Polynomial Past Paper Questions Polynomial Past Paper Questions

Polynomial Past Paper Questions

Polynomial Past Paper Questions 2001 P 2 Q 1. Given (x + 2) is a factor of 2 x 3 + x 2 + kx + 2, find the value of k. Hence solve the equation 2 x 3 + x 2 + kx + 2= 0, when k takes this value. 3 2 2 x 3 + x 2 + kx + 2 2 -2 1 -4 2 – 3 k 6 2 -2 k – 12 K+6 -2 k – 10 If a (x + 2) is a factor the remainder, R = 0 - 2 k – 10 = 0 - 2 k = 10 k=-5 3

Polynomial Past Paper Questions 2001 P 2 Q 1. Given (x + 2) is a factor of 2 x 3 + x 2 + kx + 2, find the value of k. Hence solve the equation 2 x 3 + x 2 + kx + 2= 0, when k takes this value. If k = -5 2 x 3 + x 2 – 5 x + 2 2 -2 2 Solving 3 2 1 -5 2 -4 6 -2 – 3 1 0 2 x 3 + x 2 – 5 x + 2 = 0 (x + 2)(2 x 2 – 3 x + 1) = 0 (x + 2)(2 x – 1)(x – 1) = 0 x = - 2; x = ½ ; x = 1 2

Polynomial Past Paper Questions 2002 WD P 1 Q 5 Given (x – 2) and (x + 3) are factors of f(x) = 3 x 3 + 2 x 2 + cx + d Find the values of c and d. 5 f(x) = 3 x 3 + 2 x 2 + cx + d 3 2 2 6 3 8 If this is a factor then: c 16 d 32+2 c 16+c 32+2 c+d 32 + 2 c + d = 0 1

Polynomial Past Paper Questions 2002 WD P 1 Q 5 Given (x – 2) and (x + 3) are factors of f(x) = 3 x 3 + 2 x 2 + cx + d Find the values of c and d. 5 f(x) = 3 x 3 + 2 x 2 + cx + d 3 -3 3 2 c -9 21 -7 If this is a factor then: d -63 -3 c 21+c -63 -3 c+d -63 – 3 c + d = 0 1

Polynomial Past Paper Questions 2002 WD P 1 Q 5 Given (x – 2) and (x + 3) are factors of f(x) = 3 x 3 + 2 x 2 + cx + d Find the values of c and d. 5 Using both equations we can solve for c and d simultaneously 32 + 2 c + d = 0 ----(1) -63 – 3 c + d = 0 ----(2) (1) – (2) 95 + 5 c = 0 5 c = -95 c = -19 If c = -19 subst into (1) to find d: 32 + 2 c + d = 0 32 + 2(-19) + d = 0 32 – 38 + d = 0 – 6+d=0 d=6 3

Polynomial Past Paper Questions 2002 WD P 2 Q 6 The graph of f(x) = 2 x 3 – 5 x 2 – 3 x + 1 has a root between 0 and 1. Find the value of this root to one decimal place. 3 f(x) = 2 x 3 – 5 x 2 – 3 x + 1 x = 0: x = 1: f(0) = 2(0)3 – 5(0)2 – 3(0) + 1 = 1 f(1) = 2(1)3 – 5(1)2 – 3(1) + 1 = – 5 As a change of sign occurs a root exists between x = 0 & x = 1 (with the root being much closer to x = 0 as f(x) is closer to zero) So try x = 0. 1: x = 0. 2: x = 0. 3: f(0. 1) = 2(0. 1)3 – 5(0. 1)2 – 3(0. 1) + 1 = 0. 652 f(0. 2) = 2(0. 2)3 – 5(0. 2)2 – 3(0. 2) + 1 = 0. 216 f(0. 3) = 2(0. 3)3 – 5(0. 3)2 – 3(0. 3) + 1 = -0. 296 As f(0. 2) = 0. 216 is closest to zero solution is x = 0. 2 to 1 d. p. 3

Polynomial Past Paper Questions 2003 P 2 Q 1 f(x) = 6 x 3 – 5 x 2 – 17 x + 6 Show (x – 2) is a factorof f(x). Express f(x) in its fully factorised form. 4 f(x) = 6 x 3 – 5 x 2 – 17 x + 6 6 2 6 -5 -17 6 12 14 -6 7 -3 0 f(x) = 6 x 3 – 5 x 2 – 17 x + 6 = (x – 2)(6 x 2 + 7 x – 3) = (x – 2)(3 x – 1)(2 x + 3) 4

Polynomial Past Paper Questions 2004 P 1 Q 2 f(x) = x 3 – x 2 – 5 x – 3 (i) Show (x + 1) is a factor of f(x) (ii) Hence or otherwise factorise f(x) fully. One of the turning points of the graph lies on the x-axis. Write down the coordinates of this turning point. 5 1 f(x) = x 3 – x 2 – 5 x – 3 1 -1 -5 -3 -1 2 3 -2 -3 0 f(x) = x 3 – x 2 – 5 x – 3 = (x + 1)(x 2 – 2 x – 3) = (x + 1)(x – 3) = (x + 1) 2 (x – 3) Turning Point rests on axis when (x + a) 2 (-1, 0) 6

Polynomial Past Paper Questions 2005 P 1 Q 8 f(x) = 2 x 3 – 7 x 2 + 9 Show (x – 3) is a factor of f(x) and factorise f(x) fully. 5 f(x) = 2 x 3 – 7 x 2 + 9 2 3 2 -7 0 9 6 -3 -9 -1 -3 0 f(x) = 2 x 3 – 7 x 2 + 9 = (x – 3)(2 x 2 – x – 3) = (x + 1)(2 x – 3)(x + 1) 5

Polynomial Past Paper Questions 2005 Paper 2 Q 11 (a) Show that x = -1 is a solution of the cubic x 3 + px 2 + px + 1 = 0 (b) Hence find the range of values of p for which all the roots are real 1 7 x 3 + px 2 + px + 1 = 0 1 -1 p -1 1 P– 1 p 1 1 0 1–p -1 As Remainder, R = 0 x = -1 is a solution & x 3 + px 2 + px + 1 = (x + 1)(1 x 2 + (p – 1)x + 1) 1

Polynomial Past Paper Questions 2005 Paper 2 Q 11 (b) Hence find the range of values of p for which all the roots are real 7 From (a) x 3 + px 2 + px + 1 = (x + 1)(1 x 2 + (p – 1)x + 1) a=1 b = (p – 1) c=1 If real then b 2 – 4 ac ≥ 0 1)2 (p – – 4(1)(1) ≥ 0 p 2 – 2 p + 1 – 4 ≥ 0 p 2 – 2 p – 3 ≥ 0 (p – 3)(p + 1) ≥ 0 Real when p ≤ -1 & p ≥ 3 y -1 3 x 7

Higher Polynomial Past Paper Questions Total = 36 Marks