Question

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

Answer 1

`n^4+2n^2w^2+w^4`

`=(n^2+w^2)^2`

`=(n-iw)^2(n+iw)^2`

Explanation:

`n^4+2n^2w^2+w^4`

`=(n^2)^2+2(n^2)(w^2)+(w^2)^2`

`=(n^2+w^2)^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=iw`.

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

So we find:

`n^2+w^2 = n^2 - (iw)^2 = (n-iw)(n+iw)`

Hence we can write:

`n^4+2n^2w^2+w^4 = (n-iw)^2(n+iw)^2`