Factorise :
(i) x³ - 2x² - x + 2
(ii) x³ - 3x² - 9x - 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² - 2y - 1

Question

Factorise : (i) x³ - 2x² - x + 2 (ii) x³ - 3x² - 9x - 5 (iii) x³ + 13x² + 32x + 20 (iv) 2y³ + y² - 2y - 1

KnowledgeBoat · Accepted Answer

(i) x³ - 2x² - x + 2 Let (x - a) is a factor, a = 1, -1, 2, -2....... Putting a = 1 x - 1 = 0 x = 1 p(x) = x³ - 2x² - x + 2 p(1) = (1)³ - 2 x (1)² - 1 + 2 = 1 - 2 - 1 + 2 = 0 Hence, (x - 1) is the factor of x³ - 2x² - x + 2 On dividing, x³ - 2x² - x + 2 by (x - 1) we get quotient = x² - x - 2 Factorising x² - x - 2, = x² - x - 2 = x² -(2 - 1)x - 2 = x² - 2x + x -2 = x( x - 2) + 1(x + 2) = (x + 1)(x - 2) ∴ x² - x - 2 = (x + 1)(x - 2) ∴ p(x) = x³ - 2x² - x + 2 = (x - 1) (x + 1) (x - 2) So, factors are (x - 1) (x + 1) (x - 2) (ii) x³ - 3x² - 9x - 5 Let (x - a) is a factor, a = 1, -1, 2, -2....... Putting a = -1 x + 1 = 0 x = -1 p(x) = x³ - 3x² - 9x - 5 p(-1) = (-1)³ - 3 x (-1)² - 9 x (-1) - 5 = -1 - 3 + 9 - 5 = -9 + 9 = 0 Hence, (x + 1) is a factor of x³ - 3x² - 9x - 5 On dividing, x³ - 3x² - 9x - 5 by (x + 1) We get quotient = x² - 4x - 5 Factorising x² - 4x - 5 x² -(5 - 1)x - 5 x² -5x + x - 5 x(x - 5) + 1(x - 5) (x + 1) (x - 5) ∴ x² - 4x - 5 = (x + 1) (x - 5) ∴ p(x) = x³ - 3x² -9x - 5 = (x + 1) (x + 1) (x - 5) So, factors are (x + 1) (x + 1) (x - 5) (iii) x³ + 13x² + 32x + 20 Let (x - a) is a factor, a = 1, -1, 2, -2....... Putting a = -1 x + 1 = 0 x = -1 p(x) = x³ + 13x² + 32x + 20 p(-1) = (-1)³ + 13 x (-1)² + 32 x (-1) + 20 = -1 + 13 - 32 + 20 = -33 + 33 = 0 Hence, (x + 1) is a factor of x³ + 13x² + 32x + 20 On dividing, x³ + 13x² + 32x + 20 by (x + 1) We get quotient = x² + 12x + 20 Factorising x² + 12x + 20 x² + 10x + 2x + 20 x(x + 10) + 2(x + 10) (x + 10)(x + 2) ∴ x² + 12x + 20 = (x + 10)(x + 2) ∴ p(x) = x³ + 13x² + 32x + 20 = (x + 1) (x + 10) (x + 2) So, factors are (x + 1) (x + 10) (x + 2) (iv) 2y³ + y² - 2y - 1 Let (y - a) is a factor, a = 1, -1, 2, -2....... Putting a = 1 y - 1 = 0 y = 1 p(y) = 2y³ + y² - 2y - 1 p(1) = 2 x (1)³ + (1)² - 2 x 1 - 1 = 2 + 1 - 2 - 1 = 0 Hence, (y - 1) is a factor of 2y³ + y² - 2y - 1 On dividing, 2y³ + y² - 2y - 1 by (y - 1) We get quotient = 2y² + 3y + 1 Factorising 2y² + 3y + 1 2y² + 2y + y +1 2y(y + 1) + 1(y + 1) (y + 1)(2y + 1) ∴ 2y² + 3y + 1 = (y + 1)(2y + 1) ∴ p(y) = 2y³ + y² - 2y - 1 = (y - 1)(y + 1)(2y + 1) So, factors are (y - 1)(y + 1)(2y + 1)

Mathematics

Factorise :
(i) x³ - 2x² - x + 2
(ii) x³ - 3x² - 9x - 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² - 2y - 1

Polynomials

Related Questions

Factorise :
(i) 12x² - 7x + 1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x - 6
(iv) 3x² - x - 4

Evaluate the following products without multiplying directly:
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96

Mathematics

Factorise :(i) x3 - 2x2 - x + 2(ii) x3 - 3x2 - 9x - 5(iii) x3 + 13x2 + 32x + 20(iv) 2y3 + y2 - 2y - 1

Polynomials

Related Questions

Factorise :(i) 12x2 - 7x + 1(ii) 2x2 + 7x + 3(iii) 6x2 + 5x - 6(iv) 3x2 - x - 4

Evaluate the following products without multiplying directly:(i) 103 × 107(ii) 95 × 96(iii) 104 × 96

Factorise :
(i) x³ - 2x² - x + 2
(ii) x³ - 3x² - 9x - 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² - 2y - 1

Factorise :
(i) 12x² - 7x + 1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x - 6
(iv) 3x² - x - 4

Evaluate the following products without multiplying directly:
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96