For x + 1 = 0, we have x = –1. ∴ The zero of x + 1 is –1. (i) p(x) = x3 + x2 + x + 1 ∴ p(–1) = (–1)3 + (–1)2 + (–1) + 1 = –1 + 1 – 1 + 1 = 0 i.e. when p(x) is divided by (x + 1), then the remainder is zero. ∴ (x + 1) is a factor of x3 + x2 + x + 1. (ii) ∵ p(x) = x4 + x3 + x2 + x + 1 ∴ p(–1) = (–1)4 + (–1)3 + (–1)2 + (–1) + 1 = 1 – 1 + 1 – 1 + 1 = 3 – 2 = 1 ∵ f(–1) ≠ 0 ∴ p(x) is not divisible by x + 1. i.e. (x + 1) is not a factor of x4 + x3 + x2 + x + 1. (iii) ∵ p(x) = x4 + 3x3 + 3x2 + x + 1 ∴ p(–1) = (–1)4 + 3(–1)3 + 3(–1)2 + (–1) + 1 = (1) + 3(–1) + 3(1) + (–1) + 1 = 1 – 3 + 3 – 1 + 1 = 1 ≠ 0 ∵ f(–1) ≠ 0 ∴ (x + 1) is not a factor of x4 + 3x3 + 3x2 + x + 1. (iv) ∵ p(x) = x3 – x2 – (2 + √2) x + √2) ∴ p(–1) = (–1)3 – (–1)2 – (2 + 2) (–1) + 2 = –1 – 1 – (–1) (2 + √2) + 2 = –1 – 1 + 1 (2 + √2) + 2 = –1 – 1 + 2 + √2 + √2) = –2 + 2 + 2√2 = 2√2 ≠ 0 Since p(–1) ≠ 0. ∴ (x + 1) is not a factor of x4 + 3x3 + 3x2 + x + 1. 2: Use the factor theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3 + x2 – 2x – 1, g(x) = x + 1 (ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2 (iii) p(x) = x3 – 4x2 + x + 6, g(x) = x – 3 (i) We have p(x) = 2x3 + x2 – 2x – 1 and g(x) = x + 1 ∴ p(–1) = 2(–1)3 + (–1)2 – 2(–1) –1 = 2(–1) + 1 + 2 – 1 = –2 + 1 + 2 – 1 = –3 + 3 = 0 ∵ p(–1) = 0 ∴ g(x) is a factor of p(x). (ii) We have p(x) = x3 + 3x2 + 3x + 1 and g(x) = x + 2 ∴ p(–2) = (–2)3 + 3(–2)2 + 3(–2) + 1 = –8 + 3(4) + (–6) + 1 = –8 + 12 – 6 + 1 = –8 – 6 + 12 + 1 = –14 + 13 = –1 ∴ p(–2) ≠ 0 Thus, g(x) is not a factor of p(x). (iii) We have p(x) = x3 – 4x2 + x + 6 and g(x) = x – 3 ∴ p(3) = (3)3 – 4 (3)2 + (3) + 6 = 27 –4(9) + 3 + 6 = 27 – 36 + 3 + 6 = 0 Since g(x) = 0 ∴ g(x) is a factor of p(x). 3: Find the value of k, if x – 1 is a factor of p(x) in each of the following cases: (i) p(x) = x2 + x + k (ii) p(x) = 2x2 + kx + √2 (iii) p(x) = kx2 – √2 x + 1 (iv) p(x) = kx2 – 3x + k (i) Here p(x) = x2 + x + k For x – 1 be a factor of p(x), p(1) should be equal to 0. We have p(1) = (1)2 + 1 + k or p(1) = 1 + 1 + k = k + 2 ∴ k + 2 = 0 ⇒ k= –2 (ii) Here, p(x) = 2x2 + kx + √2 For x – 1 be a factor of p(x), p(1) = 0 Since, p(1) = 2(1)2 + k(1) + √2 = 2 + k + √2 ∵ p(1) must be equal to 0. ∴ k + 2 + √2 = 0 ⇒ k = –2 – √2 or k = – (2 + √2) . (iii) Here p(x) = kx2 – √2 x + 1 and g(x) = x – 1 ∴ For (x – 1) be a factor of p(x), p(1) should be equal to 0. Since p(1) = k(1)2 – √2 (1) + 1 or p(1) = k – √2 + 1 or p(1) = k – √2 + 1 ∴ k – √2 +1 = 0 ⇒ k= √2 – 1 (iv) Here p(x) = kx2 – 3x + k and g(x) = x – 1 For g(x) be a factor of p(x), p(1) should be equal to 0. Since p(1) = k(1)2 – 3(1) + k = k – 3 + k = 2k – 3 ∴ 2k – 3 = 0 ⇒ k = 3/2 4: Factorise: (i) 12x2 – 7x + 1 (ii) 2x2 + 7x + 3 (iii) 6x2 + 5x – 6 (iv) 3x2 – x – 4 Ans: (i) 12x2 – 7x + 1 Here co-efficient of x2 = 12 Co-efficient of x = –7 and constant term = 1 ∴ a = 12, b = –7, c = 1 Now, l + m = –7 and lm = ac = 12 x 1 ∴ We have l = (–4) and m = (–3) i.e. b = –7 = (–4 – 3). Now, 12x2 – 7x + 1 = 12x2 – 4x – 3x + 1 = 4x(3x – 1) – 1(3x – 1) = (3x – 1)(4x – 1) Thus, 12x2 – 7x + 1 = (3x – 1)(4x – 1) (ii) 2x2 + 7x + 3 Here, a = 2, b = 7 and c = 3 ∴ l + m = 7 and lm = 2 x 3 = 6 i.e. 1 + 6 = 7 and 1 x 6 = 6 ∴ l = 1 and m = 6 We have 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) <!--[if !supportLineBreakNewLine]-->= (2x + 1)(x + 3) Thus, 2x2 + 7x + 3 = (2x + 1)(x + 3) (iii) 6x2 + 5x – 6 We have a = 6, b = 5 and c = –6 ∴ l + m = 5 and lm = ac = 6 x (–6) = –36 ∴ l + m = 9 + (–4) ∴ 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6 = 3x(2x + 3) – 2(2x + 3) = (2x + 3)(3x – 2) Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2) (iv)3x2 – x – 4 We have a = 3, b = –1 and c = –4 ∴ l + m = –1 and lm = 3 x (–4) = –12 ∴ l = –4 and m = 3 Now, 3x2 – x – 4 = 3x2 – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1) Thus, 3x2 – x – 4 = (3x – 4)(x + 1) 5: Factorise: (i) x3 – 2x2 – x + 2 (ii) x3 – 3x2 – 9x – 5 (iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 – 2y – 1 (i) x3 – 2x2 – x + 2 Rearranging the terms, we have x3 – 2x2 – x + 2 = x3 – x – 2x2 + 2 = x(x2 – 1) – 2(x2 – 1) = (x2 – 1)(x – 2) = [(x)2 – (1)2] [x – 2] = (x – 1)(x + 1)(x – 2) [∵ a2 – b2 = (a + b)(a – b)] Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2) (ii) x3 – 3x2 – 9x – 5 We have p(x) = x3 – 3x2 – 9x – 5 By trial, let us find: p(1) = (1)3 – 3(1)2 – 9(1) – 5 = 3 – 3 – 9 – 5 = –14 ≠ 0 Now p(–1) = (–1)3 – 3(–1)2 – 9(–1) –5 = –1 – 3(1) + 9 – 5 = –1 – 3 + 9 – 5 = 0 ∴ By factor theorem, [x – (–1)] is a factor of p(x). ∴ x2 – 3x2 – 9x – 5 = (x + 1)(x2 – 4x – 5) = (x + 1)[x2 – 5x + x – 5] [Splitting –4 into –5 and +1] = (x + 1) [x(x – 5) + 1(x – 5)] = (x + 1) [(x – 5) (x + 1)] = (x + 1)(x – 5)(x + 1) (iii) x3 + 13x2 + 32x + 20 We have p(x) = x3 + 13x2 + 32x + 20 By trial, let us find: p(1) = (1)3 + 13(1)2 + 32(1) + 20 = 1 + 13 + 32 + 20 = 66 ≠ 0 Now p(–1) = (–1)3 + 13(–1)2 + 32(–1) + 20 = –1 + 13 – 32 + 20 = 0 ∴ By factor theorem, [x – (–1)], i.e. (x + 1) is a factor p(x). or x3 + 13x2 + 32x + 20 = (x + 1)(x2 + 12x + 20) = (x + 1)[x2 + 2x + 10x + 20] [Splitting the middle term] = (x + 1)[x(x + 2) + 10(x + 2)] = (x + 1)[(x + 2)(x + 10)] = (x + 1)(x + 2)(x + 10) (iv) 2y3 + y2 – 2y – 1 We have p(y) = 2y3 + y2 – 2y – 1 By trial, we have p(1) = 2(1)3 + (1)2 – 2(1) – 1 = 2(1) + 1 – 2 – 1 = 2 + 1 – 2 – 1 = 0 ∴ By factor theorem, (y – 1) is a factor of p(y). ∴ 2y3 – y2 – 2y – 1 = (y – 1)(2y2 + 3y + 1) = (y – 1)[2y2 + 2y + y + 1] [Splitting the middle term] = (y – 1)[2y(y + 1) + 1(y + 1)] = (y – 1)[(y + 1)(2y + 1)] = (y – 1)(y + 1)(2y + 1)