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

1 Answer
Jan 2, 2016

Answer:

#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#