Sec 3: Polynomial

University:
Stanford University
Course:
Introduction to Mathematical Thinking
Academic year:

2023
Views:

499

Pages:

3
Author:

MAlphen

Polynomial 1 The polynomial 2𝑥3 − 3𝑎𝑥2 − 2𝑎𝑥 + 𝑏 has a factor 𝑥 − 12 and leaves a remainder of –8 when divided by 𝑥 − 1. (a) Find the values of a and of b. (b) Using the values of a and b in part (a), factorise the polynomial completely. [4] [3] 2 The term containing the highest power of x in the polynomial 𝑓(𝑥) 𝑖𝑠 3𝑥4. Two of the roots of the equation 𝑓(𝑥) = 0 𝑎𝑟𝑒 𝑥 = −1 𝑎𝑛𝑑 𝑥 = 𝑘 where k in an integer. Given that 𝑥2 − 2𝑥 + 6 is a quadratic factor of𝑓(𝑥) 𝑎𝑛𝑑 𝑓(𝑥) leaves a remainder of – 36 when divided by 𝑥, (a) show that 𝑘 = 2. Hence, (b) determine the number of real roots of the equation 𝑓 (𝑥) = 0. [3] [3] 3 A polynomial, 𝑃(𝑥), 𝑖𝑠 2𝑥2 + 𝑎𝑥2 + 6𝑥 + 27 , where a is a constant. The quadratic expression 𝑥2 + 𝑏𝑥 +9 is a factor of 𝑃(𝑥) , where 𝑏 is a constant. (a) Find the remaining factor of 𝑃(𝑥). (b) Show that the equation 𝑃(𝑥) = 0 has only one real root. 1 Solutions: 1 1 3 1 3 1 (a) 𝑓 ( ) = 2 ( ) − 3𝑎 ( ) − 2𝑎 ( ) (b) 1 4 2 3𝑎 − 4 2 2 2 −𝑎+𝑏=0 7𝑎 − 4𝑏 = 1 --- (1) 𝑓(1) = 2 − 3𝑎 − 2𝑎 + 𝑏 = −8 5𝑎 − 𝑏 = 10 𝑏 = 5𝑎 − 10 --- (2) Sub (2) in (1): 7𝑎 − 4(5𝑎 − 10) = 1 7𝑎 − 20𝑎 + 40 = 1 −13𝑎 + 39 = 0 𝑎=3 𝑏 = 5(3) − 10 =5 Answer: (a) 5 (b) (2𝑥 − 1)(𝑥 − 5)(𝑥 − 1) [2] [3] 𝑥2 − 4𝑥 5 3 2 2𝑥 − 1 2𝑥 − 9𝑥 − 6𝑥 + 5 − 2𝑥3 −𝑧2 − 8𝑥2 − 6𝑥 + 5 8𝑥2 + 4𝑥 −10𝑥 + 5 − −10𝑥 + 5 0 2 Solutions: (a) f(𝑥) = 3(𝑥2 − 2x + 6)(𝑥 + 1)(𝑥 − 𝑘) f(0) = −36 3(6)(−1)(𝑘) = −36 𝑘 = 2 (shown) (b) 𝑥2 − 2𝑥 + 6 = 0 Discriminant = (−2)2 − 4(6) < 0 ∴ no real roots When 𝑓(𝑥) = 0, there are 2 real roots Answer: (a) 𝑘 = 2 Shown (b) When 𝑓(𝑥) = 0, there are 2 real roots. 3 Solutions: (a) Let 2𝑥3 + 𝑎𝑥2 + 6𝑥 + 27 = (𝑥2 + 𝑏𝑥 + 9)(𝑐𝑥 + 𝑑) Compare coefficient of 𝑥3, 𝑐=2 Compare constant, 9𝑑 = 27 𝑑=3 𝑃(𝑥) = 2𝑥 + 3 (b) 2𝑥3 + 𝑎𝑥2 + 6𝑥 + 27 = (𝑥2 + 𝑏𝑥 + 9)(2𝑥 + 3) Compare coefficient of 𝑥2, 𝑎 = 3 + 2𝑏 ------------------ (1) Compare coefficient of x, 6 = 3𝑏 + 18 𝑏 = −4 (2) Sub (2) into (1), 𝑎 = 3 + 2(−4) 𝑎 = −5 2𝑥3 + 5𝑥2 + 6𝑥 + 27 = 0 (𝑥2 − 4𝑥 + 9)(2𝑥 + 3) = 0 𝑥2 − 4𝑥 + 9 = 0 or 2𝑥 + 3 = 0 For 𝑥2 = 4𝑥 + 9 = 0, Discriminant = (−4)2 − 4(1)(9) = −20 Hence there is no solution for 𝑥2 = 4𝑥 + 9 = 0. 2𝑥 + 3 = 0 3 𝑥=− 2 Hence the equation 𝑃(𝑥) = 0 has only one real root Answer: (a) 𝑃(𝑥) = 2𝑥 + 3 (b) 𝑥 = − 3 2