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

Page 224 Ex 2AQuestions 1 to 7, 10, 12 & 13

Q1. Polynomials & Synthetic Division Show (x + 3) is a factor of f(x) = 2x3 + 7x2 + 2x – 3 and factorise the expression fully. 2 7 2 –3 -3 -6 -3 3 2 1 -1 As there is no remainder, R= 0  (x + 3) is a factor of f(x) f(x) = 2x3 + 7x2 + 2x – 3 = (x + 3)(2 x2 + 1 x– 1) = (x + 3)(2x – 1)(x + 1)

Q2. Polynomials & Synthetic Division If (x + 3) is a factor of f(x) = x3 – x2 + px + 15 Find the value of p. 1 -1 p 15 -3 -3 12 -36 – 3p -4 1 12+p – 21 – 3p If this is a factor then there is no remainder, set = 0: – 21 – 3p = 0 – 3p = 21 p = –7

Q3. Polynomials & Synthetic Division If x = 2 is a solution of f(x) = 2x3 – px2 – 17x + 30 Find the value of p, and the other roots of the equation. 2 -p -17 30 2 4 – 18 – 4p 8 – 2p 2 4 – p -9 – 2p 12 – 4p If this is a factor then there is no remainder, set = 0 : 12 – 4p = 0 – 4p = – 12 p = 3

Q3. Polynomials & Synthetic Division If x = 2 is a solution of f(x) = 2x3 – px2 – 17x + 30 Find the value of p and the OTHER ROOTS of the equation. If p = 3  f(x) = 2x3–3x2 – 17x + 30 2 -3 -17 30 From before x = 2 is a solution 2 4 – 30 2 2 1 -15 0 Roots when f(x) = 2x3 – 3x2 – 17x + 30 = 0 (x – 2)(2x2 + 1x – 15) = 0 (x – 2)(2x– 5)(x + 3) = 0 x = 2x= 5/2 x = –3

Q4. Polynomials & Synthetic Division The curve y = 3x3 + x2 – 5x + d cuts the x-axis at (1, 0) Find the value of d. If cuts the x-axis at (1, 0)  x = 1 is a solution 3 1 -5 d 1 3 4 -1 4 3 -1 d – 1 If this is a factor then there is no remainder, set = 0: d – 1 = 0 d = 1

Q5. Polynomials & Synthetic Division Solve 2x3 + 5x2 – x – 6 = 0  Find the ROOTS of the equation. 2 5 -1 -6 1 2 6 7 As ends in -6 try ±1, ± 2 or ± 3? 2 7 6 0 As there is no remainder, R= 0  x = 1 is a solution  (x - 1) Roots when 2x3+ 5x2 – x – 6 = 0 (x – 1)(2x2 + 7x + 6) = 0 (x – 1)(2x+ 3)(x + 2) = 0 x = 1x= -3/2x = –2

Q6. Polynomials & Synthetic Division Find the roots of x4 – 2x3 – 4x2 + 2x + 3 = 0  Factorise and solve 1 -2 -4 2 3 • 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 1 1 -5 -1 -3 As ends in 3 try ±1or ± 3? 0 1 -1 -5 -3 3 3 6 3 As ends in -3 try ±1or ± 3? 1 2 1 0 Roots when x4 – 2x3 – 4x2 + 2x + 3 = 0 (x – 1)(x – 3)(1x2+ 2x + 1) = 0 (x – 1)(x– 3)(x + 1)(x + 1) = 0 x = 1x= 3 x = 1

Q7. Polynomials & Synthetic Division Find where the curve y = 2x3 + 3x2– 18x + 8 cuts the x-axis  Factorise and set = 0 to find the ROOTS of the equation. 2 3 -18 8 2 4 -8 14 As ends in 8 try ±1, ± 2 or ± 4? 2 7 -4 0 As there is no remainder, R= 0  x = 2 is a solution  (x - 2) Roots when 2x3 + 3x2 – 18x + 8 = 0 (x – 2)(2x2 + 7x – 4) = 0 (x – 2)(2x– 1)(x + 4) = 0 x = 2x= 1/2x = –4 So cuts axis at (2, 0), (1/2 , 0) & (-4, 0)

Q10. Polynomials & Synthetic Division Find the values of k for which the equation 2x2 + 4x + k = 0 has real roots Real roots  b2 – 4ac ≥ 0 2x2 + 4x + k = 0  a = 2, b = 4 & c = k b2 – 4ac ≥ 0 (4)2 – 4 x (2) x (k) ≥ 0 16 – 8k ≥ 0 16 ≥ 8k 2 ≥ k k ≤ 2

Q12. Polynomials & Synthetic Division If x – 2 is a factor of 3x3 – kx2 + 4. Find The value of k. The other factors of 3x3 – kx2 + 4 for this value k. 3 -k 0 4 As x – 2 is a factor, x = 2 is a solution. 2 6 24 – 4k 12 – 2k 3 6 - k 12 – 2k 28 – 4k As this is a factor the remainder, R = 0  set = 0 and solve 28 – 4k = 0 28 = 4k k = 7

Q12. Polynomials & Synthetic Division If x – 2 is a factor of 3x3 – kx2 + 4. Find The value of k. The other factors of 3x3 – kx2 + 4 for this value k. If we factorise 3x3– 7x2+ 4 we can find the other factors If k = 7 3 -70 4 2 6 – 4 – 2 3 - 1 – 2 0 (x – 2)(3x2– 1x – 2) Thus the factors of 3x3– 7x2 + 4 are (x – 2)(3x+ 2)(x – 1)

Q13. Polynomials & Synthetic Division Solve x2 – 5x – 6 ≥ 0 • Need to facorise • Sketch Quadratic • When is graph positive if ≥ 0 Consider x2 – 5x – 6 = 0 (x – 6)(x + 1) = 0 x = 6 & x = -1 -1 6 So for x2 – 5x – 6 ≥ 0 Graph is positive when x ≤ -1 and x ≥ 6

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