Answer
$-\displaystyle \frac{2}{x^{2}+x}$
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:
$6x+2=2(3x+1)$
$1-x=-(x-1)$
$x^{2}-1=(x+1)(x-1)$
$3x^{2}+x=x(3x+1)$
The problem becomes
$...=\displaystyle \frac{2(3x+1)\cdot[-(x-1)]}{(x+1)(x-1)\cdot x(3x+1)}\qquad$ ... divide out the common factors
$=\displaystyle \frac{-2\fbox{$(3x+1)$}\cdot\fbox{$(x-1)$}}{(x+1)\fbox{$(x-1)$}\cdot x\fbox{$(3x+1)$}}\qquad$
$=\displaystyle \frac{-2}{(x+1)x}$
= $-\displaystyle \frac{2}{x^{2}+x}$
