Question

How do you factor `9z^2 + 6z - 8`?

Answer 1

`9z^2+6z-8 = (3z-2)(3z+4)`

Explanation:

Given:

`9z^2+6z-8`

This example can be factored using an AC method:

Find a pair of factors of `AC = 9*8=72` which differ by `B=6`

The pair `12, 6` works.

Use this pair to split the middle term, then factor by grouping:

`9z^2+6z-8 = 9z^2+12z-6z-8`

`color(white)(9z^2+6z-8) = (9z^2+12z)-(6z+8)`

`color(white)(9z^2+6z-8) = 3z(3z+4)-2(3z+4)`

`color(white)(9z^2+6z-8) = (3z-2)(3z+4)`

Answer 2

`9z^2+6z-8 = (3z-2)(3z+4)`

Explanation:

Given:

`9z^2+6z-8`

We can factor this quadratic by completing the square.

We will also use the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=(3z+1)` and `b=3`

Note that:

`31^2 = 961`

So:

`(3z+1)^2 = 9z^2+6z+1`

[Think what happens when `z=10` to see why]

Hence:

`9z^2+6z-8 = 9z^2+6z+1-9`

`color(white)(9z^2+6z-8) = (3z+1)^2-3^2`

`color(white)(9z^2+6z-8) = ((3z+1)-3)((3z+1)+3)`

`color(white)(9z^2+6z-8) = (3z-2)(3z+4)`