Question #dbe68

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Answer 1 · 2016-08-22T19:54:12

$=(x+8)/(x-2)$

Excluded values are $x+4$

$(x^2+12x+32)/(x^2+2x-8)$

$=(x^2+8x+4x+32)/(x^2+4x-2x-8)$

$=(x(x+8)+4(x+8))/(x(x+4)-2(x+4))$

$=((x+8)(x+4))/((x-2)(x+4))$

$=((x+8)cancel((x+4)))/((x-2)(cancel(x+4)))$

$=(x+8)/(x-2)$

Excluded values are $x+4$