Question

How do you factor completely `15a^2 + 55a - 20`?

Answer 1

`15a^2+55a-20 = 5(3a-1)(a+4)`

Explanation:

First separate out the common scalar factor `5` to find:

`15a^2+55a-20 = 5(3a^2+11a-4)`

Next use an AC Method to factor `3a^2+11a-4`:

Find a pair of factors of `AC = 3*4 = 12` which differ by `B=11`.

The pair `12, 1` works, in that `12*1 = 12` and `12-1 = 11`.

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

`3a^2+11a-4 = 3a^2+12a-a-4`

`color(white)(3a^2+11a-4) = (3a^2+12a)-(a+4)`

`color(white)(3a^2+11a-4) = 3a(a+4)-1(a+4)`

`color(white)(3a^2+11a-4) = (3a-1)(a+4)`

So:

`15a^2+55a-20 = 5(3a-1)(a+4)`