Answer
$y^2-4y-96 = (y-12)~(y+8)$
Work Step by Step
Let's consider a trinomial in this form: $~x^2+bx+c$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = c$
Then we can factor the trinomial as follows:
$~x^2+bx+c = (x+r)~(x+s)$
Let's consider this trinomial: $~y^2-4y-96$
To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = -4$ and $r\times s = -96$. We can see that $(-12)+(8) = -4~$ and $(-12)\times (8) = -96$
We can factor the trinomial as follows:
$y^2-4y-96 = (y-12)~(y+8)$
