1. Page 224 Ex 2A Questions 1 to 7, 10, 12 & 13 Page 224 Ex 2A Questions 1 to 7, 10, 12 & 13

2. Q1. Polynomials & Synthetic Division Show (x + 3) is a factor of f(x) = 2x 3 + 7x 2 + 2x – 3 and factorise the expression fully. 2 7 2 –3 -3 -6 12 -3 3 f(x) = 2x 3 + 7x 2 + 2x – 3 = (x + 3)(2 x 2 + 1 x – 1) = (x + 3)(2x – 1)(x + 1) As there is no remainder, R= 0  (x + 3) is a factor of f(x)

3. Q2. Polynomials & Synthetic Division If (x + 3) is a factor of f(x) = x 3 – x 2 + px + 15 Find the value of p. 1 -1 p 15 -3 -4 1 12 12+p -36 – 3p If this is a factor then there is no remainder, set = 0: – 21 – 3p = 0 – 3p = 21 p = –7 – 21 – 3p

4. Q3. Polynomials & Synthetic Division If x = 2 is a solution of f(x) = 2x 3 – px 2 – 17x + 30 Find the value of p, and the other roots of the equation. 2 -p -17 30 2 4 4 – p2 8 – 2p -9 – 2p – 18 – 4p If this is a factor then there is no remainder, set = 0 : 12 – 4p = 0 – 4p = – 12 p = 3 12 – 4p

5. Q3. Polynomials & Synthetic Division If x = 2 is a solution of f(x) = 2x 3 – px 2 – 17x + 30 Find the value of p and the OTHER ROOTS of the equation. If p = 3  f(x) = 2x 3 – 3x 2 – 17x + 30 2 -3 -17 30 2 4 12 2 -15 – 30 0 From before x = 2 is a solution Roots when f(x) = 2x 3 – 3x 2 – 17x + 30 = 0 (x – 2)(2x 2 + 1x – 15) = 0 (x – 2)(2x – 5)(x + 3) = 0 x = 2 x = 5 / 2 x = –3

6. Q4. Polynomials & Synthetic Division The curve y = 3x 3 + x 2 – 5x + d cuts the x-axis at (1, 0) Find the value of d. 3 1 -5 d 1 3 4 3 4 If this is a factor then there is no remainder, set = 0: d – 1 = 0 d = 1 d – 1 If cuts the x-axis at (1, 0)  x = 1 is a solution

7. Q5. Polynomials & Synthetic Division Solve 2x 3 + 5x 2 – x – 6 = 0  Find the ROOTS of the equation. 2 5 -1 -6 1 2 72 7 6 6 0 As ends in -6 try ±1, ± 2 or ± 3? Roots when 2x 3 + 5x 2 – x – 6 = 0 (x – 1)(2x 2 + 7x + 6) = 0 (x – 1)(2x + 3)(x + 2) = 0 x = 1 x = -3 / 2 x = –2 As there is no remainder, R= 0  x = 1 is a solution  (x - 1)

8. Q6. Polynomials & Synthetic Division Find the roots of x 4 – 2x 3 – 4x 2 + 2x + 3 = 0  Factorise and solve 1 -2 -4 2 3 1 1 1 -5 -3 As ends in 3 try ±1 or ± 3? Roots when x 4 – 2x 3 – 4x 2 + 2x + 3 = 0 (x – 1)(x – 3)(1x 2 + 2x + 1) = 0 (x – 1)(x – 3)(x + 1)(x + 1) = 0 x = 1 x = 3 x = 1 As R= 0 x = 1 is a solution  (x - 1) is a factor Don’t stop here keep going with the fact this result ends in -3 -3 0 As ends in -3 try ±1 or ± 3? 3 1 3 2 6 1 3 0

9. Q7. Polynomials & Synthetic Division Find where the curve y = 2x 3 + 3x 2 – 18x + 8 cuts the x-axis  Factorise and set = 0 to find the ROOTS of the equation. 2 3 -18 8 2 4 72 14 -4 -8 0 As ends in 8 try ±1, ± 2 or ± 4? Roots when 2x 3 + 3x 2 – 18x + 8 = 0 (x – 2)(2x 2 + 7x – 4) = 0 (x – 2)(2x – 1)(x + 4) = 0 x = 2 x = 1 / 2 x = –4 As there is no remainder, R= 0  x = 2 is a solution  (x - 2) So cuts axis at (2, 0), ( 1 / 2, 0) & (-4, 0)

10. Q10. Polynomials & Synthetic Division Find the values of k for which the equation 2x 2 + 4x + k = 0 has real roots 2x 2 + 4x + k = 0  a = 2, b = 4 & c = k Real roots  b 2 – 4ac ≥ 0 b 2 – 4ac ≥ 0 (4) 2 – 4 x (2) x (k) ≥ 0 16 – 8k ≥ 0 16 ≥ 8k 2 ≥ k k ≤ 2

11. Q12. Polynomials & Synthetic Division If x – 2 is a factor of 3x 3 – kx 2 + 4. Find (a)The value of k. (b)The other factors of 3x 3 – kx 2 + 4 for this value k. 3 -k 0 4 2 6 6 - k3 12 – 2k 24 – 4k 28 – 4k As x – 2 is a factor, x = 2 is a solution. As this is a factor the remainder, R = 0  set = 0 and solve 28 – 4k = 0 28 = 4k k = 7

12. Q12. Polynomials & Synthetic Division If x – 2 is a factor of 3x 3 – kx 2 + 4. Find (a)The value of k. (b)The other factors of 3x 3 – kx 2 + 4 for this value k. If we factorise 3x 3 – 7x 2 + 4 we can find the other factors (x – 2)(3x 2 – 1x – 2) Thus the factors of 3x 3 – 7x 2 + 4 are (x – 2)(3x + 2)(x – 1) If k = 7  3 -7 0 4 2 6 - 13 – 2 – 4 0

13. Q13. Polynomials & Synthetic Division Solve x 2 – 5x – 6 ≥ 0  Need to facorise  Sketch Quadratic  When is graph positive if ≥ 0 x = 6 & x = -1 Graph is positive when x ≤ -1 and x ≥ 6 Consider x 2 – 5x – 6 = 0 So for x 2 – 5x – 6 ≥ 0 (x – 6)(x + 1) = 0 6