## Introductory Algebra for College Students (7th Edition)

$\displaystyle \frac{1}{4}$
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 $ax^{2}+bx+c$ by searching for two factors of $ac$ whose sum is $b,$ ... and, if they exist, rewrite $bx$ and factor in pairs. $2y^{2}-9y+9= \qquad$...$(ac=18, b=-9)$, factors: $-6$ and $-3$ $=2y^{2}-6y-3y+9$ $= 2y(y-3)-3(y-3)=(y-3)(2y-3)$ $3y-y^{2}=-(y^{2}-3y)=-y(y-3)$ $8y-12=4(2y-3)$ The problem becomes $...=\displaystyle \frac{2y\cdot[-(y-3)(2y-3)]}{-y(y-3)\cdot 4(2y-3)}\qquad$ ... divide out the common factors $=\displaystyle \frac{\fbox{$-y$}\cdot\fbox{$(y-3)$}\fbox{$(2y-3)$}}{\fbox{$-y$}\fbox{$(y-3)$}\cdot 4\fbox{$(2y-3)$}}\qquad$ = $\displaystyle \frac{1}{4}$