1. Find the value of the polynomial 5x – 4x2 + 3 at (i) x = 0 (ii) x = –1 (iii) x = 2 (i) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3 ∴ p(0) = 5(0) – 4(0) + 3 = 0 – 0 + 3 = 3 Thus, the value of 5x – 4x2 + 3 at x = 0 is 3. (ii) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3 ∴ p(–1) = 5(–1) – 4(–1)2 + 3 = –5 – 4(1) + 3 = –5 – 4 + 3 = –9 + 3 = –6 ∴ The value of 5x – 4x2 + 3 at x = –1 is –6. (iii) ∵ p(x) = 5x – 4x2 + 3 = 5(x) – 4(x)2 + 3 ∴ p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3 = 10 – 16 + 3 = –3 Thus the value of 5x – 4x2 + 3 at x = 2 is –3. 2. Find p(0), p(1) and p(2) for each of the following polynomials: (i) p(y) = y2 – y + 1 (ii) p(t) = 2 + t + 2t2 – t3 (iii) p(x) = x3 (iv) p(x) = (x – 1) (x + 1) (i) p(y) = y2 – y + 1 ∵ p(y) = y2 – y + 1 = (y)2 – y + 1 ∴ p(0) = (0)2 – (0) + 1 = 0 – 0 + 1 = 1 p(1) = (1)2 – (1) + 1 = 1 – 1 + 1 = 1 p(2) = (2)2 –2 + 1 = 4 – 2 + 1 = 3 (ii) p(t) = 2 + t + 2t2 – t3 ∵ p(t) = 2 + t + 2t2 – t3 = 2 + t + 2(t)2 – (t)3 ∴ p(0) = 2 + (0) + 2(0)2 – (0)3 = 2 + 0 + 0 – 0 = 2 p(1) = 2 + (1) + 2 (1)2 – (1)3 = 2 + 1 + 2 – 1 = 4 p(2) = 2 + 2 + 2(2)2 – (2)3 = 2 + 2 + 8 – 8 = 4 (iii) p(x) = x3 ∵ p(x) = x3 = (x)3 ∴ p(0) = (0)3 = 0 p(1) = (1)3 = 1 p(2) = (2)3 = 8 [∵ 2x2 x2 = 8] (iv) p(x) = (x – 1)(x + 1) ∵ p(x) = (x – 1)(x + 1) ∴ p(0) = (0 – 1)(0 + 1) = –1 x 1 = –1 p(1) = (1 – 1)(1 + 1) = (0)(2) = 0 p(2) = (2 – 1)(2 + 1) = (1)(3) = 3 3. Verify whether the following are zeros of the polynomial, indicated against them. (iii) Since, p(x) = x2 – 1 ∴ p(1) = (1)2 – 1 = 1 – 1 = 0 Since, p(1) = 0, ∴ x = 1 is a zero of x2 – 1. Also p(–1) = (–1)2 – 1 = 1 – 1 = 0 i.e. p(–1) = 0, ∴ x = –1 is also a zero of x2 – 1. (iv) We have p(x) = (x + 1)(x – 2) ∴ p(–1) = (–1 + 1)(–1 – 2) = (0)(–3) = 0 Since p(–1) = 0, ∴ x = –1 is a zero of (x + 1) (x – 1). Also, p(2) = (2 + 1)(2 – 2) = (3)(0) = 0 Since p(2) = 0, ∴ x = 2 is also a zero of (x + 1)(x – 1). (v) We have p(x) = x2 ∴ p(0) = (0)2 = 0 Since p(0) = 0, ∴ 0 is a zero of x2. 4. Find the zero of the polynomial in each of the following cases: (i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5 (iv) p(x) = 3x – 2 ( v ) p(x) = 3x (vi) p(x) = ax, a ≠ 0 (vii) p(x) = cx + d, c ≠ 0, c, d are real numbers. (i) We have p(x) = x + 5 ∴ p(x) = 0 ⇒ x + 5 = 0 or x = –5 Thus, a zero of x + 5 is (–5). (ii) We have p(x) = x – 5 ∴ p(x) = 0 ⇒ x – 5 = 0 or x = 5 Thus, a zero of x – 5 is 5. (iii) We have p(x) = 2x + 5 ∴ p(x) = 0 ⇒ 2x + 5 = 0 or 2x = –5 or x = -5/2 Thus, a zero of 2x + 5 is = -5/2 (iv) Since p(x) = 3x – 2 ∴ p(x) = 0 ⇒ 3x – 2 = 0 or 3x = 2 or x = 2/3 Thus, a zero of 3x – 2 is 2/3 (v) Since p(x) = 3x ∴ p(x) = 0 ⇒ 3x = 0 or x = 0/3 or 0 Thus, a zero of 3x is 0. (vi) Since, p(x) = ax, a ≠ 0 ⇒ p(x) = 0 ⇒ ax = 0 or x = 0 a = 0 Thus, a zero of ax is 0. (vii) Since, p(x) = cx + d ∴ p(x) = 0 ⇒ cx + d = 0 or c x = –d or x = – dc Thus, a zero of cx + d is – dc.