Factored completely, the expression 16k^2 + 16k + 4 is equivalent to
a) 4(2k + 1)(2k + 1)
b) 4(4k - 1)(4k + 1)
c) 4(2k - 1)(2k + 1)
d) 4(2k - 1)(2k - 1)

16k^2 + 16k + 4

We can first start off by dividing every term by 4:

4 (4k^2 + 4k + 1)

to factor, you find which two number you can multiply with each other to get the middle term. So 2 and 2 to get 4. Substitute that into the equation

4 (4k^2 + 2k + 2k + 1)

the first two terms become a group and the last 2 terms become a group. Factor like terms out of the two groups.

4 (2k (2k + 1) + 1 (2k + 1))

4 (2k + 1)(2k + 1)

Ibelieve the answer your looking for is d.