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.
