simplify the fraction (4t^2-16/18)/(t-2/6)?

Question

a.(t-2)(t+2)(t-2)/3
b.3(t+2)
c.3(t-2)
d.(t-2)(t+2)(t-2)/12

Answer 1

None of the answers listed. The answer is [4(t+2)]/3...so unless one of those statements is equivalent....I'll work it out:
[(4t^2-16)/18]/[(t-2)/6]

[(4t^2-16)/18] * [6/(t-2)]
the 18 and 6 cancel leaving a 1 in place of the 6 and a 3 in place of 18:
[(4t^2-16)/3] * [1/(t-2)]
4t^2-16=4(t+2)(t-2)
So you have:
{[4(t-2)(t+2)]/3} * [1/(t-2)]
the (t-2) factors cancel out:
[4(t+2)]/3

Answer 2

4(t^2-4) / 18 / (t-2) / 6

4(t-2)(t+2) / 18 / (t-2) / 6

4(t-2)(t+2) /18 * 6 / (t-2)

3(t+2)