Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.1 73 Chapter 5 Factorization: Writing an algebraic expression as the product of two or more algebraic expressions is called factorization Example 1: i). Factorize 2 15 10 5 x x Solution: Given 2 15 10 5 x x 2 5 3 5 2 5 x x Taking 5 as a common factors 2 5 3 2x x Example 1: Ii). Factorize 2 2 3 12 20 x y x y Solution: Given 2 2 3 12 20 x y x y Taking 2 4x y as a common factors = 2 4 3 5 x y y x Example 2: Factorize 2 a ab a b 3 3 Solution: Given 2 a ab a b 3 3 3 3 a a b a b a b a Example 3: i). Factorize 2 x x 8 16 Solution: Given 2 x x 8 16 2 2 2 2 4 4 4 x x x Example 3: ii). Factorize 2 25 30 9 y y Solution: Given 2 25 30 9 y y 2 2 2 5 2 5 3 3 5 3 y y y Example 4: a). Factorize 2 x 16 Solution: Given 2 x 16 2 2 4 4 4 x x x Example 4: b). Factorize 2 9 25 a Solution: Given 2 9 25 a 2 2 3 5 3 5 3 5 a a a Example 4: c). Factorize 4 4 6 6 x y Solution: Given 4 4 6 6 x y 4 4 2 2 2 2 6 6 x y x y 2 2 2 2 6 x y x y 2 2 6 x y x y x y Example 5: Factorize 2 2 2 a ab b c 4 4 Solution: Given 2 2 2 a ab b c 4 4 2 2 2 2 2 2 2 2 2 2 2 a a b b c a b c a b c a b c Example 6: Factorize 2 2 a b b 2 1 Solution: Given 2 2 a b b 2 1 2 2 2 2 2 2 2 2 1 2 1 1 1 1 1 1 1 a b b a b b a b a b a b a b a b Rule 1 Common Rule 2: Pair Common Rule 3: Rule 4: Rule 5: Exercise 5.1 Q1. Factorize 3 2 3 2 2 9 15 3 s t s t s t Solution: Given 3 2 3 2 2 9 15 3 s t s t s t 2 2 3 3 5 s t s t t Q2. Factorize 2 3 4 3 2 2 4 3 2 10 15 30 a b c a b c a b c Sol: Given 2 3 4 3 2 2 4 3 2 10 15 30 a b c a b c a b c 2 2 2 2 2 5 2 3 6 a b c bc a a b Q3. Factorize ax a x 1 Solution: Given ax a x 1 1 1 1 1 1 a x x a x Q4. Factorize 2 3 2 x y xy xy 2 2 Solution: Given 2 3 2 x y xy xy 2 2 Rearranging 2 2 3 x xy xy y 2 2 2 2 2 2 x x y y x y x y x y Q5. Factorize 2 2 1 4 4 x x Solution: Given 2 2 1 4 4 x x 2 2 2 1 1 2 2 2 1 2 x x x x x x Q6. Factorize 2 2 4 20 25 x y x y z z Solution: Given 2 2 4 20 25 x y x y z z 2 2 2 2 2 5 5 x y x y z z 2 2 5 x y z 2 2 2 a 2ab b a b 2 2 a b a b a b 2 2 a b c a b c a b c
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.2 74 Q7. Factorize 4 4 4 4 x y y x Solution: Given 4 4 4 4 x y y x 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 x y y x x y x y y x y x x y x y y x y x x y x y x y y x y x y x Q8. Factorize 2 2 288 x Solution: Given 2 2 288 x = 2 2 2 2 144 2 12 2 12 12 x x x x Q9. Factorize 2 2 1 2 uv u v Solution: Given 2 2 1 2 uv u v Rearranging 2 2 1 2 u v uv 2 2 2 2 1 2 1 1 1 1 1 u v uv u v u v u v u v u v Q10. Factorize 2 2 2 2 25 20 4 16 a b abc c d Solution: Given 2 2 2 2 25 20 4 16 a b abc c d 2 2 2 2 2 5 2 5 2 2 4 5 2 4 5 2 4 5 2 4 ab ab c c d ab c d ab c d ab c d Example 7: Factorize 4 4 a b 4 Sol: Given 4 4 a b 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 2 2 a b a b a b a b a b a b ab a b ab a b ab Example 8: Factorize 4 2 2 4 a a b b Sol: Given 4 2 2 4 a a b b 4 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a b a b a b a b a b 2 2 2 2 2 2 2 a b a b a b 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a b a b ab a b ab a b ab Example 9: Factorize 4 2 x x 7 1 Sol: Given 4 2 x x 7 1 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 7 1 2 1 2 1 7 1 2 7 1 9 1 3 1 3 1 3 x x x x x x x x x x x x x x x x x Rule 6 Exercise 5.2 Q1. Factorize Solution: Given Using formula Using Q2. Factorize Solution: Given Using formula Q3. Factorize 4 2 2 4 a a b b Sol: Given 4 2 2 4 a a b b 4 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a b a b a b a b a b a b a b a b a b a b 2 2 4 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a b 2a b 2a b a b c c 2a b a b c a b c 4 x 64 4 x 64 2 2 2 2 2 2 2 2 x 8 x 8 2 x 8 2 x 8 2 2 2 a 2ab b a b 2 2 2 2 2 2 2 2 2 x 8 2 x 8 x 8 16x x 8 4x 2 2 a b a b a b 2 2 2 2 x 8 4x x 8 4x x 4x 8 x 4x 8 4 4x 81 4 4x 81 2 2 2 2 2 2 2 2 2x 9 2x 9 2 2x 9 2 2x 9 2 2 2 a 2ab b a b 2 2 2 2 2 2 2 2 2 2x 9 2 2x 9 2x 9 36x 2x 9 6x 2 2 2 2 2x 9 6x 2x 9 6x 2x 6x 9 2x 6x 9
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.2 75 2 2 2 2 2 2 2 2 a b ab a b ab a b ab Q4. Factorize Solution: Given Using formula Using Q5: Sol: Given Using formula Using Using Using Q6. Factorize Solution: Given Using formula Q7. Factorize Solution: Given Using Using Q8. Factorize Solution: Given Using formula 4 2 x x 1 4 2 x x 1 4 2 2 2 2 2 2 2 2 2 2 2 x 1 x x 1 x x 1 2 x 1 2 x 1 x 2 2 2 a 2ab b a b 2 2 2 2 2 2 2 2 2 2 2 2 2 2 x 1 2 x 1 x x 1 2x x x 1 x x 1 x 2 2 a b a b a b 2 2 2 2 x 1 x x 1 x x x 1 x x 1 8 4 x x 1 8 4 x x 1 8 4 2 2 4 4 2 2 4 4 4 4 x 1 x x 1 x x 1 2 x 1 2 x 1 x 2 2 2 a 2ab b a b 2 4 4 4 2 4 4 4 2 4 4 2 2 4 2 x 1 2 x 1 x x 1 2x x x 1 x x 1 x 2 2 a b a b a b 4 2 4 2 2 2 4 2 2 2 2 2 4 2 2 2 2 2 x 1 x x 1 x x x 1 x 1 x x x 1 x 1 2 x 1 2 x 1 x 2 2 2 a 2ab b a b 2 4 2 2 2 2 2 4 2 2 2 2 2 4 2 2 2 2 2 4 2 2 x x 1 x 1 2 x 1 x x x 1 x 1 2x x x x 1 x 1 x x x 1 x 1 x 2 2 a b a b a b 4 2 2 2 4 2 2 2 x x 1 x 1 x x 1 x x x 1 x x 1 x x 1 4 4 1 x 7 x 4 4 1 x 7 x 2 2 2 2 2 2 2 2 2 2 2 2 1 x 7 x 1 1 1 x 2 x 2 x 7 x x x 2 2 2 a 2ab b a b 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 x 2 x 7 x x 1 x 2 7 x 1 x 9 x 1 x 3 x 2 2 2 2 1 1 x 3 x 3 x x 4 4 1 81x 14 81x 4 4 1 81x 14 81x 2 2 2 2 2 2 2 2 2 2 2 2 1 9x 14 9x 1 1 1 9x 2 9x 2 9x 14 9x 9x 9x 2 2 2 a 2ab b a b 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 9x 2 9x 14 9x 9x 1 9x 2 14 9x 1 9x 16 9x 1 9x 4 9x 2 2 a b a b a b 2 2 2 2 1 1 9x 4 9x 4 9x 9x 4 2 2 4 4x 4x y 64y 4 2 2 4 4x 4x y 64y 4 4 2 2 4 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4x 64y 4x y 4 x 16y x y 4 x 4y x y 4 x 4y 2 x 4y 2 x 4y x y 2 2 2 a 2ab b a b 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 x 4y 2 x 4y x y 4 x 4y 8x x 4 x 4y 9x y 4 x 4y 3xy
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.3 76 Using Q9. Factorize Solution: Given Using formula Using Q10. Factorize Solution: Given Using 2 2 2 2 xy x y xy x y xy 2 3 2 3 Ex 10: Factorize 2 y y 7 12 Sol: Given 2 y y 7 12 12 is not a perfect square 2 4 3 12 4 3 4 4 3 y y y y y y y y + 12 - - 6 1 5 2 4 3 Ex 11: Factorize 2 x x 10 21 Sol: Given 2 x x 10 21 21 is not a perfect square 2 7 3 21 7 3 7 3 7 x x x x x x x x + 21 + + 9 1 8 2 7 3 Example 12: Find an expression of the perimeter of a rectangle whose area is given by 2 x x 13 90 Solution: Area = L x w Area 2 x x 13 90 90 is not a perfect square 2 18 5 90 18 5 18 x x x x x x - 90 + - 14 1 15 2 16 3 17 4 18 5 Area of rectangle x x 18 5 Here L= x 18 and B= x 5 Perimeter of Rectangle = 2L+2B Putting the values Perimeter = 2 18 2 5 x x 2 36 2 10 4 26 x x x Example 13: Factorize 2 2 7 4 x x Sol: Given 2 2 2 7 4 2 8 4 x x x x x 2 4 1 4 2 1 4 x x x x x Rule 7: any term is missing Consider Multiply with signs + Ac Sign of b Answer If sign of ac is +ve x If sign of ac is –ve Reduce 1 from b x Increase 1 in b Exercise 5.3 Q1. Factorize 2 x x 7 12 Solution: Given 2 x x 7 12 12 is not a perfect square 2 4 3 12 4 3 4 3 4 x x x x x x x x + 12 - - 6 1 5 2 4 3 Q2. Factorize 2 x x 12 Solution: Given 2 x x 12 12 is not a perfect square 2 4 3 12 4 3 4 3 4 x x x x x x x x - 12 + - 2 1 3 2 4 3 Q3. Factorize 2 20 x x Solution: Given 2 20 x x 20 is not a perfect square 2 20 5 4 5 4 4 5 4 x x x x x x x x - 20 - + 2 1 3 2 4 3 5 X 4 Q4. Factorize 2 2 7 3 y y Solution: Given 2 2 7 3 y y 3 is not a perfect square 2 2 a b a b a b 2 2 2 2 2 2 2 2 4 x 4y 3xy x 4y 3xy 4 x 3xy 4y x 3xy 4y 4 2 2 4 16m 4m n n 4 2 2 4 16m 4m n n 4 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 16m n 4m n 4m n 4m n 4m n 2 4m n 2 4m n 4m n 2 2 2 a 2ab b a b 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4m n 2 4m n 4m n 4m n 8m n 4m n 4m n 4m n 4m n 2mn 2 2 a b a b a b 2 2 2 2 2 2 2 2 4m n 2mn 4m n 2mn 4m 2mn n 4m 2mn n 5 3 3 5 4x y 11x y 9xy 5 3 3 5 4x y 11x y 9xy 4 2 2 4 4 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 xy 4x 11x y 9y xy 4x 9y 11x y xy 2x 3y 11x y xy 2x 3y 2 2x 3y 2 2x 3y 11x y 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 xy 2x 3y 2 2x 3y 11x y xy 2x 3y 12x y 11x y xy 2x 3y x y xy 2x 3y xy 2 2 a b a b a b 2 2 a 2ab b 2 ax bx c
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.3 77 2 2 6 1 3 2 3 1 3 3 2 1 y y y y y y y y + 6 - - 6 1 Q5. Factorize 2 4 8 3 x x Solution: Given 2 4 8 3 x x 3 is not a perfect square 2 4 6 2 3 2 2 3 1 2 3 2 3 2 1 x x x x x x x x + 12 + + 7 1 6 2 Q6: Factorize R. Work Solution: Given - 10 - + 4 1 5 2 Q7Factorize R. Work Sol: Given - 6 - + 6 1 Q8: Factorize R. Work So: Given - 12 + - 5 1 6 2 Q9: R. Work Sol: Given - 24 + - 6 1 7 2 8 3 Q10: R. Work Sol: + 24 + + 9 1 8 2 7 3 6 4 Q11. Factorize 2 x x 1 3 1 2 Solution: Given 2 x x 1 3 1 2 2 is not a perfect square 2 1 2 1 1 1 2 1 1 2 1 1 2 1 1 1 2 2 3 x x x x x x x x x x + 2 + + 2 1 Q12. Factorize 8 10 5 7 2 4 4 40 84 x y x y x y Solution: Given 8 10 5 7 2 4 4 40 84 x y x y x y 2 4 6 6 3 3 4 10 21 x y x y x y 21 is not a perfect square 2 4 6 6 3 3 3 3 2 4 3 3 3 3 3 3 2 4 3 3 3 3 4 7 3 21 4 7 3 7 4 7 3 x y x y x y x y x y x y x y x y x y x y x y Q13. Factorize Find an expression for perimeter of a rectangle with area given by 2 x x 24 81 Solution: Area = L x w Area 2 x x 24 81 81 is not a perfect square 2 27 3 81 27 3 27 x x x x x x - 81 + - 25 1 26 2 27 3 Area of rectangle x x 27 3 Here L= x 27 and B= x 3 Perimeter of Rectangle = 2L+2B Putting the values Perimeter = 2 27 2 3 x x 2 57 2 6 4 51 x x x Ex 14: Factorize 2 2 x x x x 7 10 7 12 1 Sol: Given 2 2 x x x x 7 10 7 12 1 Here two terms are same Let 2 y x x 7 so, 2 2 2 2 2 10 12 1 12 10 120 1 22 121 2. .11 11 11 y y y y y y y y y y Putting back value of 2 y x x 7 = 2 2 x x 7 11 Exp15: Factorize 2 2 3 11 2 3 11 3 12 x x x x Sol: Given 2 2 3 11 2 3 11 3 12 x x x x Here two terms are same Let 2 y x x 3 11 2 2 2 3 12 3 2 6 12 5 6 y y y y y y y 2 6 1 6 6 1 6 1 6 y y y y y y y y Putting back value of 2 y x x 3 11 2 2 2 2 2 2 3 11 1 3 11 6 3 11 1 3 9 2 6 3 11 1 3 3 2 3 3 11 1 3 2 3 x x x x x x x x x x x x x x x x x x 2 10y 3y 1 2 10y 3y 1 2 10y 5y 2y 1 5y 2y 1 1 2y 1 5y 1 2y 1 3 2 6x 15x 9x 3 2 6x 15x 9x 3 2 2 2 6x 15x 9x 3x 2x 5x 3 3x 2x 6x 1x 3 3x 2x x 3 1 x 3 3x 2x 1 x 3 2 2xy 8xy 24x 2 2xy 8xy 24x 2 2 2x y 4y 12 2x y 6y 2y 12 2x y y 6 2 y 6 2x y 2 y 6 2 2 5t 12t 2 2 5t 12t 2 2 8t 3t 12t 2 1 4t 3t 1 4t 2 3t 1 4t 3 2 2 3 16x y 20x y 6xy 3 2 2 3 16x y 20x y 6xy 2 2 2 2 2xy 8x 10xy 3y 2xy 8x 6xy 4xy 3y 2xy 2x 4x 3y y 4x 3y 2xy 2x y 4x 3y
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.4 78 Example 16: x x x x 1 2 3 4 1 Sol: Given x x x x 1 2 3 4 1 Here 1 4 2 3 , so rearranging 2 2 2 2 1 4 2 3 1 4 1 4 3 2 6 1 5 4 5 6 1 x x x x x x x x x x x x x x Here two terms are same Let y 2 x x 5 2 2 2 2 2 4 6 1 6 4 24 1 10 25 2. .5 5 5 y y y y y y y y y y Putting back value of y 2 x x 5 2 2 x x5 5 Exp17: Factorize 2 2 2 x x x x x 5 6 5 6 2 Sol: Given 2 2 2 x x x x x 5 6 5 6 2 2 2 2 2 2 3 2 6 3 2 6 2 3 2 3 3 2 3 2 2 3 2 3 2 x x x x x x x x x x x x x x x x x x x 2 x x x x x 2 2 3 3 2 2 2 2 x x x 4 9 2 4 2 2 2 4 2 4 2 2 2 2 2 2 2 9 4 36 2 15 36 12 3 36 12 3 12 12 3 x x x x x x x x x x x x x x Exp 18: Factorize 3 2 2 3 8 36 54 27 a a b ab b Sol: Given 3 2 2 3 8 36 54 27 a a b ab b 3 2 2 3 3 2 3 2 3 3 2 3 3 2 3 a a b a b b a b Exp 19: Factorize 3 2 2 3 27 27 9 x x y xy y Sol: Given 3 2 2 3 27 27 9 x x y xy y 3 2 2 3 3 3 3 3 3 3 3 x x y x y y x y Rule8: Suppose same terms Rule 9 Multiply those factors which Rule 10: Exercise 5.4 Q1:Factorize Sol: Given Here two terms are same Let 2 2 2 2 2 7 15 16 15 7 105 16 22 121 2. .11 11 11 y y y y y y y y y y 2 2 4 16 11 x x Since Q2:Factorize Sol; Given Here two terms are same Let Q3: Sol: Given Rearranging accordingly 4 + 6 = 2 + 8 Here two terms are same Let 2 2 2 2 2 10 15 2. .5 5 10 15 5 x x x x x x x Q4: Sol: Given Rearranging accordingly 0 + 3 = 1 + 2 2 2 ax bx c ax bx d e 2 y ax bx x a x b x c x d e a b c d 3 3 2 2 3 a 3a b 3ab b a b 2 2 4x 16x 7 4x 16x 15 16 2 2 4x 16x 7 4x 16x 15 16 2 y 4x 16x 2 y 4x 16x 2 2 9x 9x 4 9x 9x 10 72 2 2 9x 9x 4 9x 9x 10 72 2 y 9x 9x 2 2 2 2 2 2 y 4 y 10 72 y y 10 4 y 10 72 y 10y 4y 40 72 y 14y 32 y 16y 2y 32 y y 16 2 y 16 y 2 y 16 y 9x 9x 9x 9x 2 9x 9x 16 x 2 x 4 x 6 x 8 9 x 2 x 4 x 6 x 8 9 2 2 2 2 x 4 x 6 x 2 x 8 9 x x 6 4 x 6 x x 8 2 x 8 9 x 6x 4x 24 x 8x 2x 16 9 x 10x 24 x 10x 16 9 2 y x 10x 2 2 2 2 2 2 y 24 y 16 9 y y 16 24 y 16 9 y 16y 24y 384 9 y 40y 375 y 25y 15y 375 y y 25 15 y 25 y 15 y 25 y x 10x x 10x 15 x 10x 25 x x 1 x 2 x 3 1 x x 1 x 2 x 3 1 2 2 2 2 x x 3 x 1 x 2 1 x x 3 x x 2 1 x 2 1 x 3x x 2x 1x 2 1 x 3x x 3x 2 1
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.5 79 Here two terms are same Let Q5: 2 x x x x x 1 2 3 6 3 Sol: Given 2 x x x x x 1 2 3 6 3 Here a b c d but 1 6 2 3 so 2 2 2 2 2 2 2 2 2 2 1 6 2 3 3 6 1 6 3 2 6 3 7 6 5 6 3 6 7 6 5 3 x x x x x x x x x x x x x x x x x x x x x x Here two terms are same Let 2 y x 6 2 2 2 2 7 5 3 5 7 35 3 y x y x x y xy xy x x 2 2 2 2 12 32 8 4 32 8 4 8 4 8 y xy x y xy xy x y y x x y x y x y x Putting back value of 2 y x 6 2 2 2 2 6 4 6 8 4 6 8 6 x x x x x x x x Q6: Sol: Given Using Q7: Sol: Given Using Q8: Sol: Given Using Q9: Sol: Given Using Q10: Sol: Given Example 20: Factorize 3 2 16 x Sol: Given 3 2 16 x 3 3 3 2 8 2 2 x x Using 3 3 2 2 a b a b a ab b 2 2 2 2 2 .2 2 2 2 2 4 x x x x x x Example 21: Factorize 3 3 a b 64 Sol: Given 3 3 a b 64 3 3 a b4 Using 3 3 2 2 a b a b a ab b 2 2 2 2 4 .4 4 4 4 16 a b a a b b a b a ab b Rule11: Exercise 5.5 Q1. Factorize 3 a 27 Solution: Given 3 a 27 3 3 a 3 Using 3 3 2 2 a b a b a ab b 2 2 2 3 .3 3 3 3 9 a a a a a a Q2. Factorize 6 6 a b Solution: Given 6 6 a b 3 3 2 2 a b 2 y x 3x 2 2 2 2 2 2 2 y y 2 1 y 2y 1 y 2.y.1 1 y 1 y x 3x x 3x 1 3 2 2 3 64x 144x y 108xy 27y 3 2 2 3 64x 144x y 108xy 27y 3 3 2 2 3 a b a 3a b 3ab b 3 2 2 3 3 4x 3 4x 3y 3 4x 3y 3y 4x 3y 3 3 a 1 1 b 2 2 a b ab 8 4 6 27 3 3 a 1 1 b 2 2 a b ab 8 4 6 27 3 3 2 2 3 a b a 3a b 3ab b 3 2 2 3 3 a a b a b b 3 3 2 2 3 2 3 3 a b 2 3 3 3 3 3 x 3x 3a a a x a x 3 3 3 3 x 3x 3a a a x a x 3 3 2 2 3 a b a 3a b 3ab b 3 2 2 3 3 x x a x a a 3 3 a a x a x x x a a x 3 2 2 3 27a 189a b 441ab 343 b 3 2 2 3 27a 189a b 441ab 343 b 3 3 2 2 3 a b a 3a b 3ab b 3 2 2 3 3 3a 3 3a 7b 3 3a 7b 7b 3a 7b 3 3 2 1 8x 4x 3x 27x 3 3 2 1 8x 4x 3x 27x 2 3 3 2 3 1 1 1 2x 3 2x 3 2x 3x 3x 3x 1 2x 3x 3 3 2 2 a b a b a ab b
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.5 80 Using 3 3 2 2 a b a b a ab b 2 2 2 2 2 2 2 2 2 2 4 2 2 4 a b a a b b a b a a b b Q3. Factorize 3 24 3 x Solution: Given 3 24 3 x 3 3 3 3 8 1 3 2 1 x x Using 3 3 2 2 a b a b a ab b 2 2 2 3 2 1 2 2 1 1 3 2 1 4 2 1 x x x x x x Q4. Factorize Solution: Given Using Q5: 3 2 128 x Solution: Given 3 2 128 x 3 3 3 2 64 2 4 x x Using 2 2 2 2 4 .4 4 2 4 4 16 x x x x x x Q6: Solution: Given Using Q7: Solution: Given Using Q8. Factorize 9 x 1 Solution: Given 9 x 1 3 3 3 x 1 Using 3 3 2 2 a b a b a ab b 2 2 3 3 3 3 3 6 3 1 1 1 1 1 x x x x x x Again Using 3 3 2 2 a b a b a ab b 2 2 6 3 2 6 3 1 .1 1 1 1 1 1 x x x x x x x x x x Q9. Factorize Solution: Given Using Using Q10. Factorize 3 3 27x y Solution: Given 3 3 27x y 3 3 3x y Using 2 2 2 2 3 3 3 3 9 3 x y x x y y x y x xy y Remainder Theorem: When a polynomial P x of degree n 1 is divided by x r gives a constant remainder then R P r Dividend = Divisor x Quotient + Remainder Or P x x r Q x R Where P x Dividend, x r = Divisor Q x Quotient, R Remainder Exp 22: Divide 3 2 3 7 6 3 x x x by x 2 & verify your answer by remainder theorem. Sol: Divide 3 2 P x x x x 3 7 6 3 by x 2 3 1 27r 3 1 27r 3 3 1 3r 3 3 2 2 a b a b a ab b 2 2 2 1 3r 1 1 3r 3r 1 3r 1 3r 9r 3 3 2 2 a b a b a ab b 5 2 4x 256x 5 2 4x 256x 2 3 3 3 2 4x x 64 4x x 4 3 3 2 2 a b a b a ab b 2 2 2 2 2 4x x 4 x x 4 4 4x x 4 x 4x 16 3 3 18 x y 144 a b 3 3 18 x y 144 a b 3 3 3 3 3 3 18 x y 8 a b 18 x y 2 a b 18 x y 2a 2b 3 3 2 2 a b a b a ab b 2 2 2 2 18 x y 2a 2b x y x y 2a 2b 2a 2b 18 x y 2a 2b x y x y 2a 2b 2a 2b 3 3 a c d 3 3 a c d 3 3 2 2 a b a b a ab b 2 2 2 2 a c d a a c d c d a c d a ac ad c d 2 2 2 a b a 2ab b 2 2 2 a c d a ac ad c 2cd d 2 2 2 a c d a c d ac ad 2cd 3 3 2 2 a b a b a ab b
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.5 81 2 3 2 3 2 2 2 3 4 2 3 7 6 3 3 6 6 2 4 3 4 8 5 x x x x x x x x x x x x x x R Now using remainder theorem Take divisor x x 2 0 2 Put in dividend 3 2 2 3 2 7 2 6 2 3 2 3 8 7 4 12 3 2 24 28 9 2 5 P P P P R Which is the same obtained by actual division Example 23: Without performing division. Find the remainder when 3 2 2 3 2 x x x is divided by x 3 Sol: Here Dividend 3 2 P x x x x 2 3 2 And take divisor x x 3 0 3 put 3 2 3 2 3 3 3 3 2 3 2 27 3 9 1 3 54 27 1 3 28 P P P P R Hence Remainder = 28 Example 24: For what value of k 3 2 3 9 3 4 2 x x k x will be exactly divisible by x 1 Solution: Dividend 3 2 P x x x k x 3 9 3 4 2 And take divisor x x 1 0 1 Given that exactly divisible means P R 1 0 3 2 1 3 1 9 1 3 4 1 2 1 3 9 3 4 2 0 P k P k 12 3 4 2 0 12 4 2 3 3 18 18 6 3 k k k k Example 25: Find the zero of the polynomial 2 P x x x 4 3 Solution: Given 2 P x x x 4 3 Let x 1 then 2 1 1 4 1 3 1 1 4 3 1 0 P P P Hence 1 is a zero of the given polynomial Similarly when x 3 , then 2 3 3 4 3 3 3 9 12 3 3 0 P P P Therefore 3 is another zero of given polynomial Factor Theorem: Let P x be a polynomial or dividend and a linear polynomial or divisor or factor of P x if and only if k is a zero of polynomial P x i.e. P k 0 Example 26: using factor theorem, prove that x 3 is a factor of 3 2 x x x 5 3 and find the other factors. Sol: Given 3 2 P x x x x 5 3 and Factor x 3 0 x 3 putting 3 2 3 3 3 5 3 3 3 27 9 15 3 3 27 27 0 P P P Therefore by factor theorem x 3 is a factor of P x to find the remaining factors 2 3 2 3 2 2 2 2 1 3 5 3 3 2 5 2 6 3 3 0 x x x x x x x x x x x x x x R Here 2 Q x x x 2 1 and R 0 Since P x x k Q x R 3 2 2 x x x x x x 5 3 3 2 1 0 2 2 2 3 2. .1 1 3 1 x x x x x Example 27: Prove that x 2 is a factor of 3 x x 7 6 and hence find the other factors. Sol: here 3 P x x x 7 6 and factor x 2 0 x 2 putting 3 P 2 2 7 2 6
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.6 82 8 14 6 14 14 0 Therefore x 2 is a factor of given polynomial To find other factors 2 3 3 2 2 2 2 3 2 7 6 2 2 7 2 4 3 6 3 6 0 x x x x x x x x x x x x x R Here 2 Q x x x 2 3 and R 0 Since P x x k Q x R 3 2 x x x x x 7 6 2 2 3 0 2 2 3 1 3 2 3 1 3 2 1 3 x x x x x x x x x x x Rule 12: & Put Then Rule 13: Given choose Put So Then Exercise 5.6 Q1 By using reminder theorem find reminder of following polynomials where i). is divided by Sol: Given and take factor Using reminder theorem i.e., put the value of r in eq (1) Therefore the reminder = 4 ii). is divided by Sol: Given & take factor Using reminder theorem i.e., put the value of r in eq (1) Therefore the reminder = - 44 iii). is divided by Sol: Given and take factor Using reminder theorem i.e., put the value of r in eq (1) Therefore the reminder = - 4 Q2. Without preforming division find value of a, when is divided by Sol: Given & the factor using factor theorem i.e. Q3. Without preforming division find value of b, when is divided by Sol: Given & the factor using factor theorem i.e., Q4. Using factor theorem, factorize following polynomials i). 3 2 P x ax bx cx d x k x k 0 x k 3 2 P k a k b k c k d R 3 2 P x ax bx cx d k 1, 1,2, 2,3, 3 3 2 P k a k b k c k d 0 2 3 2 Q x Ax Bx C x k ax bx cx d 3 2 2 ax bx cx d x k Ax Bx C 3 2 x 6x 8x 11 x 1 3 2 P x x 6x 8x 11 1 x r x 1 0 r 1 P r R 3 2 P r 1 6 1 8 1 11 R 1 6 8 11 R 15 11 R 4 3 2 2x 4x 7x 5 x 3 3 2 P x 2x 4x 7x 5 1 x r x 3 0 r 3 P r R 3 2 P r 2 3 4 3 7 3 5 P 3 2 27 4 9 21 5 R 54 36 26 R 54 10 R 44 3 3x x 200 x 4 3 P x 3x x 200 1 x r x 4 0 r 4 P r R 3 P 4 3 4 4 200 P r 3 64 4 200 R 192 4 200 R 196 200 R 4 3 2 2x ax 2ax 3x 2 x 1 3 2 P x 2x ax 2ax 3x 2 x 1 0 x 1 P r 0 R 0 3 2 P 1 2 1 a 1 2a 1 3 1 2 0 2 1 a 1 2a 3 2 0 2 a 2a 1 0 a 3 0 a 3 3 2 x 4x bx 2 x 1 3 2 P x x 4x bx 2 x 1 0 x 1 P r 0 R 0 3 2 P 1 1 4 1 b 1 2 0 1 4 b 2 0 b 2 3 0 b 5 0 b 5 3 2 x 2x 5x 6
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.6 83 Sol: Given Using factor theorem to find other factors we take where Take r = 1 Since , therefore by factor theorem is a factor of . To find the other factors divide by as Here and Since Putting the values we get ii). Sol: Given Using factor theorem to find other factors we take where Take r = 1 Now take r = - 1 Since , therefore by factor Theorem is a factor of . To find other factors divide by as under Here and Since Putting the values we get iii). Sol: Given Using factor theorem to find other factors we take where Take r = 1 , therefore by factor theorem x 1 is a factor of . To find other factors divide by as under Here and Since Putting the values we get iv). Sol: Given Using factor theorem to find the other factors we take where Take r = 1 3 2 P x x 2x 5x 6 P r 0, r 1, 2, 3,... 3 2 P 1 1 2 1 5 1 6 P r 1 2 5 6 R 7 7 R 0 P 1 0 x 1 P x P x x 1 2 3 2 3 2 2 2 x x 6 x 1 x 2x 5x 6 x x x 5x x x 6x 6 6x 6 0 Q x x x 6 2 R 0 P x x r Q x R 3 2 2 2 x 2x 5x 6 x 1 x x 6 x 1 x 3x 2x 6 x 1 x x 3 2 x 3 x 1 x 2 x 3 3 2 x x 4x 4 3 2 P x x x 4x 4 P r 0, r 1, 2, 3,... 3 2 P 1 1 1 4 1 4 P r 1 1 4 4 R 2 8 R 6 0 3 2 P 1 1 1 4 1 4 P r 1 1 4 4 R 5 5 R 0 P 1 0 x r x 1 x 1 P x P x x 1 2 3 2 3 2 x 4 x 1 x x 4x 4 x x 4x 4 4x 4 0 Q x x 4 2 R 0 P x x r Q x R 3 2 2 2 2 x x 4x 4 x 1 x 4 x 1 x 2 x 1 x 2 x 2 3 x 7x 6 3 P x x 7x 6 P r 0, r 1, 2, 3,... 3 P 1 1 7 1 6 P r 1 7 6 R 7 7 R 0 P 1 0 P x P x x 1 2 3 2 3 2 2 2 x x 6 x 1 x 0.x 7x 6 x x x 7x x x 6x 6 6x 6 0 Q x x x 6 2 R 0 P x x r Q x R 3 2 2 x 7x 6 x 1 x x 6 x 1 x 3x 2x 6 x 1 x x 3 2 x 3 x 1 x 2 x 3 3 2 x 9x 23x 15 3 2 P x x 9x 23x 15 P r 0, r 1, 2, 3,...
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Exercise 5.6 84 , therefore by factor theorem is a factor of . To find the other factors divide by as under Here and Since Putting the values we get v). Sol: Given Using factor theorem to find other factors we take where Take r = 1 Take r = - 1 Take r = 2 Take r = - 2 Since , therefore by factor theorem is a factor of . To find the other factors divide by as under Here and Since Putting the values we get vi). Sol: Given Using factor theorem to find other factors we take where Take r = 1 Take r = - 1 Since , therefore by factor theorem is a factor of . To find other factors divide by as Here and Since Putting the values we get 3 2 P 1 1 9 1 23 1 15 P r 1 9 23 15 R 1 23 9 15 R 24 24 0 P 1 0 x 1 P x P x x 1 2 3 2 3 2 2 2 x 8x 15 x 1 x 9x 23x 15 x x 8x 23x 8x 8x 15x 15 15x 15 0 Q x x x 6 2 R 0 P x x r Q x R 3 2 2 2 x 9x 23x 15 x 1 x 8x 15 x 1 x 5x 3x 15 x 1 x x 5 3 x 5 x 1 x 3 x 5 3 2 x 4x 3x 18 3 2 P x x 4x 3x 18 P r 0, r 1, 2, 3,... 3 2 P 1 1 4 1 3 1 18 P r 1 4 3 18 R 1 18 4 3 R 19 7 12 0 3 2 P 1 1 4 1 3 1 18 P r 1 4 3 18 R 5 21 16 0 3 2 P 1 2 4 2 3 2 18 P r 8 16 6 18 R 8 18 16 6 R 26 22 4 0 3 2 P 1 2 4 2 3 2 18 P r 8 16 6 18 R 24 24 0 P 2 0 x r x 2 x 2 P x P x x 2 2 3 2 3 2 2 2 x 6x 9 x 2 x 4x 3x 18 x 2x 6x 3x 6x 12x 9x 18 9x 18 0 Q x x 6x 9 2 R 0 P x x r Q x R 3 2 2 2 2 2 x 4x 3x 18 x 2 x 6x 9 x 2 x 2.x.3 3 x 2 x 3 3 2 x 2x 19x 20 3 2 P x x 2x 19x 20 P r 0, r 1, 2, 3,... 3 2 P 1 1 2 1 19 1 20 P r 1 2 19 20 R 3 39 R 36 0 3 2 P 1 1 2 1 19 1 20 P r 1 2 19 20 R 21 21 R 0 P 1 0 x 1 P x P x x 1 2 3 2 3 2 2 2 x x 20 x 1 x 2x 19x 20 x x x 19x x x 20x 20 20x 20 0 Q x x x 20 2 R 0 P x x r Q x R
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Review Exercise 5 85 vii). Sol: Given Using factor theorem to find other factors we take where Take r = 1 Take r = - 1 Take r = 2 Since , therefore by factor theorem is a factor of . To find the other factors divide by as under Here and Since Putting the values we get viii). Sol: Given Using factor theorem to find the other factors we take where Take r = 1 Take r = - 1 Take r = 2 Take r = - 2 Since , therefore by factor theorem is a factor of . To find the other factors divide by as under Here and Since Putting the values we get Review Exercise 5 Q1. Ture and false questions. Read the following sentences carefully i). 2 x x x x 11 30 6 5 T ii). 3 2 x x x x 27 3 3 9 T iii). 3 2 x x x x 8 2 2 4 T iv). 2 4 4 x y x y x y x y F v). 6 6 3 3 3 3 x y x y x y F vi). 5 5 5 x y x y F Q2. Complete the following sentences i). 2 4 2 2 9 16 3 4 3 4 x y x y x y ii). 3 3 2 2 x y x y x xy y 27 3 3 9 iii). 2 x x x x 5 6 2 3 3 2 2 2 x 2x 19x 20 x 1 x x 20 x 1 x 5x 4x 20 x 1 x x 5 4 x 5 x 1 x 4 x 5 3 2 x x 14x 24 3 2 P x x x 14x 24 P r 0, r 1, 2, 3,... 3 2 P 1 1 1 14 1 24 P r 1 1 14 24 R 10 0 3 2 P 1 1 1 14 1 24 P r 1 1 14 24 R 2 38 36 0 3 2 P 2 2 2 14 2 24 P r 8 4 28 24 R 8 24 4 28 R 32 32 0 P 2 0 x r x 2 P x P x x 2 2 3 2 3 2 2 2 x x 12 x 2 x x 14x 24 x 2x x 14x x 2x 12x 24 12x 24 0 Q x x x 12 2 R 0 P x x r Q x R 3 2 2 2 x x 14x 24 x 2 x x 12 x 2 x 4x 3x 12 x 2 x x 4 3 x 4 x 2 x 3 x 4 3 2 x 6x 32 3 2 P x x 6x 32 P r 0, r 1, 2, 3,... 3 2 P 1 1 6 1 32 P r 1 6 32 R 5 32 27 0 3 2 P 1 1 6 1 32 P r 1 6 32 R 7 32 25 0 3 2 P 2 2 6 2 32 P r 8 24 32 R 8 8 16 0 3 2 P 2 2 6 2 32 P r 8 24 32 R 32 32 0 P 2 0 x r x 2 x 2 P x P x x 2 2 3 2 3 2 2 2 x 8x 16 x 2 x 6x 0.x 32 x 2x 8x 0.x 8x 16x 16x 32 16x 32 0 Q x x 8x 16 2 R 0 P x x r Q x R 3 2 2 2 2 2 x 6x 32 x 2 x 8x 16 x 2 x 2.x.4 4 x 2 x 4
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Review Exercise 5 86 iv). 2 2 2 a b a b ab 2 v). 3 3 2 2 8 2 4 2 b c b c b bc c Q3. Choose the correct answer i). Factors of 2 x x 2 24 are a). x x 4, 6 b). x x 4, 6 c). x x 3, 8 d). x x 8, 3 ii). Factorization of 2 2 2 x xy y z 2 is a). x y z x y z b). x y z x y z c). x y z x y z d). x y z x y z iii). Factorization of 3 3 a b is a). 2 2 a b a ab b b). 2 2 a b a ab b c). 2 2 a b a ab b d). 2 2 a b a ab b iv). Factors of 3 3 8y z are a). 2 2 2y z,4y 2yz z b). 2y z,2y z,2y z c). 2 2 2y z,4y 2yz z d). 2 2 2y z,4y 2yz z v). In simplified form 2 2 1 ...... b a b a b a). 2 2 b 1 a b b). 2 2 a a b c). 2 2 b a b d). 2 2 b a a b vi). Factorization of 2 2 a b b 10 25 is a). a b a b 5 5 b). a b a b 5 5 c). a b a b 5 5 d). a b a b 5 5 vii). Completely factorize 4 16 81 x a). 2 2 3 2 3 4 9 x x x b). 2 2 4 9 4 9 x x c). 2 3 2 3 2 3 x x x d). 2 4 9 2 3 2 3 x x x Q4: i). Factorize 3 2 3 3 18 x x x Solution: Given 3 2 3 3 18 x x x 3 2 2 2 3 3 18 3 6 3 3 2 6 3 3 2 3 3 2 3 x x x x x x x x x x x x x x x x x Q4: ii). Factorize 3 64 27 x Solution: Given 3 64 27 x 3 3 2 2 2 4 3 4 3 4 4 3 3 4 3 16 12 9 x x x x x x x Q4: iii).Factorize 6 6 x y Solution: Given 6 6 x y 2 2 3 3 3 3 3 3 2 2 2 2 2 2 2 2 x y x y x y x y x xy y x y x xy y x y x y x xy y x xy y Q4: iv).Factorize 3 3 343 125 5 7 a b b a Solution: Given 3 3 343 125 5 7 a b b a Using Q5. Factorize x x x x 1 2 3 4 120 Sol: Given x x x x 1 2 3 4 120 Here 1 4 2 3 , so rearranging 2 2 2 2 1 4 2 3 120 4 1 4 3 2 6 120 5 4 5 6 120 x x x x x x x x x x x x x x Here two terms are same Let y 2 x x 5 2 2 4 6 120 6 4 24 120 10 96 y y y y y y y 2 y y y 16 6 96 y y y 16 6 16 y y 6 16 Putting back value of y 2 x x 5 2 2 2 2 2 2 5 6 5 16 6 1 6 5 16 6 1 6 5 16 1 6 5 16 x x x x x x x x x x x x x x x x x x Q6: i). Using factor theorem factorize 3 x x 17 26 Solution: Given 3 P x x x 17 26 Put x 2 3 P 2 2 17 2 26 2 8 34 26 2 34 34 2 0 P P P Hence x 2 is a factor so 3 3 3 3 343a 125b 7a 5b 7a 5b 1 7a 5b 3 3 2 2 a b a b a ab b 3 3 2 2 2 2 2 2 7a 5b 1 7a 5b 7a 5b 7a 7a 5b 5b 1 7a 5b 7a 5b 49a 35ab 25b 1 7a 5b 7a 5b 49a 35ab 25b 1
Chapter 4 Khalid Mehmood M-Phil Applied Mathematics Review Exercise 5 87 2 3 3 2 2 2 2 13 2 17 26 2 2 17 2 4 13 26 13 26 0 x x x x x x x x x x x x x R Here 2 Q x x x 2 13 and R 0 Since P x x k Q x R 3 2 x x x x x 17 26 2 2 13 0 2 x x x 2 2 13 Q6: ii). Using factor theorem factorize 3 2 x x x 39 124 84 Solution: Given 3 2 x x x 39 124 84 Using factor theorem to find the other factors we take where Take r = - 1 Since , therefore by factor theorem is a factor of . To find the other factors divide by as under Here and Since Putting the values we get Q7. Factorize 4 2 2 4 a a b b Sol: Given 4 2 2 4 a a b b 4 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b a b a b a b a b a b a b a b a b a b a b a b ab a b ab a b ab Q8. By using long division method, find the quotient and the remainder when 3 2 2 4 6 x x is divided by x 3 Solution: Given dividend 3 2 2 4 6 x x Divisor x 3 2 3 2 3 2 2 2 2 2 6 3 2 4 6 2 6 2 6 2 6 6 6 6 18 24 x x x x x x x x x x x x R Here Q(x) = 2 2 2 6 x x and R = -24 Q9: i. Find the remainder when 3 2 x x x 2 3 1 is divided by x 1 Solution: Given 3 2 P x x x x 2 3 1 And divisor x 1 0 x 1 3 2 1 1 2 1 3 1 1 1 1 2 3 1 1 1 P P P Hence remainder R = 1 Q9: i. Find the remainder when 3 2 x x x 2 3 1 is divided by x 2 Solution: Given 3 2 P x x x x 2 3 1 And divisor x 2 0 x 2 3 2 2 2 2 2 3 2 1 2 8 8 6 1 2 7 P P P Hence remainder R = 7 Q10. By using remainder theorem find remainder when 72 48 x x 7 1 is divided by x 1 Solution: Given 72 48 P x x x 7 1 And divisor x 1 0 x 1 72 48 1 1 7 1 1 1 1 7 1 1 5 P P P Hence remainder R = - 5 P r 0, r 1, 2, 3,... 3 2 3 2 x 39x 124x 84 P 1 1 39 1 124 1 84 P r 1 39 124 84 R 40 40 0 P 1 0 x r x 1 x 1 P x P x x 1 2 3 2 3 2 2 2 x 40x 84 x 1 x 39x 124x 84 x x 40x 124x 40x 40x 84x 84 84x 84 0 Q x x 40x 84 2 R 0 P x x r Q x R 3 2 2 2 x 39x 124x 84 x 1 x 40x 84 x 1 x 42x 2x 84 x 1 x x 42 2 x 42 x 1 x 2 x 42