Question

How do you factor given that `f(-2)=0` and `f(x)=9x^3+10x^2-17x-2`?

Answer 1

Use synthetic division to get `(9x+1)(x-1)(x+2)`.

Explanation:

Factor given that `f(-2)=0`

`f(x)=9x^3+10x^2-17x-2`

If `f(-2)=0`, one of the factors is `(x+2)`

To find the other factors, divide the original polynomial by `(x+2)`. This could be accomplished by long division, but synthetic division is quicker.

`-2|9color(white)(aaa)10color(white)(a)-17color(white)(a)-2`
`color(white)(aaa^2)darr-18color(white)(aaa)16color(white)(aa^2a)2`
`color(white)(aa^2aa)color(red)9color(white)(aa)color(red)(-8)color(white)(aaa)color(red)(-1)color(white)(aaa^2)0`

The red numbers in the quotient are the coefficients of a polynomial.

`color(red)9x^2-color(red)8xcolor(red)(-1)`

Factor:
`(9x+1)(x-1)`

The complete factorization is `(9x+1)(x-1)(x+2)`.

Answer 2

`f(x) = (x + 2)(9x + 1)(x - 1)`

Explanation:

Given: `f(-2) = 0`

Then `x + 2` is a factor

When you divide by `(x + 2)`, the remaining quadratic is:

`9x^2 - 8x - 1`

Check the discriminant:

`b^2 - 4(a)(c) = (-8)^2 - 4(9)(-1) = 100`

`r_1 = {8 - sqrt(100)}/(2(9)) = -1/9`

`r_2 = {8 + sqrt(100)}/(2(9))` = 1

This makes the factors of the quadratic:

`(9x + 1)(x - 1)`

Multiply by `(x +2)` for factors of the cubic:

`f(x) = (x + 2)(9x + 1)(x - 1)`