## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

The simplified form of the rational expression $\frac{{{x}^{2}}+4x+3}{{{x}^{2}}+x-2}\cdot \frac{{{x}^{2}}+3x+2}{{{x}^{2}}+2x-3}$ is$\frac{{{\left( x+1 \right)}^{2}}}{{{\left( x-1 \right)}^{2}}}$.
$\frac{{{x}^{2}}+4x+3}{{{x}^{2}}+x-2}\cdot \frac{{{x}^{2}}+3x+2}{{{x}^{2}}+2x-3}$ $\frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C}$ So, multiply the divisor, $\frac{{{x}^{2}}+4x+3}{{{x}^{2}}+x-2}\cdot \frac{{{x}^{2}}+3x+2}{{{x}^{2}}+2x-3}=\frac{\left( {{x}^{2}}+4x+3 \right)\left( {{x}^{2}}+3x+2 \right)}{\left( {{x}^{2}}+x-2 \right)\left( {{x}^{2}}+2x-3 \right)}$ Factor the numerator as: \begin{align} & \left( {{x}^{2}}+4x+3 \right)\left( {{x}^{2}}+3x+2 \right)=\left( {{x}^{2}}+x+3x+3 \right)\left( {{x}^{2}}+x+2x+2 \right) \\ & =\left( x\left( x+1 \right)+3\left( x+1 \right) \right)\left( x\left( x+1 \right)+2\left( x+1 \right) \right) \\ & =\left( x+3 \right)\left( x+1 \right)\cdot \left( x+2 \right)\left( x+1 \right) \end{align} Factor the denominator as: \begin{align} & \left( {{x}^{2}}+x-2 \right)\left( {{x}^{2}}+2x-3 \right)=\left( {{x}^{2}}+2x-x-2 \right)\left( {{x}^{2}}+3x-x-3 \right) \\ & =\left( x\left( x+2 \right)-\left( x+2 \right) \right)\left( x\left( x+3 \right)-\left( x+3 \right) \right) \\ & =\left( x+2 \right)\left( x-1 \right)\cdot \left( x+3 \right)\left( x-1 \right) \end{align} So, the rational expression becomes, $\frac{{{x}^{2}}+4x+3}{{{x}^{2}}+x-2}\cdot \frac{{{x}^{2}}+3x+2}{{{x}^{2}}+2x-3}=\frac{\left( x+3 \right)\left( x+1 \right)\cdot \left( x+2 \right)\left( x+1 \right)}{\left( x+2 \right)\left( x-1 \right)\cdot \left( x+3 \right)\left( x-1 \right)}$ Regroup and remove the factor equal to 1, \begin{align} & \frac{{{x}^{2}}+4x+3}{{{x}^{2}}+x-2}\cdot \frac{{{x}^{2}}+3x+2}{{{x}^{2}}+2x-3}=\frac{\left( x+3 \right)\left( x+2 \right)\left( x+1 \right)\left( x+1 \right)}{\left( x+3 \right)\left( x+2 \right)\left( x-1 \right)\left( x-1 \right)} \\ & =\frac{\left( x+3 \right)\left( x+2 \right)}{\left( x+3 \right)\left( x+2 \right)}\cdot \frac{\left( x+1 \right)\left( x+1 \right)}{\left( x-1 \right)\left( x-1 \right)} \\ & =\frac{{{\left( x+1 \right)}^{2}}}{{{\left( x-1 \right)}^{2}}} \end{align}