Question

How do you factor `12x^2+8x-16`?

Answer 1

`12x^2+8x-16 = 4/3(3x+1-sqrt(13))(3x+1+sqrt(13))`

Explanation:

Given:

`12x^2+8x-16`

We can factor this by completing the square then using the difference of squares identity:

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

with `a=(3x+1)` and `b=sqrt(13)` as follows.

First note that all of the terms are divisible by `4` and that in order to make the leading coefficient into a square number without making the other coefficients into fractions, we need to multiply by `3`. So for simplicity, start by multiplying by `3/4`...

`3/4(12x^2+8x-16) = 9x^2+6x-12`

`color(white)(3/4(12x^2+8x-16)) = (3x)^2+2(3x)+1-13`

`color(white)(3/4(12x^2+8x-16)) = (3x+1)^2-(sqrt(13))^2`

`color(white)(3/4(12x^2+8x-16)) = ((3x+1)-sqrt(13))((3x+1)+sqrt(13))`

`color(white)(3/4(12x^2+8x-16)) = (3x+1-sqrt(13))(3x+1+sqrt(13))`

Then multiplying both ends by `4/3`, we find:

`12x^2+8x-16 = 4/3(3x+1-sqrt(13))(3x+1+sqrt(13))`