134k views
4 votes
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.



1 Answer

0 votes

Answer:

see explanation

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) + 2c

For the 2 sides to be equal then like terms must equate, that is

5ax³ = 5x³ ⇒ 5a = 5 ⇒ a = 1

2c = - 6 ⇒ c = - 3

5b + 2a = k

5c + 2b = - 7 ← substitute c = - 3

- 15 + 2b = - 7 ⇒ 2b = 8 ⇒ b = 4

Substitute a = 1, b = 4 into 5b + 2a = k

20 + 2 = k ⇒ k = 22

The required values are

a = 1, b = 4, c = - 3 and k = 22

User Zhi Lv
by
5.6k points