# How do you factor 54c^3d^4+9c^4d^2?

Jun 16, 2018

$9 {c}^{3} {d}^{2} \left(6 {d}^{2} + c\right)$

#### Explanation:

We check it by expanding:
$9 {c}^{3} {d}^{2} \left(6 {d}^{2} + c\right) = 54 {c}^{3} {d}^{4} + 9 {c}^{4} {d}^{2}$

Aug 9, 2018

$9 {c}^{3} {d}^{2} \left(6 {d}^{2} + c\right)$
Look at what power each $c$ and $d$ is raised to and take out the most that you can out of both of them:
there are at least three c's in both terms so you can pull out a ${c}^{3}$ and there are at least 2 d's in both terms so pull out ${d}^{2}$. Finally there is a common factor 9 in both terms so we'll pull that out as well:
$9 {c}^{3} {d}^{2} \left(6 {d}^{2} + c\right)$