Question

Factor completely. M^4 - n^4

Answer 1

Answer:

(m^2+n^2)(m+n)(m-n)

Step-by-step explanation:

Answer 2

Answer:

$(M^2+n^2)(M-n)(M+n)$

Step-by-step explanation:

This is a difference of squares polynomial of the form $a^2-b^2=(a+b)(a-b)$. Since $M^4$ and $n^4$ are both perfect squares then take the square root of each so a=$M^2$ and b=$n^2$.

$(M^2+n^2)(M^2-n^2)$

We have another difference of squares and repeat the process.

$(M^2+n^2)(M-n)(M+n)$

Notice that a sum of squares is not factorable.