab = 3a(a – b) Hence, LCM = 2 × 3 × a × (a – b)(a + b) = 6a(a2 – b2 ). Example : 3 Find the LCM of x2 – 3x + 2 and x2 – 4x + 3. Solution: Here, 1st Exp = x2 – 3x + 2 = x2 – (2 + 1) x + 2 = x2 – 2x – x + 2 = x(x – 2) – 1 ( x – 2) = ( x – 1)(x – 2)

172 The Leading Maths - 8 2nd Exp = x2 – 4x + 3 = x2 – (3 + 1) x + 3 = x2 – 3x – x + 3 = x(x – 3) – 1 (x – 3) = (x – 1)(x – 3) So, LCM = (x – 1)(x – 2)(x – 3)(x – 6). Example : 4 Find the LCM of; 3a3 – 18a2 x + 27ax2 and 4a4 + 24a3 x + 36a2 x2 . Solution: Here, 1st Exp. = 3a3 – 18a2 x + 27ax2 = 3a(a2 – 6ax + 9x2 ) = 3a{(a2 – 2 . 3x + (3x)2 } = 3a(a – 3x)2 2nd Exp. = 4a4 + 24a3 x + 36a2 x2 = 4a2 (a2 + 6ax + 9x2 ) = 4a2 {(a2 + 2 .a .3x + (3x)2 } = 2 × 2a2 (a + 3x)2 Then, LCM = 3 × 2 × 2 × a2 (a + 3x)2 (a – 3x)2 = 12a2 (a + 3x)2 (a – 3x)2 . Note: Regarding the Highest Common Factor and the Lowest Common Multiple of two expressions, it will be of interest to note that the product of the HCF and the LCM is equal to the product of the two expressions. i.e. HCF × LCM = 1st expression × 2nd expression. EXERCISE 9.5 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) Write the full form of LCM and define it. (b) Define common multiples.

ALGEBRA 173 (c) What is the LCM of 3x2 y and 8x3 y2 ? (d) Write the relation between LCM and HCF for two expressions? (e) What is the LCM of 4x2 + y2 and 4x2 – y2 ? 2. Circle ( ) the correct answers. (a) LCM is the product of .................... i. common factors ii. remaining factors iii. common and remaining factors iv. the given expressions (b) What is the LCM of 6a3 b2 c and 9a2 bc3 ? i. 3a2 b2 c ii. 18a3 b2 c3 iii. 54a5 b3 c4 iv. 27a5 b3 c3 (c) What is the LCM of 4xy and 9ab? i. 36abxy ii. xyab iii. 6xy iv. 6ab (d) For LCM and HCF of the expressions, ............ i. LCM = HCF ii. LCM × HCF = common factors iii. LCM × HCF = product of expressions iv. LCM HCF = Remaining factors (e) The LCM of a2 – b2 and (a – b)2 is .................. i. (a – b) (a + b)2 ii. (a + b) (a – b)2 ii. (a + b) (a – b) iv. (a2 – b2 )(a + b) 3. Find the Least Common Multiple of the following expressions and show them in the Venn diagram. (a) ab and a2 b2 (b) xy2 and xy3 (c) x2 y3 , x3 y4 and x3 y3 (d) 3a2 x7 and 4a3 x5 (e) 7ax3 y4 z, 21bxz2 and 28abxyz (f) a2 – 4 and a2 + 2a (g) (x – 2)2 and x3 – 8 (h) 3ab(a – b)3 and 2a2 b(a – b)2 (i) x4 – x and x3 + x2 + x (j) a2 – a – 2 and a2 + a – 6 (k) x2 – 21x + 108 and x2 – 81

174 The Leading Maths - 8 4. Find the LCM of the following expressions and show them in the Venn diagram. (a) x2 – 18x + 45, x2 – 26x + 165 and x2 – 14x + 33 (b) 24(y3 – x3 ), 25x2 (y2 – x2 ) and 60x5 (c) x2 + x + 1, x6 – 1 and x3 – 1 (d) ax4 – 8ax, a2 x4 + 8a2 x and ax4 – 4ax2 . 5. Find the HCF and LCM of the following expressions. (a) x2 + 6x + 9 and x2 – x – 12 (b) x2 + x – 20 and x2 + 6x + 5 (c) 2x2 + 3x – 2 and 2x2 – 3x + 1 (d) – 3x2 – 10x + 8 and – 6 + 11x – 3x2 (e) 3 + 4x + x2 , x2 + x – 6 and x2 + 2x – 3 (f) x2 – 4, x2 – 3x + 2 and – x2 + 5x – 6 ANSWERS 3 (a) a2 b2 (b) xy3 (c) x3 y4 (d) 12 a3 x7 (e) 84 abx3 y4 z2 (f) a (a2 – 4) (g) (x – 2) (x3 – 8) (h) 6a2 b (a – b)3 (i) x (x3 – 1) (j) (a + 1) (a – 2) (a + 3) (k) (x + 12) (x2 – 81) 4. (a) (x – 3) (x – 11) ( x – 15) (b) 1800x5 (x + y) (y3 – x3 ) (c) x6 – b (d) a2 x2 (x6 – 64) 5. (a) (x + 3), (x + 3) (x – 4) (b) (x + 5), (x + 5) (x – 4) (x + 1) (c) (2x – 1), (2x – 1) (x + 2) (x – 1) (d) (2 – 3x), (2 – 3x) (x + 4) (x – 3) (e) (x + 3), (x + 3) (x – 2) (x + 1) (x – 1) (f) (x – 2), (x – 2) (x + 2) (x +1) (3 – x)

ALGEBRA 175 9.6 Rational Algebraic Expression At the end of this topic, the student will be able to: ¾ Simplify the rational algebraic fractions. Learning Objectives ” What are fraction and ratio? ” What is rational number? ” What are the areas of a rectangular rooms A and B in the given picture? Area of the room A (A1 ) = ......................... Area of the room B (A1 ) = ........................ ” Which room has more space? How can we find it? ” How can we compare their areas? ” What is the ratio of their areas? Ratio of the rooms A and B = (2x2 + 4x) sq. units (x2 – 4) sq. units = 2x2 + 4x 7x2 – 4 ” What are the expressions in numerator and denominator of the ratio? This is called algebraic fraction or algebraic rational expression. Room A Room B x + 2 x + 2 x – 2 2x WARM-UP I. Introduction A rational algebraic expression (An algebraic fraction) is formed when an algebraic expression is divided by another non-zero algebraic expression. A class of fractions may have polynomials only in their numerators and denominators. Algebraic fractions can be grouped together. They can be added. One can be subtracted from another. Two or more such fractions can be multiplied. We can divide one by another non-zero algebraic expression also. In short, the four fundamental rules of arithmetic can be extended to the case of algebraic fractions. To simplify algebraic expression is to apply these rules and put the expression in the reduced form or lowest term. II. Reduction to the lowest term In order to reduce a given fraction to its lowest or simplest form, we often shall have to apply the following fundamental principle of fractions:

176 The Leading Maths - 8 “ A fraction remains unchanged if its numerator and denominator are multiplied or divided by the same non-zero expression.” In other words, if F G denotes an algebraic fraction, then F G = F × H G × H = F/H G/H Where, G ≠ 0 and H ≠ 0. Working Rule: 1. Factorize both numerator and denominator. 2. Apply the fundamental principle in the above statement. CLASSWORK EXAMPLES Example : 1 Reduce to the lowest term: (a) 6 9 (b) 6ax3 y 9bxy3 Solution: (a) 6 9 = 3 × 2 3 × 3 = 2 3, (b) 6ax3 y 9bxy3 = 2.3.a.x.x.x.y 3.3.b.x.y.y.y = 2ax2 3by2 [ Dividing the numerator and denominator by 3xy.] Example : 2 Simplify: (a) ax – ay bx – by (b) a2 – b2 a2 – 2ab + b2 Solution: (a) ax – ay bx – by = a(x – y) b(x – y) = a b (b) a2 – b2 b2 – 2ab + b2 = (a + b)(a – b) (a – b)2 = a + b a – b [ Dividing the numerator and denominator by 3 or canceling the common factor 3.] [ Dividing the numerator and denominator by x - y] [ Dividing the numerator and denominator by a - b]

ALGEBRA 177 Example : 3 Simplify: 8a3 b – 8a2 b2 + 2ab3 2ab Solution: 8a3 b – 8a2 b2 + 2ab3 2ab = 8a3 b 2ab + 8a2 b2 2ab + 2ab3 2ab = 4a2 – 4ab + b2 . In dealing with one or more fractions, it is often advisable to reduce a given fraction to its simplest form as shown above. III. Addition and Subtraction of Fractions The next step is to find the algebraic sum of two or more fractions. (i) The basic procedure is as indicated below: a d + b d + c d + ... + m d = a + b + c + .. + m d , provided that d ≠ 0. Let us see how this procedure helps in simplification: 3x 3x + y + y 3x + y = 3x + y 3x + y = 1 Thus, we have seen how fractions with the same or common denominator are added. Here the numerator is the algebraic sum of the numerators while the common denominator is retained in the sum of the fractions. (ii) To add fractions having different denominators, we first express the given fractions as equivalent fraction having the same denominator and then proceed as explained above. It is desirable, but not essential, to use the lowest common multiplies of denominators of all the fractions involved. For instances, 1 + a b = b b + a b = b + a b a b – c d = ad bd – bc bd = ad – bc bd ; LCM of b and d = bd Note: To add two or more fractions, find the LCM of the denominators, this will be used as the common denominator; divide it by the denominator of the first fraction, and multiply its numerator by the quotient so obtained; the same process is repeated for the other fractions also. The sum of the numerators so obtained with the common denominator is the required sum of the fractions. Here, we have a polynomial divided by a non-zero monomial, we apply the distributive property (i.e., we simplify by computing the ratio of each term of the numerator to the monomial and then taking the sum).

178 The Leading Maths - 8 IV. Multiplication and Division of Fractions Multiplication and division of two fractions is based upon the following principles: (i) a b × c d = ac bd, b ≠ 0, d ≠ 0; That is, the product of two or more fractions is a fraction whose numerator is the product of the numerators of the original fractions and whose denominator is the product of the denominators. (ii) a b ÷ c d = a b × d c, b ≠ 0, c ≠ 0, d ≠ 0. To divide a fraction by another, we multiply the dividend (numerator) by the reciprocal of the divisor (denominator). CLASSWORK EXAMPLES Example : 1 Express as a single fraction: x bc + y ac + z ab. Solution: Here, x bc + y ac + z ab = x bc + a a + y ac × b b + z ab × c c = ax abc + by abc + cz abc = ax + by + cz abc . Example : 2 Express as a single fraction: 1 a + b + 1 a – b. Solution: Here, 1 a + b + 1 a – b = a – b (a + b) (a – b) + a + b (a + b)(a – b) = a – b + a + b (a + b)(a – b) = 2a a2 – b2 . Example : 3 Simplify: a x2 y + b y2 z – c z Solution: Here, a x2 y + b x2 z – c z2 x = ayz2 x2 y2 z2 + bzx2 x2 y2 z2 + cxy2 x2 y2 z2 = ayz2 + bzx2 + cxy2 x2 y2 z2 LCM of bc, ac and ab equals with the expression each fraction as an equivalent with denominator bc, ac and ab is abc. The LCM of the denominators x2 y, y2 z and z2 x is x2 y2 z2 . Divide LCM = x2 y2 z2 by each denominator and multiply to respective numerator. The denominators of the two terms are different and their LCM = (a + b)(a - b).

ALGEBRA 179 Example : 4 Simplify : 2 a + b – 2 a – b + 4a a2 – b2. Solution: Here, 2 a + b – 2 a – b + 4a a2 – b2 = 2(a – b) – 2(a + b) + 4a a2 – b2 = 2a – 2b – 2a – 2b + 4a a2 – b2 = –4b + 4a a2 – b2 = 4(a – b) (a + b)(a – b) = 4 a + b. Alternatively 2 a + b – 2 a – b + 4a a2 – b2 = 2 (a – b) – 2 (a + b) (a + b) (a – b) + 4a a2 – b2 = 2a – 2b – 2a – 2b a2 – b2 + 4a a2 – b2 = – 4 b a2 – b2 + 4a a2 – b2 = – 4b + 4a a2 – b2 = 4 (a – b) (a + b) (a – b) = 4 a + b Example : 5 Combine into a single fraction: 1 x2 – 3x + 2 – 1 x2 – 4x + 3 Solution: Here, 1 x2 – 3x + 2 – 1 x2 – 4x + 3 = 1 x2 – (2 + 1) x + 2 – 1 x2 – (3 + 1) x + 3 = 1 x2 – 2x – x + 2 – 1 x2 – 3x – x + 3 LCM of a + b, a – b and a2 – b2 is a2 – b2 . a2 – b2 a + b = a – b and 2 × (a – b) a2 – b2 a – b = a + b and 2 × (a + b) a2 – b2 a – b = 1 and 4a × 1 Factories the denominators first.

180 The Leading Maths - 8 = 1 x(x – 2) – (x – 2) – 1 x(x – 3) – 1(x – 3) = 1 (x – 1)(x – 2) – 1 (x – 1) (x – 3) = (x – 3) – (x – 2) (x – 1) (x – 2) (x – 3) = x – 3 – x + 2 (x – 1) (x – 2) (x – 3) = –1 (x – 1)(x – 2)(x – 3) = 1 (1 – x) (x – 2) (x – 3) Example : 6 Simplify : 1 1 – x + 1 1 + x – 2 1 + x2 Solution: Here, 1 1 – x + 1 1 + x – 2 1 + x2 = 1 + x + 1 – x 1 – x2 – 2 1 + x2 = 2 1 – x2 – 2 1 + x2 = 2(1 + x2 ) – 2(1 – x2 ) (1 – x2 ) (1 + x2 ) = 2 + 2x2 – 2 + 2x2 12 – (x2 )2 = 4x2 1 – x4 EXERCISE 9.6 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) Define algebraic rational expression with an example. (b) What is the lowest term of 2x2 y 4xy ? (c) What is the lowest term of x2 – y2 x – y ? LCM of the denominators is (x – 1) (x – 2) (x – 3). Multiply to the numerator by quotient. Denominators of first two terms are different. So, cross multiply.

ALGEBRA 181 (d) What is the value of x x + y + y x + y ? (e) What is the value of 1 a + b + 1 a – b ? 2. Circle ( ) the correct answers. (a) Which is rational expression? i. x + 2 x , x ∈ W ii. x + 2 x – 2, x ∈ N iii. x – 4 x + 1, x ∈ N iv. 2x + 1 x + 3 , x ∈ Z (b) Which is the lowest term of 3xy2 6x ? i. 3y2 x ii. y2 2 iii. xy2 2 iv. 3y2 6 (c) Which is the lowest expression of x2 – y2 (x – y)2? i. x – y x + y ii. x + y x – y iii. x + y2 x + y iv. 1 (d) 1 2x – 1 2y = ....................... i. y – x 2xy ii. x – y 2xy iii. y – x 4xy iv. x – y 4xy (e) y x – y – x x – y = ...................... i. 0 ii. 1 iii. –1 iv. x – y x + y 3. Simplify by reducing to a single fraction. (a) x 2 + x 2 (b) 1 2x + 1 3x (c) 3 – 4 x (d) 2a 3b – 3b 2a (e) 1 xy + 1 yz (f) 1 bc + 1 ca + 1 ab (g) a x2 y + b y2 z – c z2 x (h) 1 a – 2 + 1 a – 3 (i) 1 x – 2 – 1 x – 3 (j) a x – a – b x – b

182 The Leading Maths - 8 (k) 1 a(a – b) + 1 a(a + b) (l) x + 2 – x – 2 x – 1 (m) 3a (a + 1)2 – 2 a + 1 (n) 5 x – 2 – 4x (x – 2)(x + 1) (o) x x – 5 – x2 x2 – 25 4. Simplify: (a) 1 x – y + 1 x + y – 2y x2 – y2 (b) 1 a2 – 6a + 9 – 1 a2 – 5a + 6 (c) 1 x2 3x + 2 + 1 x2 + 3x – 10 (d) 1 x + 3 – 1 x + 2 + 1 x2 + 5x + 6 (e) a a – b + b a + b – b2 a2 + b2 (f) x 4(1 + x) – x 4(1 – x) + 3 2(1 + x) (g) 1 a – x – 1 a + x – 2x a2 + x2 (h) a x2 – x + a x2 – x3 – a x2 – 1 (i) a (a – c) (a – c) + a (b – c)(b – a) + a (c – a) (c – b) ANSWERS 3. (a) x (b) 5 6x (c) 3x – 4 x (d) 4a2 – 9b2 6ab (e) x + z xyz (f) a + b + c abc (g) ayz2 + bx2 z – cxy2 x2 y2 z2 (h) 2a – 5 (a – 2) (a –3) (i) 1 3 – x (j) (a – b) x (x – a) (x – b) (k) 2 a2 – b2 (l) x2 x – 1 (m) a – 2 a + 1 (n) x + 5 (x – 2) (x + 1) (o) 5x x2 – 25 4. (a) 2 x + y (b) 1 (a – 2) (a – 3)2 (c) 2 (x + 2) (x – 1) (x – 2) (x + 5) (d) 0 (e) a2 – 2ab a2 + b2 (f) x2 + 3x – 3 2 (x2 – 1) (g) 4x3 a4 – x4 (h) a x2 (1 – x2 ) (i) 0

ALGEBRA 183 CHAPTER 10 EQUATION, INEQUALITY AND GRAPH Lesson Topics Pages 10.1 Solving simultaneous, Linear Equation in two variables 184 10.2 Quadratic Equation 189 ” What is the value of measuring cylinder in the beam balance? ” What represents the above number line? ” How much milk do you drink in a day? ” How much water do you drink in a day? ” How many members in your family? ” What is the sign of equal? ” If x is equal to 3, write it in a mathematical sentence. ” If the sum of two numbers is equal to 5, write it in a mathematical sentence. ” What is the sign of unequal? ” x is never equal to 6. Write this statement in mathematics form. ” When two quantities are not equal, what possible cases can arise? ” Observe the following tables. x + y = 5 x 2 3 4 y ? ? ? 2x – y = 3 x ? ? ? y 1 3 5 Table - 1 Table - 2 (i) In the table - 1, what will be the values of y by using given equation? (ii) In the table - 2, what will be the values of x by using given equation? WARM-UP

184 The Leading Maths - 8 10.1 Solving Simultaneous At the end of this topic, the student will be able to: ¾ solve the simultaneous linear equation of two variables. Learning Objectives I. Review ACTIVITY - 1 If the sum of a pencil and an eraser is Rs. 6 and the cost of an eraser is twice of that of pencil, what are the cost of pencil and eraser? Let, x and y be the cost of a pencil and an eraser respectively. Then by question, Consider the equation: x + y – 6 = 0 ...........(i) 2x – y = 0 ............(ii) Making a table of values for these, we get x + y – 6 = 0 x 0 6 y 6 0 And 2x – y = 0 x 0 2 0 0 4 We get a graph as shown above. The point P(2, 4) lies on both lines x + y – 6 = 0 and 2x – y = 0, and hence satisfies both equations x + y – 6 = 0 and 2x – y = 0 simultaneously. Therefore, x = 2 and y = 4. Hence, we have the definition. (6, 0) (2, 4) 2x – y = 0 x + y – 6 = 0 (0, 6) (0, 0) Y O X Y' X'

ALGEBRA 185 Two linear equations a1 x, + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 are said to be simultaneous equation if there exists a unique solution satisfying both equations. This solution may be obtained by finding the coordinates of the point of intersection of the graph of these equations. II. Solving simultaneous linear Equations by Substitution Method Consider two equations 2x – y + 1 = 0 ............(i) x – 2y + 8 = 0 ..........(ii) From eqn (i), y = 2x + 1 ............. (iii) Substituting the value of y in eqn (ii), x – 2(2x + 1) + 8 = 0 or, x – 4x – 2 + 8 = 0 or, – 3x + 6 = 0 or, 3x = 6 x = 6 3 = 2 From eqn (iii), y = 2 × 2 + 1 = 5 \ x = 2 and y = 5 CLASSWORK EXAMPLES Example : 1 Solve the given simultaneous linear equations by graphical method: 3x + y = 17 and 2x – y = 3 Solution: Here, 3x + 7 = 17 ................. (i) 2x – y = 3 ................... (ii) Now, making a table For eqn (i), For eqn (ii), (2, 5) 2x – y + 1 = 0 x – 2y + 8 = 0 (0, 1) (–2, 3) Y O X Y' X' (0, 4) x 4 5 6 y 5 2 – 1 x 2 3 4 y 1 3 5

186 The Leading Maths - 8 Drawing a graph (4, 5) 3x + y = 17 2x – y = 3 Y O X Y' X' In the above graph, the straight lines obtained from the given equations intersect at (4, 5). So, x = 4 and y = 4. Example : 2 The present age of a mother is 3 times her daughter and the sum of their ages is 48 now. (a) Express the given statement into mathematical equation. (b) What are the present ages of the mother and her daughter? Find graphically. (c) What was the age of the mother when her daughter was born? Solution: (a) Let the present ages of a mother and her daughter be x years and y years respectively. Then, by equation, x = 3y ................... (i) x + y = 48 ............ (ii) (b) Making tables, For eqn (i), For eqn (ii), x 0 30 24 y 0 10 8 x 32 28 23 y 16 20 25

ALGEBRA 187 Now, drawing a graph, x = 3y (36, 12) x + y = 48 Y O 5 10 10 15 15 20 20 25 25 30 30 35 35 5 X Y' X' In the above graph, the two straight lines intersect at (36, 12). So, x = 36 and y = 12. Hence, their present ages are 36 years and 12 years respectively. (c) Again, x – y = 36 – 12 = 24 years. Hence, the age of the mother was 24 years when her daughter was born. EXERCISE 10.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is linear equation? Define with suitable example. (b) Define simultaneous linear equations with example. (c) What is the value of y in x + y = 2 if x = 1? (d) What is the value of x in 2x – y = 4 if y = 2? (e) What is the value of x when solving the equations x + y = 5 and x – y = 7? 3. Solve the following system of simultaneous equation graphically. (a) 2x – y = 4 and 2x + 3y = 4 (b) 3x – 4y = – 7 and 3x + y = – 2 (c) 2x + 4y = 7 and 4x – 3y = 3 (d) 2x – 5y = 14 and x – 3y = 9

188 The Leading Maths - 8 (e) 4x – 5y = 22 and 7x + 3y = 15 (f) 3x – 2y = 8 and 4x + y = 7 (g) x – 5y + 14 = 0 and 2x – y + 1 = 0 (h) x – 5y + 14 = 0 and x – 2y + 8 = 0 (i) 2x + 3y = 13 and 9x – 2y = 12 (j) 3x – 4y = 18 and 7x + 2y = 8 (k) 3y = x + 18 and y + 3x = 26 (l) 2x + y = 7 and 3x – 5y – 4 = 0 4. Solve the following problems by using graph. (a) If the sum of the present ages of two sisters is 25 years and that of difference is 5 years, i. Express the given information into mathematical statements. ii. Find their present ages. iii. How many years older is the elder sister than her younger sister? (b) The present age of a father is double of his son's age. The sum of their ages is 60 years now. i. Express the given information into mathematical statements. ii. What are the present ages of the father and his son? Find. iii. How many years older is the father than his son's present age? Find. (c) The cost of 2kg of potatoes and 3 kg of tomatoes is Rs. 280 and that of 3 kg of potatoes and 2 kg of tomatoes is Rs. 270. i. Express the given information into mathematical statements. ii. Find the rate of potatoes and tomatoes per kg. iii. Which vegetable is expensive and how much percentage? ANSWERS 3. (a) (2, 0) (b) (– 1, 1) (c) (1.5, 1) (d) (3, 1) (e) (3, – 2) (f) (2, – 1) (g) (1, 3) (h) (– 4, 2) (i) (2, 3) (j) (2, –3) (k) (6, 8) (l) (3, 1) 4. (a) i. 15 yrs, 10 yrs ii. 5 yrs (b) i. Rs. 50, Rs. 60 ii. Tomatoes, 2%

ALGEBRA 189 10.2 Quadratic Equation At the end of this topic, the student will be able to: ¾ solve the quadratic equation or by factorization Learning Objectives I. Introduction ax2 + bx + c = 0, where a ≠ 0, b and c are constants is a quadratic equation in x. Here by definition, x2 – 4 = 0 ......................(i) → Pure quadratic equation (ax2 + b = 0) 3x2 – 2x – 15 = 0 ..........(ii) → Complete quadratic equation (ax2 + bx + c = 0) 2x2 – 5x = 0 ................(iii) → Incomplete quadratic equation (ax2 + bx = 0) are quadratic equations in the variable x. We can solve these equations by factorization method, adopting the agreement that if a × b = 0 either a = 0 or b = 0 CLASSWORK EXAMPLES Example : 1 Solve: (a) x2 = 9 (b) 2x2 + 3x = 0 (c) 2x2 – 3x – 5 = 0 Solution: (a) Here, x2 = 9 or, x2 – 9 = 0 or, x2 – 32 = 0 or, (x + 3) (x – 3) = 0 Either x + 3 = 0 ∴ x = – 3 or, x – 3 = 0 ∴ x = 3 ∴ x = ± 3 is the required solution. 'Next Method' x2 = 9 or, x2 = 32 ∴ x = ± 3. (b) 2x2 + 3x = 0 or, x(2x + 3) = 0 Either, x = 0

190 The Leading Maths - 8 OR, 2x + 3 = 0 or, 2x = – 3 \ x = – 3 2 \ x = 0, – 3 2 as required. (c) 2x2 – 3x – 5 = 0 or, 2x2 – (5 – 2)x – 5 = 0 or, 2x2 – 5x + 2x – 5 = 0 or, x(2x – 5) + 1(2x – 5) = 0 or, (x + 1) (2x – 5) = 0 Either, x + 1 = 0 \ x = –1 OR, 2x – 5 = 0 or, 2x = 5 \ x = 5 2 \ x =1 and 5 2 are the required solutions. Example : 2 A rectangular court-yard has length 4 m more than the breadth and its area is 96 m2 . Find the length and breathe of the court-yard. Solution: Let breadth of the rectangular court yard (b) = x. Then, its length (l) = x + 4. Area (A) = 96 m2 or, l × b = 96 or, (x + 4) x = 96 or, x2 + 4x = 96 or, x2 + 4x – 96 = 0 or, x2 + (12 – 8) x – 96 = 0 or, x2 + 12x – 8x – 96 = 0 or, x(x + 12) – 8(x + 12) = 0 or, (x + 12) (x – 8) = 0 Either, x + 12 = 0 \ x = – 12 [This value does not exist. Why?] OR, x – 8 = 0 \ x = 8 Hence, breadth (b) = x = 8 m Length (l) = x + 4 = (8 + 4) m = 12 m x + 4 x

ALGEBRA 191 EXERCISE 10.2 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do 1. Find the value of x for which (a) x2 – 16 = 0 (b) x2 – 4 = 0 (c) x2 – 25 = 0 (d) x2 = 36 (e) x2 – 64 = 0 (f) x2 = 49 (g) x2 – 81 = 0 (h) x2 = 2 2. Solve the following quadratic equations for x. (a) x2 – 2x = 0 (b) x2 – 3x = 0 (c) x2 + 4x = 0 (d) x2 + 5x = 0 (e) 2x2 – 3x = 0 (f) 3x2 + 5x = 0 (g) 3x2 = 5x (h) 4x2 = 7x 3. Solve the following quadratic equations for x: (a) x2 + x – 2 = 0 (b) x2 – x – 6 = 0 (c) x2 + 2x + 1 = 0 (d) x2 + 8x – 33 = 0 (e) 2x2 + 3x + 1 = 0 (f) 3x2 + 4x + 1 = 0 (g) 4x2 + 5x + 1 = 0 (h) 2x2 + 9x – 5 = 0 (i) 2x2 + x – 15 = 0 (j) 2x2 – 5x – 3 = 0 (k) (2x + 3) (3x – 7) = 0 (l) 25x (x+1) + 4 = 0 4. (a) The length of a rectangular garden is 5 m more than its breadth. If its area is 500 m2 , find its length and breadth. (b) The present are of father is 3 times the age of his son. The product of their ages before 10 years was 500. Find their present ages. ANSWERS 1. (a) ± 4 (b) ± 2 (c) ± 5 (d) ± 6 (e) ± 8 (f) ± 7 (g) ± 9 (h) ± 2 2. (a) 0,2 (b) 0, 3 (c) 0, – 4 (d) 0, –5 (e) 0, 3 2 (f) 0, – 5 3 (g) 0, 5 2 (h) 0, 7 4 3. (a) 1, – 2 (b) 3, – 2 (c) – 1 (d) 3, – 11 (e) – 1, – 1 2 (f) 1, – 11 (g) – 1, – 1 4 (h) 1 2, – 5 (i) – 3, 5 2 (j) 3, – 1 2 (k) – 3 2 , 7 3 (l) – 4 5 , – 1 5 4. (a) 25 m, 20 m (b) 60 yrs, 20 yrs.

192 The Leading Maths - 8 1. (a) Write the value of (7xy)0 . (b) Simplify: xa – b × xb – c × x c – a 2. (a) Compare which one is greater between 230 and 320. (b) Prove that:   xa xb   a + b ×   xb xc   b + c ×   xc xa  a + c = 1 3. (a) What is the value of 32x × 33x × 1 3x? (b) Simplify: ap(x – y) × ap(y – z) × ap(z – x) 4. (a) Factorize: 3xy – 6y2 (b) Simplify: 1 x – y – 1 x + y 5. (a) Factorize: 4x2 – 9y2 (b) Simplify: x + y x – y + x – y x + y 6. (a) Factorize: x2 – 5x + 6 (b) Find the LCM of: x2 + 4x + 4 and x2 – 4. 7. (a) Find the HCF of: x2 – 5x + 4 and 4x – 16 (b) Simplify: 3a + 2 + 3a 3a × 10 8. (a) Simplify: 4x2 + y2 4x2 – y2 – 2x – y 2x + y (b) Find the HCF of: x2 – 3x – 18 and x2 + 5x + 6 9. (a) Solve the equations graphically: 2x + y = 5 and 3x – y = 5 (b) Solve: x2 + 5x + 14 = 0 10. The sum of two positive numbers is 18 and their difference is 4. (a) Make two equations with letting x and y for two numbers. (b) Solve the required equations graphically. MIXED PRACTICE–IV

ALGEBRA 193 11. The present ages of father and his son are 42 years and 16 years respectively. (a) Write the ages of father and son before x years. (b) If, x years ago, the product of their ages was 272, find the value of x. 12. Two expressions are given below. x2 + 5x + 6 and x2 + 7x + 12 (a) Find the LCM of the above expressions. (b) Find the HCF of the above expressions. (c) For what values of x, x2 + 5x + 6 = 0 (d) Prove that: 1 x2 + 7x + 12 + 1 x2 + 5x + 6 = 2 x2 + 6x + 8 13. A rectangle has length 4 cm more than its breadth. (a) Find the value of length if its breadth is x. (b) If the area of a rectangle is 96 cm2 , find the length and breadth of the rectangle. [Use : A = l × b] ANSWERS 1. (a) 1 (b) 1 2. (a) 320 3. (a) 34x (b) 1 4. (a) 3y(x – 2y) (b) 2y x2 – y2 5. (a) (2x + 3y) (2x – 3y) (b) 2(x2 + y2 ) x2 – y2 6. (a) (x – 2) (x – 3) (b) (x + 4)2 (x – 4) 7. (a) x – 4 (b) 1 8. (a) 4xy 4x2 – y2 (b) x + 3 9. (a) 2, 1 (b) 2, – 7 10. (a) x + y = 18, x – y = 4 (b) 11, 7 11. (a) 42 – x, 16 – x (b) 8 12. (a) (x + 2) (x + 3) (x + 4) (b) x + 3 (c) – 2, – 4 13. (a) x + 4 (b) 12 cm, 8 cm

194 The Leading Maths - 8 FM : 21 Time : 40 Min. CONFIDENCE LEVEL TEST IV ALGEBRA 1. (a) What is the value of (3ab)0 ? Write it. [1] (b) Factorize: 6xy2 – 4xy [1] (c) Prove that:   xa xb  a + b ×   xb xc   b + c ×   xc xa   a + c = 1 [2] 2. (a) Simplify: a – b a + b – a + b a – b [1] (b) Factorize: x2 – 5x + 6 [2] (c) Compare which one is greater between 230 and 320. [1] 3. (a) Write full form of LCM. [1] (b) Find the HCF of x2 – 5x + 4 and 4x – 4. [2] (c) Simplify: 3a + 2 + 3a 3a × 10 [1] 4. (a) Solve the equations graphically: 3x + 2y = 13 and x – y = 1 [1] (b) The sum of a number and its square is 12. (i) Make an equation to represent the given statement. [1] (ii) Find the numbers. [2] 5. The sum of two positive numbers is 12 and their difference is 8. (a) Make two equations with supposing x and y for two numbers. [1] (b) Solve the required equations by graphical method. [2] (c) Write the relation between these two numbers. [1] 6. (a) Prove that: 1 x2 – 5x + 6 + 1 x2 – 7x + 12 = 2 x2 – 6x + 8 [2] (b) Find the LCM of x2 – 2x and 3x – 6. Attempt all the questions.