# How do you factor completely: 2u^3 w^4 -2u^3?

Jul 24, 2015

$2 {u}^{3} {w}^{4} - 2 {u}^{3} = 2 {u}^{3} \left(w - 1\right) \left(w + 1\right) \left({w}^{2} + 1\right)$

#### Explanation:

First separate out the common factor $2 {u}^{3}$ to get:

$2 {u}^{3} {w}^{4} - 2 {u}^{3} = 2 {u}^{3} \left({w}^{4} - 1\right)$

Then use the difference of squares identity (twice):

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

$2 {u}^{3} \left({w}^{4} - 1\right)$

$= 2 {u}^{3} \left({\left({w}^{2}\right)}^{2} - {1}^{2}\right)$

$= 2 {u}^{3} \left({w}^{2} - 1\right) \left({w}^{2} + 1\right)$

$= 2 {u}^{3} \left({w}^{2} - {1}^{2}\right) \left({w}^{2} + 1\right)$

$= 2 {u}^{3} \left(w - 1\right) \left(w + 1\right) \left({w}^{2} + 1\right)$

${w}^{2} + 1$ has no linear factors with real coefficients since ${w}^{2} + 1 \ge 1 > 0$ for all $w \in \mathbb{R}$