Question 77522
{{{((y^2-36)/(y^2-4))*((y^2+6y+8)/(y^2-2y-24))}}}

Factor the first numerator

*[invoke factoring_quadratics 1, 0, -36]


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Factor the first denominator

*[invoke factoring_quadratics 1, 0, -4]


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Factor the second numerator

*[invoke factoring_quadratics 1, 6, 8]

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Factor the second denominator

*[invoke factoring_quadratics 1, -2, -24]





So the whole expression becomes


{{{(((y-6)(y+6))/((y-2)(y+2)))*(((y+2)(y+4))/((y+4)(y-6)))}}}



{{{((cross((y-6))(y+6))/((y-2)cross((y+2))))*((cross((y+2))cross((y+4)))/(cross((y+4))cross((y-6))))}}}  Cancel like terms

So the whole expression reduces to 

{{{(y+6)/(y-2)}}}

So the answer is A)