Answer
$\displaystyle \frac{y+5}{y-12}$
Work Step by Step
Step by step multiplication of rational expressions:
1. Factor completely what you can
2. Reduce (divide) numerators and denominators by common factors.
3. Multiply the remaining factors in the numerators and
multiply the remaining factors in the denominators. $(\displaystyle \frac{P}{Q}\cdot\frac{R}{S}=\frac{PR}{QS})$
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Factor what we can:
... factor $x^{2}+bx+c $ by searching for two factors of $c$ whose sum is $b$.
$y^{2}-4y-32=(y-8)(y+4)$
$y^{2}-8y-48=(y-12)(y+4)$
... factor $ax^{2}+bx+c $ by searching for two factors of $ac$ whose sum is $b,$
... and, if they exist, rewrite $bx$ and factor in pairs.
$ 3y^{2}+17y+10=\qquad$ ... factors of $3(10)=30$ whose sum is $17$ ... are $15$ and $2$
$=3y^{2}+15y+2y+10=3y(y+5)+2(y+5)=(y+5)(3y+2)$
$ 3y^{2}-22y-16=\qquad$
... factors of $3(-16)=-48$ whose sum is $-22$ ... are $-24$ and $+2$
$=3y^{2}-24y+2y-16=3y(y-8)+2(y-8)=(y-8)(3y+2)$
The problem becomes
$...=\displaystyle \frac{(y+5)(3y+2)\cdot(y-8)(y+4)}{(y-8)(3y+2)\cdot(y-12)(y+4)}\qquad $... divide out the common factors
$=\displaystyle \frac{(y+5)\fbox{$(3y+2)$}\cdot\fbox{$(y-8)$}\fbox{$(y+4)$}}{\fbox{$(y-8)$}\fbox{$(3y+2)$}\cdot(y-12)\fbox{$(y+4)$}}\qquad $..
= $\displaystyle \frac{y+5}{y-12}$
