# How do you factor fully, if possible 21c^4d^3-28c^2d^5+7cd^3?

Nov 14, 2016

$7 c {d}^{3} \left(3 {c}^{3} - 4 c {d}^{2} + 1\right)$

#### Explanation:

You find the lowest form of each constant and variable which is present in each term and factor it out. In this case it would be $7$, ${c}^{1} = c$ and ${d}^{3}$ giving:

$7 c {d}^{3} \left(3 {c}^{4 - 1} {d}^{3 - 3} - 4 {c}^{2 - 1} {d}^{5 - 3} + 1 {c}^{1 - 1} {d}^{3 - 3}\right)$

$7 c {d}^{3} \left(3 {c}^{3} {d}^{0} - 4 c {d}^{2} + 1 {c}^{0} {d}^{0}\right)$

$7 c {d}^{3} \left(3 {c}^{3} - 4 c {d}^{2} + 1\right)$