Answer
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$\frac{{{\left( x+1 \right)}^{2}}}{{{\left( x-1 \right)}^{2}}}$.
Work Step by Step
$\frac{{{x}^{2}}+4x+3}{{{x}^{2}}+x-2}\cdot \frac{{{x}^{2}}+3x+2}{{{x}^{2}}+2x-3}$
$\frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C}$
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,
$\begin{align}
& \frac{{{x}^{2}}+4x+3}{{{x}^{2}}+x-2}\cdot \frac{{{x}^{2}}+3x+2}{{{x}^{2}}+2x-3}=\frac{\left( x+3 \right)\left( x+2 \right)\left( x+1 \right)\left( x+1 \right)}{\left( x+3 \right)\left( x+2 \right)\left( x-1 \right)\left( x-1 \right)} \\
& =\frac{\left( x+3 \right)\left( x+2 \right)}{\left( x+3 \right)\left( x+2 \right)}\cdot \frac{\left( x+1 \right)\left( x+1 \right)}{\left( x-1 \right)\left( x-1 \right)} \\
& =\frac{{{\left( x+1 \right)}^{2}}}{{{\left( x-1 \right)}^{2}}}
\end{align}$
