factoring algebraic expressions assistance

Question

We were given this algebraic expression: $m^6+4m^2n^4-4m^4n^2-16n^4$. I already tried grouping the terms first but couldnt arrive at a specific answer. can anyone explain how i can solve this expression?

Answer 1

hint: $m^6+4m^2n^4-4m^4n^2 = m^2(m^4-4m^2n^2 + 4n^4) = m^2(m^2-2n^2)^2 = (m(m^2-2n^2))^2$, and $16n^4 = (4n^2)^2$. Does the identity $A^2-B^2 = (A+B)(A-B)$ sound familiar here?

Answer 2

The direct solution has been hinted already, so here is a fallback if you don't "see" that one offhand.

Consider the polynomial as a quadratic in $x = n^2\,$:

$$m^6+4m^2n^4-4m^4n^2-16n^4 \;=\; (4m^2-16) \cdot x^2 - 4m^4 \cdot x + m^6$$

Its reduced discriminant is:

$$\require{cancel} \frac{1}{4}\Delta = \cancel{4 m^8} - (\cancel{4m^2}-16)m^6 = 16 m^6 = \left(4m^3\right)^2 $$

Therefore its roots are $\displaystyle x_{1,2}=\frac{2m^4 \pm 4m^3}{4m^2-16}$, and the factorization follows from $(n^2-x_1)(n^2-x_2)$.