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 (<u><em>if</em></u> it was t^-9 then the answer would be (t+3)/(t+4))