# How do you factor completely n^4 + 2n^2w^2+ w^4?

Jan 2, 2016

${n}^{4} + 2 {n}^{2} {w}^{2} + {w}^{4}$

$= {\left({n}^{2} + {w}^{2}\right)}^{2}$

$= {\left(n - i w\right)}^{2} {\left(n + i w\right)}^{2}$

#### Explanation:

${n}^{4} + 2 {n}^{2} {w}^{2} + {w}^{4}$

$= {\left({n}^{2}\right)}^{2} + 2 \left({n}^{2}\right) \left({w}^{2}\right) + {\left({w}^{2}\right)}^{2}$

$= {\left({n}^{2} + {w}^{2}\right)}^{2}$

In addition, if we allow Complex coefficients then ${n}^{2} + {w}^{2}$ can be treated as a difference (!) of squares, being of the form ${a}^{2} - {b}^{2}$ with $a = n$ and $b = i w$.

The difference of squares identity can be written:

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

So we find:

${n}^{2} + {w}^{2} = {n}^{2} - {\left(i w\right)}^{2} = \left(n - i w\right) \left(n + i w\right)$

Hence we can write:

${n}^{4} + 2 {n}^{2} {w}^{2} + {w}^{4} = {\left(n - i w\right)}^{2} {\left(n + i w\right)}^{2}$