# How do you factor a^4 - 8a^3 + 16^2?

Jun 15, 2017

Assuming you mean ${a}^{4} - 8 {a}^{3} + 16 {a}^{2}$, note the ${a}^{2}$ in common:

${a}^{2} \left({a}^{2} - 8 a + 16\right)$

Then, we look for factors of $16$ that when added together in a certain way give $8$. As it turns out, this is a perfect square.

$= \textcolor{b l u e}{{a}^{2} {\left(a - 4\right)}^{2}}$

If you multiply this out using the "FOIL" method, you should get the same thing back that you started with:

${a}^{2} {\left(a - 4\right)}^{2} = {a}^{2} \left({a}^{2} - 4 a - 4 a + 16\right)$

$= {a}^{2} \left({a}^{2} - 8 a + 16\right)$

$= {a}^{4} - 8 {a}^{3} + 16 {a}^{2}$