Question
Simplify $\frac{b^2+3 b-28}{b^2+4 b+4} \div \frac{b^2-49}{b^2-5 b-14}$
✓
Answer
$b^2+3 b-28=(b+7)(b-4)$
$b^2+4 b+4=(b+2)(b+2)$
$b^2-49=b^2-7^2$
$=(b+7)(b-7)$
$b^2-5 b-14=(b-7)(b+2)$
$\begin{aligned} & \frac{b^2+3 b-28}{b^2+4 b+4} \div \frac{b^2-49}{b^2-5 b-14} \\ & =\frac{(b+7)(b-4)}{(b+2)(b+2)} \div \frac{(b+7)(b-7)}{(b-7)(b+2)} \\ & =\frac{(b+7)(b-4)}{(b+2)(b+2)} \times \frac{(b+2)}{(b+7)} \\ & =\frac{(b-4)}{(b+2)}\end{aligned}$
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