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

(i) Given x³ - x² - x + 2
= x²(x - 1) - 1(x - 2) [taking x² and -1 as common]
= (x - 1)(x² - 1) [Taking (x - 1) as common]
= (x - 1)(x + 1)(x - 1) [since, a² - b² = (a - b)(a + b)]
= (x - 1)²(x + 1)
(ii) Given x³ - x² - x - 1
Here -x² = -2x² + x² = x³ - 2x² + x² - x - 1
= x²(x - 2) + x(x - 1) - 1 (x + 1)
= x²(x - 1) - 1(x + 1)
= (x - 1)(x² - 1)
= (x - 1)(x + 1)(x -1)
(iii) Given x³ + 13x² + 32x + 20
Here, 13x² = 2x² + 11x², 32x = 22x + 10x
Thus, x³ + 13x² + 32x + 20 = x³ + 2x² + 11x² + 22x + 10x + 20
= x²(x + 2) + 11x(x + 2) + 10(x + 2) [taking x², 11x and 10 as common]
= (x + 2)(x² + 11x + 10)
= (x + 2)(x² + x + 10x + 10) [Here 11x = 10x + 1x]
= (x + 2)[x(x + 1) + 10(x + 1)] [Taking x, 10 as a common]
= (x + 2)(x + 1)(x + 10)
(iv) Given 2y³ + y² - y - 1
Here, y² = 2y² - y² and -y = -y - y
Thus, 2y³ + y² - y - 1 = 2y³ + 2y² - y² - y - 1
= 2y²(y + 1) - y(y + 1) - 1(y + 1)
= (y + 1)(2y² - y - 1) [Taking 2y², -y and -1 as common]
= (y + 1)(2y² - 2y + y - 1) [Here -y = -2y + y]
= (y + 1)[2y(y - 1) + 1(y - 1)] [Taking 2y and 1 as a common]
= (y + 1)(y - 1)(2y + 1)