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lindseyharrison
x+4/x^2+x-12 times x+4/x^2+8x+16
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\[\frac{ x+4 }{ x ^{2}+x - 12 }* \frac{ x+4 }{ x ^{2}+8x+16 }\] \[\frac{ x+4 }{ (x+4)*(x-3) } * \frac{ x+4 }{ (x+4)*(x+4) }\] Cancel the (x+4) and \[\frac{ 1 }{ x+3 }*\frac{ 1 }{ x+4 } = \frac{ 1 }{ x ^{2}+7x + 12 }\] That is your final answer.
a) 1/(x-4)^2 b) 1/(x+3)(x+4) c) 1/x+4)^2 d) 1/(x-3)(x+4)