Question

If
$ax {}^{3 } + bx {}^{2} + x - 6$
has x+2 as a factor and leaves a remainder 4 when divided by x-2,find the values of a and b

Answer 1

Hey there!

x + 2 is a factor, which means, it leaves the remainder zero.

let us take x + 2 = 0

Then, x = -2

Let's substitute ⇒

-8a + 4b - 2 - 6 = 0

-8a + 4b - 8 = 0

4b - 8a = 8

When divided with x - 2 , leaves the remainder is 4

x - 2 = 0

x = 2

Let us substitute,

8a + 4b + 2 - 6 = 4

8a + 4b - 4 = 4

8a + 4b = 8

Lets put the two equations,

4b - 8a = 8

4b + 8a = 8

If we add them, we get,

8b = 16

b = 2

Substitute b and we get a

8 + 8a = 8

8a = 0

a = 0

A = 0 AND B = 2

Hope my answer helps!