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

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Answer 1 · 2016-05-09T19:18:57

$14r^4-378rs^3 = 14r(r-3s)(r^2+3rs+9s^2)$

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

$14r^4-378rs^3 = 14r(r^3-27s^3)$

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

$a^3-b^3=(a-b)(a^2+ab+b^2)$

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

$r^3-27s^3$

$=r^3-(3s)^3$

$=(r-3s)(r^2+r(3s)+(3s)^2)$

$=(r-3s)(r^2-3rs+9s^2)$

Putting it all together:

$14r^4-378rs^3 = 14r(r-3s)(r^2+3rs+9s^2)$

How do you factor 14r^4-378rs^314r^4-378rs^3?