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

Nov 28, 2015

Factor as a quadratic in $\left(x - 4\right)$ then simplify at the last stage to find:

$3 {\left(x - 4\right)}^{2} + 16 \left(x - 4\right) - 12 = \left(x + 2\right) \left(3 x - 14\right)$

#### Explanation:

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

$3 {\left(x - 4\right)}^{2} + 16 \left(x - 4\right) - 12$

$= 3 {\left(x - 4\right)}^{2} - 2 \left(x - 4\right) + 18 \left(x - 4\right) - 12$

$= \left(3 {\left(x - 4\right)}^{2} - 2 \left(x - 4\right)\right) + \left(18 \left(x - 4\right) - 12\right)$

$= \left(x - 4\right) \left(3 \left(x - 4\right) - 2\right) + 6 \left(3 \left(x - 4\right) - 2\right)$

$= \left(\left(x - 4\right) + 6\right) \left(3 \left(x - 4\right) - 2\right)$

$= \left(x + 2\right) \left(3 x - 14\right)$

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

$3 {\left(x - 4\right)}^{2} + 16 \left(x - 4\right) - 12$

$= 3 {t}^{2} + 16 t - 12$

$= 3 {t}^{2} - 2 t + 18 t - 12$

$= \left(3 {t}^{2} - 2 t\right) + \left(18 t - 12\right)$

$= t \left(3 t - 2\right) + 6 \left(3 t - 2\right)$

$= \left(t + 6\right) \left(3 t - 2\right)$

$= \left(\left(x - 4\right) + 6\right) \left(3 \left(x - 4\right) - 2\right)$

$= \left(x + 2\right) \left(3 x - 14\right)$

Nov 29, 2015

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

$3 {\left(x - 4\right)}^{2} + 16 \left(x - 4\right) - 12 = \left(x + 2\right) \left(3 x - 14\right)$

#### Explanation:

$3 {\left(x - 4\right)}^{2} + 16 \left(x - 4\right) - 12$

$= 3 \left({x}^{2} - 8 x + 16\right) + 16 \left(x - 4\right) - 12$

$= 3 {x}^{2} - 24 x + 48 + 16 x - 64 - 12$

$= 3 {x}^{2} - 8 x - 28$

Look for a pair of factors of $3 \times 28 = 84$ with difference $8$.

The pair $6$, $14$ works.

So:

$3 {x}^{2} - 8 x - 28$

$= 3 {x}^{2} - 14 x + 6 x - 28$

$= \left(3 {x}^{2} - 14 x\right) + \left(6 x - 28\right)$

$= x \left(3 x - 14\right) + 2 \left(3 x - 14\right)$

$= \left(x + 2\right) \left(3 x - 14\right)$