Question

Factor the expression.

4r^2 – 64

1)(4r + 5)(r – 16)
2)4(r – 4)(r – 4)
3)(4r – 5)(r + 16)
4)4(r + 4)(r – 4)

Answer 1

4r^2-64
4(r^2-16)
4(r+4)(r-4)
First factor out the gcf. Because (r^2-16) has perfect squares, you can factor it out one more time.
Final answer: D

Answer 2

Hey!

First, let's write the problem,
$4r^2-64$
Factor out the common term, which is 4,
$=4\left(r^2-16\right)$
Then, let's factor this part: $r^2-16$
We are going to factor it using the difference of squares rule.
$=\left(r+4\right)\left(r-4\right)$
Our final factored answer would be,
$=4\left(r+4\right)\left(r-4\right)$

Thanks!
-TetraFish