Introductory Algebra for College Students (7th Edition)

$\displaystyle \frac{y+2}{y+4}$
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. $(\displaystyle \frac{P}{Q}\cdot\frac{R}{S}=\frac{PR}{QS})$ --- 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 = $\displaystyle \frac{y+2}{y+4}$