Answer
$\displaystyle \frac{y+2}{y+4}$
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 $b$.
$y^{2}-7y-30=(y-10)(y+3)$
$y^{2}-6y-40=(y-10)(y+4)$
... 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.
$ 2y^{2}+5y+2=\qquad$ ... factors of $2(2)=4$ whose sum is 5 ... are 4 and 1
$=2y^{2}+4y+y+2=2y(y+2)+(y+2)=(y+2)(2y+1)$
$ 2y^{2}+7y+3=\qquad$ ... factors of $2(3)=6$ whose sum is $7$ ... are $6$ and $1$
$=2y^{2}+6y+y+3=2y(y+3)+(y+3)=(y+3)(2y+1)$
The problem becomes
$...=\displaystyle \frac{\fbox{$(y-10)$}\fbox{$(y+3)$}\cdot(y+2)\fbox{$(2y+1)$}}{\fbox{$(y-10)$}(y+4)\cdot\fbox{$(y+3)$}\fbox{$(2y+1)$}}\qquad $... divide out the common factors
= $\displaystyle \frac{y+2}{y+4}$
