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Polynomial Past Paper Questions

Polynomial Past Paper Questions. 2x 3 + x 2 + kx + 2 2 1 k 2 -2 If a (x + 2) is a factor the remainder , R = 0  - 2k – 10 = 0 - 2k = 10 k = - 5 . Polynomial Past Paper Questions. -4. 6. -2k – 12 . 2001 P2 Q1.

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Polynomial Past Paper Questions

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  1. PolynomialPast Paper Questions

  2. 2x3 + x2 + kx+ 2 2 1 k 2 -2 If a (x + 2) is a factor the remainder, R = 0  - 2k – 10 = 0 - 2k = 10 k = - 5 Polynomial Past Paper Questions -4 6 -2k – 12 2001 P2 Q1. Given (x + 2) is a factor of 2x3 + x2 + kx + 2, find the value of k. 3 Hence solve the equation 2x3 + x2 + kx + 2= 0, when k takes this value. 2 2 –3 K+6 -2k – 10 3

  3. If k = -52x3 + x2– 5x + 2 2 1 -5 2 -2 Solving 2x3 + x2 – 5x + 2 = 0 (x + 2)(2x2 – 3x + 1) = 0 (x + 2)(2x– 1)(x – 1)= 0 x = - 2; x = ½ ; x = 1 Polynomial Past Paper Questions -4 6 -2 2001 P2 Q1. Given (x + 2) is a factor of 2x3 + x2 + kx + 2, find the value of k. 3 Hence solve the equation 2x3 + x2 + kx + 2= 0, when k takes this value. 2 2 –3 1 0 2

  4. f(x) = 3x3 + 2x2 + cx+ d 3 2 c d 2 If this is a factor then: 32 + 2c + d = 0 Polynomial Past Paper Questions 6 16 32+2c 2002 WD P1 Q5 Given (x – 2) and (x + 3) are factors of f(x) = 3x3 + 2x2 + cx + d Find the values of c and d.5 3 8 16+c 32+2c+d 1

  5. f(x) = 3x3 + 2x2 + cx+ d 3 2 c d -3 If this is a factor then: -63 – 3c + d = 0 Polynomial Past Paper Questions -9 21 -63-3c 2002 WD P1 Q5 Given (x – 2) and (x + 3) are factors of f(x) = 3x3 + 2x2 + cx + d Find the values of c and d.5 3 -7 21+c -63-3c+d 1

  6. Using both equations we can solve for c and d simultaneously 32 + 2c + d = 0 ----(1) -63 – 3c + d = 0 ----(2) (1) – (2)  95 + 5c = 0 5c = -95 c = -19 If c = -19 subst into (1) to find d: 32 + 2c + d = 0 32 + 2(-19) + d = 0 32 – 38 + d = 0 – 6 + d = 0 d = 6 Polynomial Past Paper Questions 2002 WD P1 Q5 Given (x – 2) and (x + 3) are factors of f(x) = 3x3 + 2x2 + cx + d Find the values of c and d.5 3

  7. f(x) = 2x3 – 5x2 – 3x + 1 x = 0: f(0) = 2(0)3– 5(0)2– 3(0)+ 1 = 1 x = 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:f(0.1) = 2(0.1)3 – 5(0.1)2 – 3(0.1) + 1 = 0.652 x = 0.2:f(0.2) = 2(0.2)3– 5(0.2)2– 3(0.2) + 1 = 0.216 x = 0.3: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. Polynomial Past Paper Questions 2002 WD P2 Q6 The graph of f(x) = 2x3 – 5x2 – 3x + 1 has a root between 0 and 1. Find the value of this root to one decimal place. 3 3

  8. f(x) = 6x3 – 5x2 – 17x + 6 6 -5 -17 6 2 f(x) = 6x3 – 5x2 – 17x + 6 = (x –2)(6x2 + 7x – 3) = (x – 2)(3x– 1)(2x + 3) Polynomial Past Paper Questions 12 14 -6 2003 P2 Q1f(x) = 6x3 – 5x2 – 17x + 6 Show (x – 2) is a factorof f(x). Express f(x) in its fully factorised form. 4 6 7 -3 0 4

  9. f(x) = x3 – x2 –5x – 3 1 -1 -5 -3 -1 f(x) = x3 – x2– 5x – 3 = (x + 1)(x2– 2x – 3) = (x + 1)(x + 1)(x– 3) = (x + 1)2(x– 3) Turning Point rests on axis when (x + a)2  (-1, 0) Polynomial Past Paper Questions -1 2 3 2004 P1 Q2 f(x) = x3 – x2 – 5x – 3 (i) Show (x + 1) is a factor of f(x) (ii) Hence or otherwise factorise f(x) fully. 5 One of the turning points of the graph lies on the x-axis. Write down the coordinates of this turning point. 1 1 -2 -3 0 6

  10. f(x) = 2x3 – 7x2 + 9 2 -7 0 9 3 f(x) = 2x3 – 7x2 + 9 = (x – 3)(2x2 – x – 3) = (x + 1)(2x– 3)(x + 1) Polynomial Past Paper Questions 6 -3 -9 2005 P1 Q8 f(x) = 2x3 – 7x2 + 9 Show (x – 3) is a factor of f(x) and factorise f(x) fully. 5 2 -1 -3 0 5

  11. x3 + px2 + px + 1 = 0 1 p p 1 -1 As Remainder, R = 0  x = -1 is a solution & x3 + px2 + px + 1 = (x + 1)(1x2 + (p – 1)x + 1) Polynomial Past Paper Questions -1 1–p -1 2005 Paper 2 Q11 Show that x = -1 is a solution of the cubic x3 + px2 + px + 1 = 0 1 Hence find the range of values of p for which all the roots are real 7 1 P–1 1 0 1

  12. From (a) x3 + px2 + px + 1 = (x + 1)(1x2 + (p – 1)x + 1) a = 1 If real then b2 – 4ac ≥ 0 b = (p – 1) c = 1 (p – 1)2 – 4(1)(1)≥ 0 p2 – 2p + 1– 4 ≥ 0 p2 – 2p – 3 ≥ 0 (p – 3)(p + 1) ≥ 0  Real when p ≤ -1 & p ≥ 3 Polynomial Past Paper Questions y 2005 Paper 2 Q11 (b) Hence find the range of values of p for which all the roots are real 7 x -1 3 7

  13. Higher PolynomialPast Paper Questions Total = 36Marks

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