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

1 Answer
Jun 15, 2017

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

#a^2(a^2 - 8a + 16)#

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.

#= color(blue)(a^2(a - 4)^2)#

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

#a^2(a - 4)^2 = a^2(a^2 - 4a - 4a + 16)#

#= a^2(a^2 - 8a + 16)#

#= a^4 - 8a^3 + 16a^2#