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Chapter 6: Factorization of Polynomials Exercise – 6.5 Question: 1 Using factor theorem, factorize of the polynomials: x3 + 6x2 + 11x + 6 Solution: Given polynomial, f(x) = x3 + 6x2 + 11x + 6 The constant term in f(x) is 6 The factors of 6 are ± 1, ± 2, ± 3, ± 6 Let, x + 1 = 0 => x = -1 Substitute the value of x in f(x) f(-1) = (−1)3 + 6(−1)2 + 11(−1) + 6 = - 1 + 6 - 11 + 6 = 12 – 12 = 0 So, (x + 1) is the factor of f(x) Similarly, (x + 2) and (x + 3) are also the factors of f(x) Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors. ∴ f(x) = k(x + 1)(x + 2)(x + 3) => x3 + 6x2 + 11x + 6 = k(x + 1)(x + 2)(x + 3) Substitute x = 0 on both the sides => 0 + 0 + 0 + 6 = k(0 +1)(0 + 2)(0 + 3) => 6 = k(1*2*3) => 6 = 6k => k = 1 Substitute k value in f(x) = k(x + 1)(x + 2)(x + 3) => f(x) = (1)(x + 1)(x + 2)(x + 3) => f(x) = (x + 1)(x + 2)(x + 3) ∴ x3 + 6x2 + 11x + 6 = (x + 1)(x + 2)(x + 3) Question: 2 Using factor theorem, factorize of the polynomials: x3 + 2x2 – x – 2 Solution: Given, f(x) = x3+ 2x2 – x – 2 The constant term in f(x) is -2 The factors of (-2) are ±1, ± 2 Let, x – 1 = 0 => x = 1 Substitute the value of x in f(x) f(1) = (1)3 + 2(1)2 – 1 – 2 = 1 + 2 – 1 – 2 = 0 Similarly, the other factors (x + 1) and (x + 2) of f(x) Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors. ∴ f(x) = k(x – 1)(x + 2)(x + 1 ) x3 + 2x2 – x – 2 = k(x – 1)(x + 2)(x + 1 ) Substitute x = 0 on both the sides 0 + 0 – 0 – 2 = k(-1)(1)(2) => – 2 = - 2k => k = 1 Substitute k value in f(x) = k(x – 1)(x + 2)(x + 1) f(x) = (1)(x – 1)(x + 2)(x + 1) => f(x) = (x – 1)(x + 2)(x + 1) So, x3 + 2x2 – x – 2 = (x – 1)(x + 2)(x + 1) Question: 3 Using factor theorem, factorize of the polynomials: x3 – 6x2 + 3x + 10 Solution: Let, f(x) = x3 – 6x2 + 3x + 10 The constant term in f(x) is 10 The factors of 10 are ± 1, ± 2, ± 5, ± 10 Let, x + 1 = 0 => x = -1 Substitute the value of x in f(x) f(-1) = (−1)3- 6(−1)2 + 3(−1) + 10 = -1 – 6 – 3 + 10 = 0 Similarly, the other factors (x – 2) and (x – 5) of f(x) Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors. ∴ f(x) = k(x + 1)(x – 2)(x – 5) Substitute x = 0 on both sides => x3– 6x2 + 3x + 10 = k(x + 1)(x – 2)(x – 5) => 0 – 0 + 0 + 10 = k(1)(-2)(-5) => 10 = k(10) => k = 1 Substitute k = 1 in f(x) = k(x + 1)(x – 2)(x – 5) f(x) = (1)(x + 1)(x – 2)(x – 5 ) so, x3 – 6x2 + 3x + 10 = (x + 1)(x – 2)(x – 5) Question: 4 Using factor theorem, factorize of the polynomials: x4 –7x3 + 9x2 + 7x –10 Solution: Given, f(x) = x4–7x3 + 9x2 + 7x – 10 The constant term in f(x) is 10 The factors of 10 are ± 1, ± 2, ± 5, ±10 Let, x – 1 = 0 => x = 1 Substitute the value of x in f(x) f(x) = 14 – 7(1)3 + 9(1)2 + 7(1) – 10 = 1 – 7 + 9 + 7 – 10 = 10 – 10 = 0 (x – 1) is the factor of f(x) Similarly, the other factors are (x + 1), (x – 2), (x – 5) Since, f(x) is a polynomial of degree 4. So, it cannot have more than four linear factor. So, f(x) = k(x – 1)(x + 1)(x – 2)(x – 5) => x4 –7x3 + 9x2 + 7x – 10 = k(x – 1)(x + 1)(x – 2)(x – 5) Put x = 0 on both sides 0 – 0 + 0 – 10 = k(-1)(1)(-2)(-5) – 10 = k(-10) => k = 1 Substitute k = 1 in f(x) = k(x – 1)(x + 1)(x – 2)(x – 5) f(x) = (1)(x – 1)(x + 1)(x – 2)(x – 5) = (x – 1)(x + 1)(x – 2)(x – 5) So, x4 – 7x3 + 9x2 + 7x – 10 = (x – 1)(x + 1)(x – 2)(x – 5) Question: 5 Using factor theorem, factorize of the polynomials: x4 – 2x3 – 7x2 + 8x + 12 Solution: Given, f(x) = x4 – 2x3 –7x2 + 8x + 12 The constant term f(x) is equal is 12 The factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12 Let, x + 1 = 0 => x = -1 Substitute the value of x in f(x) f(-1) = (−1)4 – 2(−1)3–7(−1)2 + 8(−1)+12 = 1 + 2 – 7 – 8 + 12 = 0 So, x + 1 is factor of f(x) Similarly, (x + 2), (x – 2), (x – 3) are also the factors of f(x) Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors. => f(x) = k(x + 1)(x + 2)(x – 3)(x – 2) => x4 – 2x3 –7x2 + 8x + 12 = k(x + 1)(x + 2)(x – 3)(x – 2) Substitute x = 0 on both sides, => 0 – 0 – 0 + 12 = k(1)(2)(- 2)(- 3) => 12 = 12K => k = 1 Substitute k = 1 in f(x) = k(x – 2)(x + 1)(x + 2)(x – 3) f(x) = (x – 2)(x + 1)(x + 2)(x – 3) so, x4 – 2x3 – 7x2 + 8x + 12 = (x – 2)(x + 1)(x + 2)(x – 3) Question: 6 Using factor theorem, factorize of the polynomials: x4 + 10x3 + 35x2 + 50x + 24 Solution: Given, f(x) = x4 + 10x3 + 35x2 + 50x + 24 The constant term in f(x) is equal to 24 The factors of 24 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24 Let, x + 1 = 0 => x = -1 Substitute the value of x in f(x) f(-1) = (-1)4 + 10(-1)3 + 35(-1)2 + 50(-1) + 24 = 1-10 + 35 - 50 + 24 = 0 => (x + 1) is the factor of f(x) Similarly, (x + 2), (x + 3), (x + 4) are also the factors of f(x) Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors. => f(x) = k(x + 1)(x + 2)(x + 3)(x + 4) => x4 + 10x3 + 35x2 + 50x + 24 = k(x + 1)(x + 2)(x + 3)(x + 4) Substitute x = 0 on both sides => 0 + 0 + 0 + 0 + 24 = k(1)(2)(3)(4) => 24 = k(24) => k = 1 Substitute k = 1 in f(x) = k(x + 1)(x + 2)(x + 3)(x + 4) f(x) = (1)(x + 1)(x + 2)(x + 3)(x + 4) f(x) = (x + 1)(x + 2)(x + 3)(x + 4) hence, x4 + 10x3 + 35x2 + 50x + 24 = (x + 1)(x + 2)(x + 3)(x + 4) Question: 7 Using factor theorem, factorize of the polynomials: 2x4–7x3–13x2 + 63x – 45 Solution: Given, f(x) = 2x4–7x3–13x2 + 63x – 45 The factors of constant term - 45 are ± 1, ± 3, ± 5, ± 9, ± 15, ± 45 The factors of the coefficient of x4 is 2. Hence possible rational roots of f(x) are ± 1, ± 3, ± 5, ± 9, ± 15, ± 45, ± 1/2, ± 3/2, ± 5/2, ± 9/2, ± 15/2, ± 45/2 Let, x – 1 = 0 => x = 1 f(1) = 2(1)4 – 7(1)3 – 13(1)2 + 63(1) – 45 = 2 – 7 – 13 + 63 – 45 = 0 Let, x – 3 = 0 => x = 3 f(3) = 2(3)4 – 7(3)3 – 13(3)2 + 63(3) - 45 = 162 – 189 – 117 + 189 – 45 = 0 So, (x – 1) and (x – 3) are the roots of f(x) => x2 – 4x + 3 is the factor of f(x) Divide f(x) with x2 – 4x + 3 to get other three factors By long division, 2x2 + x – 15 x2 – 4x + 3 2x4 – 7x3 – 13x2 + 63x – 45 2x4 – 8x3 + 6x2 (-) (+) (-) x3 – 19x2 + 63x x3 – 4x2 + 3x (-) (+) (-) -15x2 + 60x - 45 -15x2 + 60x - 45 (+) (-) (+) 0 => 2x4 – 7x3 – 13x2 + 63x – 45 = (x2– 4x + 3) (2x2+ x – 15) => 2x4 - 7x3– 13x2 + 63x – 45 = (x – 1) (x – 3) (2x2+ x – 15) Now, 2x2 + x – 15 = 2x2 + 6x – 5x –15 = 2x(x + 3) – 5 (x + 3) = (2x – 5) (x + 3) So, 2x4 – 7x3 – 13x2 + 63x – 45 = (x – 1)(x – 3)(x + 3)(2x – 5) Question: 8 Using factor theorem, factorize of the polynomials: 3x3 - x2 – 3x + 1 Solution: Given, f(x) = 3x3 - x2 – 3x + 1 The factors of constant term 1 is ± 1 The factors of the coefficient of x2 = 3 The possible rational roots are ±1, 1/3 Let, x – 1 = 0 => x = 1 f(1) = 3(1)3 - (1)2 - 3(1) + 1 = 3 – 1 – 3 + 1 = 0 So, x – 1 is the factor of f(x) Now, divide f(x) with (x – 1) to get other factors By long division method, 3x2 + 2x – 1 x – 1 3x3 – x2 – 3x + 1 3x3 – x2 (-) (+) 2x2 – 3x 2x2 – 2x (-) (+) - x + 1 - x + 1 (+) (-) 0 => 3x3- x2 – 3x + 1 = (x – 1)( 3x2 + 2x – 1) Now, 3x2 + 2x -1 = 3x2 + 3x – x - 1 = 3x(x + 1) -1(x + 1) = (3x – 1)(x + 1) Hence, 3x3- x2- 3x + 1= (x – 1) (3x – 1)(x + 1) Question: 9 Using factor theorem, factorize of the polynomials: x3- 23x2 + 142x - 120 Solution: Let, f(x) = x3- 23x2 + 142x - 120 The constant term in f(x) is -120 The factors of -120 are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 8, ± 10, ± 12, ± 15, ± 20, ± 24, ± 30, ± 40, ± 60, ± 120 Let, x - 1 = 0 => x = 1 f(1) = (1)3- 23(1)2 + 142(1) - 120 = 1 - 23 + 142 - 120 = 0 So, (x – 1) is the factor of f(x) Now, divide f(x) with (x – 1) to get other factors By long division, x2 – 22x + 120 x – 1 x3 – 23x2 + 142x – 120 x3 – x2 (-) (+) - 22x2 + 142x - 22x2 + 22x (+) (-) 120x – 120 120x – 120 (-) (+) 0 => x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120) Now, x2 – 22x + 120 = x2 – 10x – 12x + 120 = x(x – 10) – 12(x – 10) = (x – 10) (x – 12) Hence, x3 – 23x2 + 142x – 120 = (x – 1) (x – 10) (x – 12) Question: 10 Using factor theorem, factorize of the polynomials: y3 – 7y + 6 Solution: Given, f(y) = y3 – 7y + 6 The constant term in f(y) is 6 The factors are ± 1, ± 2, ± 3, ± 6 Let, y – 1 = 0 => y = 1 f(1) = (1)3 – 7(1) + 6 = 1 – 7 + 6 = 0 So, (y – 1) is the factor of f(y) Similarly, (y – 2) and (y + 3) are also the factors Since, f(y) is a polynomial which has degree 3, it cannot have more than 3 linear factors => f(y) = k(y – 1)( y – 2)(y + 3) => y3 – 7y + 6 = k(y – 1)( y – 2)(y + 3) —– 1 Substitute k = 0 in eq 1 => 0 – 0 + 6 = k(-1)(-2)(3) => 6 = 6k => k = 1 y3 – 7y + 6 = (1)(y – 1)( y – 2)(y + 3) y3 – 7y + 6 = (y – 1)( y – 2)(y + 3) Hence, y3–7y + 6 = (y – 1)( y – 2)(y + 3) Question: 11 Using factor theorem, factorize of the polynomials: x3 – 10x2 – 53x – 42 Solution: Given, f(x) = x3–10x2 – 53x – 42 The constant in f(x) is - 42 The factors of - 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42 Let, x + 1 = 0 => x = - 1 f(-1) = (−1)3 –10(−1)2 – 53(−1) – 42 = -1 – 10 + 53 – 42 = 0 So., (x + 1) is the factor of f(x) Now, divide f(x) with (x + 1) to get other factors By long division, x2 – 11x – 42 x + 1 x3 – 10x2 – 53x – 42 x3 + x2 (-) (-) -11x2 – 53x -11x2 – 11x (+) (+) - 42x – 42 - 42x – 42 (+) (+) 0 => x3 – 10x2 – 53x – 42 = (x + 1) (x2 – 11x – 42) Now, x2 – 11x – 42 = x2 – 14x + 3x – 42 = x(x – 14) + 3(x – 14) = (x + 3)(x – 14) Hence, x3 – 10x2 – 53x – 42 = (x + 1) (x + 3)(x – 14) Question: 12 Using factor theorem, factorize of the polynomials: y3 – 2y2 – 29y – 42 Solution: Given, f(x) = y3 – 2y2 – 29y – 42 The constant in f(x) is - 42 The factors of -42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42 Let, y + 2 = 0 => y = – 2 f(-2) = (−2)3 – 2(−2)2–29(−2) – 42 = -8 -8 + 58 – 42 = 0 So, (y + 2) is the factor of f(y) Now, divide f(y) with (y + 2) to get other factors By, long division y2 – 4y – 21 y + 2 y3 – 2y2 – 29y – 42 y3 + 2y2 (-) (-) - 4y2 – 29y - 4y2 – 8y (+) (+) - 21y – 42 - 21y – 42 (+) (+) 0 => y3 – 2y2 – 29y – 42 = (y + 2) (y2 – 4y – 21) Now, y2 – 4y – 21 = y2 – 7y + 3y – 21 = y(y – 7) +3(y – 7) = (y – 7)(y + 3) Hence, y3 – 2y2 – 29y – 42 = (y + 2) (y – 7)(y + 3) Question: 13 Using factor theorem, factorize of the polynomials: 2y3 – 5y2 – 19y + 42 Solution: Given, f(x) = 2y3 – 5y2 – 19y + 42 The constant in f(x) is + 42 The factors of 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42 Let, y – 2 = 0 => y = 2 f(2) = 2(2)3 – 5(2)2 – 19(2) + 42 = 16 – 20 – 38 + 42 = 0 So, (y – 2) is the factor of f(y) Now, divide f(y) with (y – 2) to get other factors By, long division method 2y2 – y – 21 y – 2 2y3 – 5y2 -19y + 42 2y3 – 4y2 (-) (+) - y2 – 19y - y2 + 2y (+) (-) - 21y + 42 - 21y + 42 (+) (-) 0 => 2y3 – 5y2 - 19y + 42 = (y – 2) (2y2 – y – 21) Now, 2y2 – y – 21 The factors are (y + 3) (2y – 7) Hence, 2y3 – 5y2 -19y + 42 = (y – 2) (y + 3) (2y – 7) Question: 14 Using factor theorem, factorize of the polynomials: x3 + 13x2 + 32x + 20 Solution: Given, f(x) = x3 + 13x2 + 32x + 20 The constant in f(x) is 20 The factors of 20 are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20 Let, x + 1 = 0 => x = -1 f(-1) = (−1)3+13(−1)2 + 32(−1) + 20 = -1 + 13 – 32 + 20 = 0 So, (x + 1) is the factor of f(x) Divide f(x) with (x + 1) to get other factors By, long division x2 + 12x + 20 x + 1, x3 + 13x2 + 32x + 20 x3 + x2 (-) (-) 12x2 + 32x 12x2 + 12x (-) (-) 20x – 20 20x – 20 (-) (-) 0 => x3 + 13x2 +32x + 20 = (x + 1)( x2 + 12x + 20) Now, x2 + 12x + 20 = x2 + 10x + 2x + 20 = x(x + 10) + 2(x + 10) The factors are (x + 10) and (x + 2) Hence, x3 + 13x2 + 32x + 20 = (x + 1)(x + 10)(x + 2) Question: 15 Using factor theorem, factorize of the polynomials: x3 – 3x2 – 9x – 5 Solution: Given, f(x) = x3 – 3x2 – 9x – 5 The constant in f(x) is -5 The factors of -5 are ±1, ±5 Let, x + 1 = 0 => x = -1 f(-1) = (−1)3 - 3(−1)2 - 9(-1) - 5 = -1 – 3 + 9 – 5 = 0 So, (x + 1) is the factor of f(x) Divide f(x) with (x + 1) to get other factors By, long division x2 – 4x – 5 x + 1 x3 – 3x2 – 9x – 5 x3 + x2 (-) (-) - 4x2 – 9x - 4x2 – 4x (+) (+) - 5x – 5 - 5x – 5 (+) (+) 0 => x3 – 3x2 – 9x – 5 = (x + 1)( x2 – 4x – 5) Now, x2 – 4x – 5 = x2 – 5x + x – 5 = x(x – 5) + 1(x – 5) The factors are (x – 5) and (x + 1) Hence, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x + 1) Question: 16 Using factor theorem, factorize of the polynomials: 2y3 + y2 – 2y – 1 Solution: Given, f(y) = 2y3 + y2 – 2y – 1 The constant term is 2 The factors of 2 are ± 1, ± 1/2 Let, y – 1= 0 => y = 1 f(1) = 2(1)3 +(1)2 – 2(1) – 1 = 2 + 1 – 2 – 1 = 0 So, (y – 1) is the factor of f(y) Divide f(y) with (y – 1) to get other factors By, long division 2y2 + 3y + 1 y – 1, 2y3 + y2 – 2y – 1 2y3 – 2y2 (-) (+) 3y2 – 2y 3y2 – 3y (-) (+) y – 1 y – 1 (-) (+) 0 => 2y3 + y2 – 2y – 1 = (y – 1) (2y2 + 3y + 1) Now, 2y2 + 3y + 1 = 2y2 + 2y + y + 1 = 2y(y + 1) + 1(y + 1) = (2y + 1) (y + 1) are the factors Hence, 2y3 + y2 – 2y – 1 = (y – 1) (2y + 1) (y + 1) Question: 17 Using factor theorem, factorize of the polynomials: x3 – 2x2 – x + 2 Solution: Let, f(x) = x3 – 2x2 – x + 2 The constant term is 2 The factors of 2 are ±1, ± 1/2 Let, x – 1= 0 => x = 1 f(1) = (1)3 – 2(1)2 – (1) + 2 = 1 – 2 – 1 + 2 = 0 So, (x – 1) is the factor of f(x) Divide f(x) with (x – 1) to get other factors By, long division x2 – x – 2 x – 1, x3 – 2x2 – y + 2 x3 – x2 (-) (+) - x2 – x - x2 + x (+) (-) – 2x + 2 – 2x + 2 (+) (-) 0 => x3 – 2x2 – y + 2 = (x – 1) (x2 – x – 2) Now, x2 – x – 2 = x2 – 2x + x – 2 = x(x – 2) + 1(x – 2) =(x – 2)(x + 1) are the factors Hence, x3 – 2x2 – y + 2 = (x – 1)(x + 1)(x – 2) Question: 18 Factorize each of the following polynomials: 1. x3 + 13x2 + 31x – 45 given that x + 9 is a factor 2. 4x3 + 20x2 + 33x + 18 given that 2x + 3 is a factor Solution: 1. x3 + 13x2 + 31x – 45 given that x + 9 is a factor let, f(x) = x3 + 13x2 + 31x – 45 given that (x + 9) is the factor divide f(x) with (x + 9) to get other factors by , long division x2 + 4x – 5 x + 9 x3 + 13x2 + 31x – 45 x3 + 9x2 (-) (-) 4x2 +31x 4x2 +36x (-) (-) -5x – 45 -5x – 45 (+) (+) 0 => x3 + 13x2 + 31x – 45 = (x + 9)( x2 + 4x – 5) Now, x2 + 4x – 5 = x2 + 5x – x – 5 = x(x + 5) -1(x + 5) = (x + 5) (x – 1) are the factors Hence, x3 + 13x2 + 31x – 45 = (x + 9)(x + 5)(x – 1) 2. 4x3 + 20x2 + 33x + 18 given that 2x + 3 is a factor let, f(x) = 4x3 + 20x2 + 33x + 18 given that 2x + 3 is a factor divide f(x) with (2x + 3) to get other factors by, long division 2x2 + 7x + 6 2x + 3, 4x3 + 20x2 + 33x + 18 4x3 + 6x2 (-) (-) 14x2 – 33x 14x2 – 21x (-) (+) 12x + 18 12x + 18 (-) (-) 0 => 4x3 + 20x2 + 33x + 18 = (2x + 3) (2x2 + 7x + 6) Now, 2x2 + 7x + 6 = 2x2 + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = (2x + 3)(x + 2) are the factors Hence, 4x3 + 20x2 + 33x + 18 = (2x + 3)(2x + 3)(x + 2)
Using factor theorem, factorize of the polynomials:
x3 + 6x2 + 11x + 6
Given polynomial, f(x) = x3 + 6x2 + 11x + 6
The constant term in f(x) is 6
The factors of 6 are ± 1, ± 2, ± 3, ± 6
Let, x + 1 = 0
=> x = -1
Substitute the value of x in f(x)
f(-1) = (−1)3 + 6(−1)2 + 11(−1) + 6
= - 1 + 6 - 11 + 6
= 12 – 12
= 0
So, (x + 1) is the factor of f(x)
Similarly, (x + 2) and (x + 3) are also the factors of f(x)
Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.
∴ f(x) = k(x + 1)(x + 2)(x + 3)
=> x3 + 6x2 + 11x + 6 = k(x + 1)(x + 2)(x + 3)
Substitute x = 0 on both the sides
=> 0 + 0 + 0 + 6 = k(0 +1)(0 + 2)(0 + 3)
=> 6 = k(1*2*3)
=> 6 = 6k
=> k = 1
Substitute k value in f(x) = k(x + 1)(x + 2)(x + 3)
=> f(x) = (1)(x + 1)(x + 2)(x + 3)
=> f(x) = (x + 1)(x + 2)(x + 3)
∴ x3 + 6x2 + 11x + 6 = (x + 1)(x + 2)(x + 3)
x3 + 2x2 – x – 2
Given, f(x) = x3+ 2x2 – x – 2
The constant term in f(x) is -2
The factors of (-2) are ±1, ± 2
Let, x – 1 = 0
=> x = 1
f(1) = (1)3 + 2(1)2 – 1 – 2
= 1 + 2 – 1 – 2
Similarly, the other factors (x + 1) and (x + 2) of f(x)
∴ f(x) = k(x – 1)(x + 2)(x + 1 )
x3 + 2x2 – x – 2 = k(x – 1)(x + 2)(x + 1 )
0 + 0 – 0 – 2 = k(-1)(1)(2)
=> – 2 = - 2k
Substitute k value in f(x) = k(x – 1)(x + 2)(x + 1)
f(x) = (1)(x – 1)(x + 2)(x + 1)
=> f(x) = (x – 1)(x + 2)(x + 1)
So, x3 + 2x2 – x – 2 = (x – 1)(x + 2)(x + 1)
x3 – 6x2 + 3x + 10
Let, f(x) = x3 – 6x2 + 3x + 10
The constant term in f(x) is 10
The factors of 10 are ± 1, ± 2, ± 5, ± 10
f(-1) = (−1)3- 6(−1)2 + 3(−1) + 10
= -1 – 6 – 3 + 10
Similarly, the other factors (x – 2) and (x – 5) of f(x)
∴ f(x) = k(x + 1)(x – 2)(x – 5)
Substitute x = 0 on both sides
=> x3– 6x2 + 3x + 10 = k(x + 1)(x – 2)(x – 5)
=> 0 – 0 + 0 + 10 = k(1)(-2)(-5)
=> 10 = k(10)
Substitute k = 1 in f(x) = k(x + 1)(x – 2)(x – 5)
f(x) = (1)(x + 1)(x – 2)(x – 5 )
so, x3 – 6x2 + 3x + 10 = (x + 1)(x – 2)(x – 5)
x4 –7x3 + 9x2 + 7x –10
Given, f(x) = x4–7x3 + 9x2 + 7x – 10
The factors of 10 are ± 1, ± 2, ± 5, ±10
f(x) = 14 – 7(1)3 + 9(1)2 + 7(1) – 10
= 1 – 7 + 9 + 7 – 10
= 10 – 10
(x – 1) is the factor of f(x)
Similarly, the other factors are (x + 1), (x – 2), (x – 5)
Since, f(x) is a polynomial of degree 4. So, it cannot have more than four linear factor.
So, f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)
=> x4 –7x3 + 9x2 + 7x – 10 = k(x – 1)(x + 1)(x – 2)(x – 5)
Put x = 0 on both sides
0 – 0 + 0 – 10 = k(-1)(1)(-2)(-5)
– 10 = k(-10)
Substitute k = 1 in f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)
f(x) = (1)(x – 1)(x + 1)(x – 2)(x – 5)
= (x – 1)(x + 1)(x – 2)(x – 5)
So, x4 – 7x3 + 9x2 + 7x – 10
x4 – 2x3 – 7x2 + 8x + 12
Given,
f(x) = x4 – 2x3 –7x2 + 8x + 12
The constant term f(x) is equal is 12
The factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12
f(-1) = (−1)4 – 2(−1)3–7(−1)2 + 8(−1)+12
= 1 + 2 – 7 – 8 + 12
So, x + 1 is factor of f(x)
Similarly, (x + 2), (x – 2), (x – 3) are also the factors of f(x)
Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors.
=> f(x) = k(x + 1)(x + 2)(x – 3)(x – 2)
=> x4 – 2x3 –7x2 + 8x + 12 = k(x + 1)(x + 2)(x – 3)(x – 2)
Substitute x = 0 on both sides,
=> 0 – 0 – 0 + 12 = k(1)(2)(- 2)(- 3)
=> 12 = 12K
Substitute k = 1 in f(x) = k(x – 2)(x + 1)(x + 2)(x – 3)
f(x) = (x – 2)(x + 1)(x + 2)(x – 3)
so, x4 – 2x3 – 7x2 + 8x + 12 = (x – 2)(x + 1)(x + 2)(x – 3)
x4 + 10x3 + 35x2 + 50x + 24
Given, f(x) = x4 + 10x3 + 35x2 + 50x + 24
The constant term in f(x) is equal to 24
The factors of 24 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24
f(-1) = (-1)4 + 10(-1)3 + 35(-1)2 + 50(-1) + 24
= 1-10 + 35 - 50 + 24
=> (x + 1) is the factor of f(x)
Similarly, (x + 2), (x + 3), (x + 4) are also the factors of f(x)
=> f(x) = k(x + 1)(x + 2)(x + 3)(x + 4)
=> x4 + 10x3 + 35x2 + 50x + 24 = k(x + 1)(x + 2)(x + 3)(x + 4)
=> 0 + 0 + 0 + 0 + 24 = k(1)(2)(3)(4)
=> 24 = k(24)
Substitute k = 1 in f(x) = k(x + 1)(x + 2)(x + 3)(x + 4)
f(x) = (1)(x + 1)(x + 2)(x + 3)(x + 4)
f(x) = (x + 1)(x + 2)(x + 3)(x + 4)
hence, x4 + 10x3 + 35x2 + 50x + 24 = (x + 1)(x + 2)(x + 3)(x + 4)
2x4–7x3–13x2 + 63x – 45
Given, f(x) = 2x4–7x3–13x2 + 63x – 45
The factors of constant term - 45 are ± 1, ± 3, ± 5, ± 9, ± 15, ± 45
The factors of the coefficient of x4 is 2. Hence possible rational roots of f(x) are
± 1, ± 3, ± 5, ± 9, ± 15, ± 45, ± 1/2, ± 3/2, ± 5/2, ± 9/2, ± 15/2, ± 45/2
f(1) = 2(1)4 – 7(1)3 – 13(1)2 + 63(1) – 45
= 2 – 7 – 13 + 63 – 45
Let, x – 3 = 0
=> x = 3
f(3) = 2(3)4 – 7(3)3 – 13(3)2 + 63(3) - 45
= 162 – 189 – 117 + 189 – 45
So, (x – 1) and (x – 3) are the roots of f(x)
=> x2 – 4x + 3 is the factor of f(x)
Divide f(x) with x2 – 4x + 3 to get other three factors
By long division,
2x2 + x – 15
x2 – 4x + 3 2x4 – 7x3 – 13x2 + 63x – 45
2x4 – 8x3 + 6x2
(-) (+) (-)
x3 – 19x2 + 63x
x3 – 4x2 + 3x
-15x2 + 60x - 45
(+) (-) (+)
0
=> 2x4 – 7x3 – 13x2 + 63x – 45 = (x2– 4x + 3) (2x2+ x – 15)
=> 2x4 - 7x3– 13x2 + 63x – 45 = (x – 1) (x – 3) (2x2+ x – 15)
Now,
2x2 + x – 15 = 2x2 + 6x – 5x –15
= 2x(x + 3) – 5 (x + 3)
= (2x – 5) (x + 3)
So, 2x4 – 7x3 – 13x2 + 63x – 45 = (x – 1)(x – 3)(x + 3)(2x – 5)
3x3 - x2 – 3x + 1
Given, f(x) = 3x3 - x2 – 3x + 1
The factors of constant term 1 is ± 1
The factors of the coefficient of x2 = 3
The possible rational roots are ±1, 1/3
f(1) = 3(1)3 - (1)2 - 3(1) + 1
= 3 – 1 – 3 + 1
So, x – 1 is the factor of f(x)
Now, divide f(x) with (x – 1) to get other factors
By long division method,
3x2 + 2x – 1
x – 1 3x3 – x2 – 3x + 1
3x3 – x2
(-) (+)
2x2 – 3x
2x2 – 2x
- x + 1
(+) (-)
=> 3x3- x2 – 3x + 1 = (x – 1)( 3x2 + 2x – 1)
3x2 + 2x -1 = 3x2 + 3x – x - 1
= 3x(x + 1) -1(x + 1)
= (3x – 1)(x + 1)
Hence, 3x3- x2- 3x + 1= (x – 1) (3x – 1)(x + 1)
x3- 23x2 + 142x - 120
Let, f(x) = x3- 23x2 + 142x - 120
The constant term in f(x) is -120
The factors of -120 are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 8, ± 10, ± 12, ± 15, ± 20, ± 24, ± 30, ± 40, ± 60, ± 120
Let, x - 1 = 0
f(1) = (1)3- 23(1)2 + 142(1) - 120
= 1 - 23 + 142 - 120
So, (x – 1) is the factor of f(x)
x2 – 22x + 120
x – 1 x3 – 23x2 + 142x – 120
x3 – x2
- 22x2 + 142x
- 22x2 + 22x
120x – 120
=> x3 – 23x2 + 142x – 120 = (x – 1) (x2 – 22x + 120)
x2 – 22x + 120 = x2 – 10x – 12x + 120
= x(x – 10) – 12(x – 10)
= (x – 10) (x – 12)
Hence, x3 – 23x2 + 142x – 120 = (x – 1) (x – 10) (x – 12)
y3 – 7y + 6
Given, f(y) = y3 – 7y + 6
The constant term in f(y) is 6
The factors are ± 1, ± 2, ± 3, ± 6
Let, y – 1 = 0
=> y = 1
f(1) = (1)3 – 7(1) + 6
= 1 – 7 + 6
So, (y – 1) is the factor of f(y)
Similarly, (y – 2) and (y + 3) are also the factors
Since, f(y) is a polynomial which has degree 3, it cannot have more than 3 linear factors
=> f(y) = k(y – 1)( y – 2)(y + 3)
=> y3 – 7y + 6 = k(y – 1)( y – 2)(y + 3) —– 1
Substitute k = 0 in eq 1
=> 0 – 0 + 6 = k(-1)(-2)(3)
y3 – 7y + 6 = (1)(y – 1)( y – 2)(y + 3)
y3 – 7y + 6 = (y – 1)( y – 2)(y + 3)
Hence, y3–7y + 6 = (y – 1)( y – 2)(y + 3)
x3 – 10x2 – 53x – 42
f(x) = x3–10x2 – 53x – 42
The constant in f(x) is - 42
The factors of - 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42
=> x = - 1
f(-1) = (−1)3 –10(−1)2 – 53(−1) – 42
= -1 – 10 + 53 – 42
So., (x + 1) is the factor of f(x)
Now, divide f(x) with (x + 1) to get other factors
x2 – 11x – 42
x + 1 x3 – 10x2 – 53x – 42
x3 + x2
(-) (-)
-11x2 – 53x
-11x2 – 11x
(+) (+)
- 42x – 42
=> x3 – 10x2 – 53x – 42 = (x + 1) (x2 – 11x – 42)
x2 – 11x – 42 = x2 – 14x + 3x – 42
= x(x – 14) + 3(x – 14)
= (x + 3)(x – 14)
Hence, x3 – 10x2 – 53x – 42 = (x + 1) (x + 3)(x – 14)
y3 – 2y2 – 29y – 42
Given, f(x) = y3 – 2y2 – 29y – 42
The factors of -42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42
Let, y + 2 = 0
=> y = – 2
f(-2) = (−2)3 – 2(−2)2–29(−2) – 42
= -8 -8 + 58 – 42
So, (y + 2) is the factor of f(y)
Now, divide f(y) with (y + 2) to get other factors
By, long division
y2 – 4y – 21
y + 2 y3 – 2y2 – 29y – 42
y3 + 2y2
- 4y2 – 29y
- 4y2 – 8y
- 21y – 42
=> y3 – 2y2 – 29y – 42 = (y + 2) (y2 – 4y – 21)
y2 – 4y – 21 = y2 – 7y + 3y – 21
= y(y – 7) +3(y – 7)
= (y – 7)(y + 3)
Hence, y3 – 2y2 – 29y – 42 = (y + 2) (y – 7)(y + 3)
2y3 – 5y2 – 19y + 42
Given, f(x) = 2y3 – 5y2 – 19y + 42
The constant in f(x) is + 42
The factors of 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42
Let, y – 2 = 0
=> y = 2
f(2) = 2(2)3 – 5(2)2 – 19(2) + 42
= 16 – 20 – 38 + 42
So, (y – 2) is the factor of f(y)
Now, divide f(y) with (y – 2) to get other factors
By, long division method
2y2 – y – 21
y – 2 2y3 – 5y2 -19y + 42
2y3 – 4y2
- y2 – 19y
- y2 + 2y
- 21y + 42
=> 2y3 – 5y2 - 19y + 42 = (y – 2) (2y2 – y – 21)
The factors are (y + 3) (2y – 7)
Hence, 2y3 – 5y2 -19y + 42 = (y – 2) (y + 3) (2y – 7)
x3 + 13x2 + 32x + 20
Given, f(x) = x3 + 13x2 + 32x + 20
The constant in f(x) is 20
The factors of 20 are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20
f(-1) = (−1)3+13(−1)2 + 32(−1) + 20
= -1 + 13 – 32 + 20
Divide f(x) with (x + 1) to get other factors
x2 + 12x + 20
x + 1, x3 + 13x2 + 32x + 20
12x2 + 32x
12x2 + 12x
20x – 20
=> x3 + 13x2 +32x + 20 = (x + 1)( x2 + 12x + 20)
x2 + 12x + 20 = x2 + 10x + 2x + 20
= x(x + 10) + 2(x + 10)
The factors are (x + 10) and (x + 2)
Hence, x3 + 13x2 + 32x + 20 = (x + 1)(x + 10)(x + 2)
x3 – 3x2 – 9x – 5
Given, f(x) = x3 – 3x2 – 9x – 5
The constant in f(x) is -5
The factors of -5 are ±1, ±5
f(-1) = (−1)3 - 3(−1)2 - 9(-1) - 5
= -1 – 3 + 9 – 5
x2 – 4x – 5
x + 1 x3 – 3x2 – 9x – 5
- 4x2 – 9x
- 4x2 – 4x
- 5x – 5
=> x3 – 3x2 – 9x – 5 = (x + 1)( x2 – 4x – 5)
x2 – 4x – 5 = x2 – 5x + x – 5
= x(x – 5) + 1(x – 5)
The factors are (x – 5) and (x + 1)
Hence, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x + 1)
2y3 + y2 – 2y – 1
Given, f(y) = 2y3 + y2 – 2y – 1
The constant term is 2
The factors of 2 are ± 1, ± 1/2
Let, y – 1= 0
f(1) = 2(1)3 +(1)2 – 2(1) – 1
= 2 + 1 – 2 – 1
Divide f(y) with (y – 1) to get other factors
2y2 + 3y + 1
y – 1, 2y3 + y2 – 2y – 1
2y3 – 2y2
3y2 – 2y
3y2 – 3y
y – 1
=> 2y3 + y2 – 2y – 1 = (y – 1) (2y2 + 3y + 1)
2y2 + 3y + 1 = 2y2 + 2y + y + 1
= 2y(y + 1) + 1(y + 1)
= (2y + 1) (y + 1) are the factors
Hence, 2y3 + y2 – 2y – 1 = (y – 1) (2y + 1) (y + 1)
x3 – 2x2 – x + 2
Let, f(x) = x3 – 2x2 – x + 2
The factors of 2 are ±1, ± 1/2
Let, x – 1= 0
f(1) = (1)3 – 2(1)2 – (1) + 2
= 1 – 2 – 1 + 2
Divide f(x) with (x – 1) to get other factors
x2 – x – 2
x – 1, x3 – 2x2 – y + 2
- x2 – x
- x2 + x
– 2x + 2
=> x3 – 2x2 – y + 2 = (x – 1) (x2 – x – 2)
x2 – x – 2 = x2 – 2x + x – 2
= x(x – 2) + 1(x – 2)
=(x – 2)(x + 1) are the factors
Hence, x3 – 2x2 – y + 2 = (x – 1)(x + 1)(x – 2)
Factorize each of the following polynomials:
1. x3 + 13x2 + 31x – 45 given that x + 9 is a factor
2. 4x3 + 20x2 + 33x + 18 given that 2x + 3 is a factor
let, f(x) = x3 + 13x2 + 31x – 45
given that (x + 9) is the factor
divide f(x) with (x + 9) to get other factors
by , long division
x2 + 4x – 5
x + 9 x3 + 13x2 + 31x – 45
x3 + 9x2
4x2 +31x
4x2 +36x
-5x – 45
=> x3 + 13x2 + 31x – 45 = (x + 9)( x2 + 4x – 5)
x2 + 4x – 5 = x2 + 5x – x – 5
= x(x + 5) -1(x + 5)
= (x + 5) (x – 1) are the factors
Hence, x3 + 13x2 + 31x – 45 = (x + 9)(x + 5)(x – 1)
let, f(x) = 4x3 + 20x2 + 33x + 18
given that 2x + 3 is a factor
divide f(x) with (2x + 3) to get other factors
by, long division
2x2 + 7x + 6
2x + 3, 4x3 + 20x2 + 33x + 18
4x3 + 6x2
14x2 – 33x
14x2 – 21x
12x + 18
=> 4x3 + 20x2 + 33x + 18 = (2x + 3) (2x2 + 7x + 6)
2x2 + 7x + 6 = 2x2 + 4x + 3x + 6
= 2x(x + 2) + 3(x + 2)
= (2x + 3)(x + 2) are the factors
Hence, 4x3 + 20x2 + 33x + 18 = (2x + 3)(2x + 3)(x + 2)
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Chapter 6: Factorization of Polynomials Exercise...