# How do you factor 8u^3-24u^2v+18uv^2?

Feb 5, 2017

The answer is $= 2 u {\left(2 u - 3 v\right)}^{2}$

#### Explanation:

We need

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

Therefore,

$8 {u}^{2} - 24 {u}^{2} v + 18 u {v}^{2}$

$= 2 u \left(4 {u}^{2} - 12 u v + 9 {v}^{2}\right)$

$= 2 u \left(2 u - 3 v\right) \left(2 u - 3 v\right)$

$= 2 u {\left(2 u - 3 v\right)}^{2}$

Feb 5, 2017

$8 {u}^{3} - 24 {u}^{2} v + 18 u {v}^{2} = 2 u {\left(2 u - 3 v\right)}^{2}$

#### Explanation:

Given:

$8 {u}^{3} - 24 {u}^{2} v + 18 u {v}^{2}$

Note that all of the terms are divisible by $2 u$, so we can separate that out as a factor first.

Then the remaining quadratic is a perfect square trinomial, being of the form:

${A}^{2} - 2 A B + {B}^{2} = {\left(A - B\right)}^{2}$

with $A = 2 u$ and $B = 3 v$ ...

$8 {u}^{3} - 24 {u}^{2} v + 18 u {v}^{2} = 2 u \left(4 {u}^{2} - 12 u v + 9 {v}^{2}\right)$

$\textcolor{w h i t e}{8 {u}^{3} - 24 {u}^{2} v + 18 u {v}^{2}} = 2 u \left({\left(2 u\right)}^{2} - 2 \left(2 u\right) \left(3 v\right) + {\left(3 v\right)}^{2}\right)$

$\textcolor{w h i t e}{8 {u}^{3} - 24 {u}^{2} v + 18 u {v}^{2}} = 2 u {\left(2 u - 3 v\right)}^{2}$