Question

How do you factor `3(x-4)^2+16(x-4) -12`?

Answer 1

Factor as a quadratic in `(x-4)` then simplify at the last stage to find:

`3(x-4)^2+16(x-4)-12 = (x+2)(3x-14)`

Explanation:

One way is to leave the `(x-4)`'s intact until the end and look for a pair of factors of `3 xx 12 = 36`, whose difference is `16`. The pair `2`, `18` works, so use that to split the middle term and factor by grouping:

`3(x-4)^2+16(x-4)-12`

`=3(x-4)^2-2(x-4)+18(x-4)-12`

`=(3(x-4)^2-2(x-4))+(18(x-4)-12)`

`=(x-4)(3(x-4)-2)+6(3(x-4)-2)`

`=((x-4)+6)(3(x-4)-2)`

`=(x+2)(3x-14)`

Another way of expressing this solution is to substitute `t = x-4` and proceed as follows:

`3(x-4)^2+16(x-4)-12`

`=3t^2+16t-12`

`=3t^2-2t+18t-12`

`=(3t^2-2t)+(18t-12)`

`=t(3t-2)+6(3t-2)`

`=(t+6)(3t-2)`

`=((x-4)+6)(3(x-4)-2)`

`=(x+2)(3x-14)`

Answer 2

Alternatively, multiply out the original quadratic and simplify before factoring to find:

`3(x-4)^2+16(x-4)-12 = (x+2)(3x-14)`

Explanation:

`3(x-4)^2+16(x-4)-12`

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

`=3x^2-24x+48+16x-64-12`

`=3x^2-8x-28`

Look for a pair of factors of `3 xx 28 = 84` with difference `8`.

The pair `6`, `14` works.

So:

`3x^2-8x-28`

`=3x^2-14x+6x-28`

`=(3x^2-14x)+(6x-28)`

`=x(3x-14)+2(3x-14)`

`=(x+2)(3x-14)`