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Polynomials & Synthetic Division Homework

Polynomials & Synthetic Division Homework. Exercise 7K Page 139 Questions 4, 5, 7, 12 & 13. Polynomials & Synthetic Division Homework. (a) Factorise f(x) = x 3 + x 2 – 16x – 16 . Try ± 1, ± 2, ± 4. 1 1 – 16 – 16 . 4. 4. 20. 16. 5. 4. 1. Exercise 7K Page 139 Q4 .

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Polynomials & Synthetic Division Homework

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  1. Polynomials & Synthetic Division Homework Exercise 7K Page 139 Questions 4, 5, 7, 12 & 13

  2. Polynomials & Synthetic Division Homework (a) Factorise f(x) = x3 + x2 – 16x – 16 Try ±1, ±2, ±4 1 1 – 16 – 16 4 4 20 16 5 4 1 Exercise 7K Page 139 Q4 f(x) = x3 + x2 – 16x – 16 = (x – 4)(1 x2 + 5 x+ 4) = (x – 4)(x + 4)(x + 1)

  3. Polynomials & Synthetic Division Homework 4(b) Write down the 4 points where f(x) cuts the axes. When x = 0 : f(0) = (0)3 + (0)2 – 16(0) – 16= – 16 Thus cuts y axis at (0, – 16) Exercise 7K Page 139 Q4 From 4(a) f(x) = x3 + x2 – 16x – 16 = (x – 4)(x + 4)(x + 1) Set f(x) = 0 to solve for roots of x when y = 0 (x – 4)(x + 4)(x + 1) = 0 Then x = 4 x = -4 x = -1 4 points are therefore (0, – 16), (4, 0), (-4, 0) & (-1, 0)

  4. Polynomials & Synthetic Division Homework (x – 1) & (x + 4) are factors of f(x) = 2x3 + 5x2 + px + q (a) Find p & q. 2 5 p q If (x – 1) is a factor (x – 1) = 0 x = 1 is a solution 1 7 2 p+7 7 p+7 p+q+7 = 0 2 Exercise 7K Page 139 Q5 2 5 p q If (x + 4) is a factor (x + 4) = 0 x = -4 is a solution -4 -8 12 -4p – 48 -4p+q-48 = 0 -3 p+12 2

  5. Polynomials & Synthetic Division Homework (x – 1) & (x + 4) are factors of f(x) = 2x3 + 5x2 + px + q (a) Find p & q. p + q + 7 = 0  p + q = –7 (1) -4p + q = 48 (2) -4p + q – 48 = 0  Exercise 7K Page 139 Q5 (1) – (2)  5p = –55 p = –11 Subst in to (1) to find q If p = -11 : p + q = –7 (1) -11 + q = –7 q = 4  p = -11 & q = 4

  6. Polynomials & Synthetic Division Homework Hence, solve f(x) = 0  Solve 2x3 + 5x2-11x + 4 = 0 2 5 – 11 4 1 2 7 -4 7 -4 2 Exercise 7K Page 139 Q5 f(x) = 2x3 + 5x2 – 11x + 4 = (x – 1)(2x2 + 7x – 4 ) = 0 = (x – 1)(2x – 1 )(x + 4) = 0 Then x – 1 = 0 2x – 1 = 0 x + 4 = 0 2x = 1 x = 1 x = ½ x = – 4

  7. Polynomials & Synthetic Division Homework f(x) = x3 – 2x2 – 5x + 6 & g(x) = x – 1 (a) Show f(g(x)) = x3 – 5x2 + 2x + 8 f(g(x)) = f(x – 1) = (x – 1)3 – 2(x – 1)2 – 5(x – 1) + 6 = (x3 – 3x2 + 3x – 1) – 2(x2 – 2x + 1) – 5(x – 1) + 6 Exercise 7K Page 139 Q7 = x3 – 3x2+ 3x – 1 – 2x2+ 4x – 2– 5x + 5 + 6 (x – 1)2 = x2 – 2x + 1  Thus f(g(x)) = x3 – 5x2 + 2x + 8 as required. (x – 1)3 = (x – 1)(x – 1)2 = (x – 1)(x2 – 2x + 1) = x3 – 2x2 + x – x2 + 2x – 1 = x3 – 3x2 + 3x – 1

  8. Polynomials & Synthetic Division Homework (b) Factorise fully f(g(x)) = x3 – 5x2 + 2x + 8 1 – 5 2 8 2 2 -6 -8 -3 -4 1 Exercise 7K Page 139 Q7 f(g(x)) = x3– 5x2+ 2x + 8 = (x – 2)(1x2– 3x – 4 ) = (x – 2)(x – 4 )(x + 1)

  9. Polynomials & Synthetic Division Homework The function k is such that k(x) = 1 . f(g(x)) For what values of x is the function not defined? Exercise 7K Page 139 Q7 From (b) f(g(x)) = x3– 5x2+ 2x + 8 = (x – 2)(x – 4 )(x + 1) k(x) = 1 = 1  k(x) is undefined f(g(x)) (x – 2)(x – 4 )(x + 1) at x = 2, 4 & -1 As cannot have zero on the denominator

  10. Polynomials & Synthetic Division Homework The diagram is a sketch of a graph of a cubic function y = f(x). If y = -16 is a tangent to the curve, find the formula for f(x). y y = k(x – 0) 2(x – 6) -16 = k(4 - 0) 2(4 – 6) -16 = k(16)(– 2) -16 = – 32 k – 32 k = -16 k = ½ y= ½ (x – 0)2(x – 6) y = ½ x2(x – 6) y = ½ x3 – 3x2 y = f(x) x o 4 6 Exercise 7K Page 139 Q12 -16

  11. Polynomials & Synthetic Division Homework (a) Show that (x – 1) is a factor of f(x) = x3– 6x2 + 9x – 4 1 -6 9 -4 If (x – 1) is a factor (x – 1) = 0 x = 1 is a solution 1 -5 1 4 -5 4 1 Exercise 7K Page 139 Q13 f (x) = x3– 6x2+ 9x –4 = (x – 1)(1x2– 5x + 4 ) = (x – 1)(x – 1)(x – 4) = (x – 1)2(x – 4)

  12. Polynomials & Synthetic Division Homework (b) Write down the co-ordinates where the graph y = f(x) meets the axes. f(x) = x3 – 6x2 + 9x – 4 = (x – 1)2(x – 4) x = 0 : f(0) = (0)3 – 6(0)2 + 9(0) – 4 = – 4  (0, -4) Exercise 7K Page 139 Q13 y = 0 f(x) = x3 – 6x2 + 9x – 4 = 0  (x – 1)2(x – 4) = 0 (x – 1)2 = 0(x – 4) = 0 x = 1 x = 4 3 points are therefore (0, – 4), (1, 0) &(4, 0)

  13. Polynomials & Synthetic Division Homework (c) Find the stationary points of y = f(x) and determine their nature. f(x) = x3 – 6x2 + 9x – 4 For Stat Pts set f’(x) = 0 f’(x) = 3x2 – 12x + 9 = 0 3(x2 – 4x + 3) = 0 Exercise 7K Page 139 Q13 x  1  3  (x – 3)(x – 1) = 0 – – – 0 + (x – 3)  2 stat pts at x = 3 & x = 1 – 0 + 0 + (x – 1) x = 3 : f(3) = (3)3 – 6(3)2 + 9(3) – 4 = -4 f’(x) + 0 – 0 + x = 1 : f(1) = (1)3 – 6(1)2 + 9(1) – 4 = 0 Slope Therefore Max (1, 0) & Min (3, -4)

  14. Polynomials & Synthetic Division Homework (d) Sketch y = f(x) For f(x) = x3 – 6x2 + 9x – 4 y y = f(x) Cuts axis at (0, – 4), (1, 0) &(4, 0) Max (1, 0) Max (1, 0) & Min (3, -4) x o 1 4 Exercise 7K Page 139 Q13 -4 Min (3, -4)

  15. Polynomials & Synthetic Division Homework (e) Use the graph to find the number of solutions of the equation f(x) = -x y y = f(x) Max (1, 0) x o 1 4 Exercise 7K Page 139 Q13 How many solutions are there when f(x) = –x is drawn on graph? -4 Min (3, -4) f(x) = x3 – 6x2 + 9x – 4 f(x) = –x Crosses at 3 points  3 solutions exist.

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