Question 78831
Does your problem look like this?


{{{((t^2-2t-8)/(4-t^2))*((t^2-5t+6)/(t^2-t-12))}}}


In order to simplify this algebraic expression, we need to factor each term


So lets factor the first numerator {{{(t^2-2t-8)}}}


*[invoke factoring_quadratics 1, -2, -8]


Now lets factor the 1st denominator {{{(4-t^2)}}}


Notice this is a difference of squares so we have 


{{{(4-t^2)=(2+t)(2-t)}}}


Now lets factor the 2nd numerator {{{(t^2-5t+6)}}}

*[invoke factoring_quadratics 1, -5, 6]


Now lets factor the 2nd denominator {{{t^2-t-12}}}


*[invoke factoring_quadratics 1, -1, -12]


So our whole expression factors to



{{{(((t+2)(t-4))/((2+t)(2-t)))*(((t-2)(t-3))/((t+3)(t-4)))}}}


{{{((cross((t+2))cross((t-4)))/(cross((2+t))(2-t)))*(((t-2)(t-3))/((t+3)cross((t-4))))}}} Notice these terms cancel


So we're left with

{{{((t-3)(t-2))/((2-t)(t+3))}}}

Now factor out a -1 to make t-2 become 2-t


{{{-((t-3)(2-t))/((2-t)(t+3))}}}


{{{-((t-3)cross((2-t)))/(cross((2-t))(t+3))}}} Notice these terms cancel

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Answer:
So the whole thing reduces to 


{{{-(t-3)/(t+3)}}}