How to factor $a^2-4b^2+4ac-8bc$

Question

I am not sure how I would factor this. Can someone explain how I would factor this?

Answer 1

Hint: Think about the terms $a^2-4b^2$. And $a-2b$ inside it.

Answer 2

I'd factor the first 2 terms which is a difference of squares one way and then the next pair with a common factoring and see if something common pops up:

$ a^2-4b^2+4ac-8bc = (a+2b)(a-2b)+4c(a-2b) = (a-2b)(a+2b+4c) $

Notice that the $a-2b$ is a common factor in the second part that allows for another step to be done.

Answer 3

Soln: $$\begin{align*} a^2 - 4b^2 + 4ac -8bc &= (a - 2b)(a + 2b) + 4c(a-2b)\\ &= (a - 2b)(a + 2b + 4c) \end{align*}$$

Answer 4

You can also do

$$a^2-4b^2+4ac-8bc=a^2+4ac+4c^2-4b^2-8bc-4c^2$$ $$=(a^2+4ac+4c^2)-4(b^2-2bc+c^2)$$