Suppose 5x^3 + kx^2 – 7x - 6 = (5x + 2)(ax^2 + bx+c) for all x. Find the values of a, b,c, and k.
Accepted Solution
A:
Answer:see explanationStep-by-step explanation:Expand the right side and compare the coefficients of like terms(5x + 2)(ax² + bx + c)= 5x(ax² + bx + c) + 2(ax² + bx + c) ← distribute parenthesis= 5ax³ + 5bx² + 5cx + 2ax² + 2bx + 2c ← collect like terms= 5ax³ + x²(5b + 2a) + x(5c + 2b) + 2cFor the 2 sides to be equal then like terms must equate, that is5ax³ = 5x³ ⇒ 5a = 5 ⇒ a = 12c = - 6 ⇒ c = - 35b + 2a = k5c + 2b = - 7 ← substitute c = - 3- 15 + 2b = - 7 ⇒ 2b = 8 ⇒ b = 4Substitute a = 1, b = 4 into 5b + 2a = k20 + 2 = k ⇒ k = 22The required values are a = 1, b = 4, c = - 3 and k = 22