# How do you factor 14r^4-378rs^3?

May 9, 2016

$14 {r}^{4} - 378 r {s}^{3} = 14 r \left(r - 3 s\right) \left({r}^{2} + 3 r s + 9 {s}^{2}\right)$

#### Explanation:

First note that both terms are divisible by $14 r$, so separate that out as a factor first.

$14 {r}^{4} - 378 r {s}^{3} = 14 r \left({r}^{3} - 27 {s}^{3}\right)$

Next note that both ${r}^{3}$ and $27 {s}^{3} = {\left(3 s\right)}^{3}$ are perfect cubes, so we can use the difference of cubes identity:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

with $a = r$ and $b = 3 s$ as follows:

${r}^{3} - 27 {s}^{3}$

$= {r}^{3} - {\left(3 s\right)}^{3}$

$= \left(r - 3 s\right) \left({r}^{2} + r \left(3 s\right) + {\left(3 s\right)}^{2}\right)$

$= \left(r - 3 s\right) \left({r}^{2} - 3 r s + 9 {s}^{2}\right)$

Putting it all together:

$14 {r}^{4} - 378 r {s}^{3} = 14 r \left(r - 3 s\right) \left({r}^{2} + 3 r s + 9 {s}^{2}\right)$