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

$-\displaystyle \frac{2}{x^{2}+3x}$
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: $8x+2=2(4x+1)$ $3-x=-(x-3)$ $x^{2}-9=(x+3)(x-3)$ $4x^{2}+x=x(4x+1)$ The problem becomes $...=\displaystyle \frac{2(4x+1)\cdot[-(x-3)]}{(x+3)(x-3)\cdot x(4x+1)}\qquad$ ... divide out the common factors $=\displaystyle \frac{-2\fbox{$(4x+1)$}\cdot\fbox{$(x-3)$}}{(x+3)\fbox{$(x-3)$}\cdot x\fbox{$(4x+1)$}}\qquad$ $=\displaystyle \frac{-2}{(x+3)x}$ = $-\displaystyle \frac{2}{x^{2}+3x}$