Question

Can someone do this aleg.2

Answer 1

So first we factor
9-t^2=difference of 2 perfect squares
a^2-b^2=(a-b)(a+b)
3^2-t^2=(3-t)(3+t)


t^2+t-12
factor by finding what 2 number add to get 1 and multiply to get -12
the numbers are -3 and 4
(t-3)(t+4)


[(3-t)(3+t)]/[(t-3)(t+4)] we can't factor out any further

if it was t^2-9 then we could go further but since it isn't this is the factored form
(if it was t^-9 then the answer would be (t+3)/(t+4))