Answer
The simplified form of the expression $\frac{-1}{{{x}^{2}}+7x+10}-\frac{3}{{{x}^{2}}+8x+15}$ is $\frac{-4x-9}{\left( x+2 \right)\left( x+3 \right)\left( x+5 \right)}$.
Work Step by Step
$\frac{-1}{{{x}^{2}}+7x+10}-\frac{3}{{{x}^{2}}+8x+15}$
$\begin{align}
& {{x}^{2}}+7x+10={{x}^{2}}+5x+2x+10 \\
& =x\left( x+5 \right)+2\left( x+5 \right) \\
& =\left( x+5 \right)\left( x+2 \right)
\end{align}$
And
$\begin{align}
& {{x}^{2}}+8x+15={{x}^{2}}+5x+3x+15 \\
& =x\left( x+5 \right)+3\left( x+5 \right) \\
& =\left( x+5 \right)\left( x+3 \right)
\end{align}$
Thus, the expression becomes:
$\frac{-1}{{{x}^{2}}+7x+10}-\frac{3}{{{x}^{2}}+8x+15}=\frac{-1}{\left( x+5 \right)\left( x+2 \right)}-\frac{3}{\left( x+5 \right)\left( x+3 \right)}$
$\left( x+5 \right)\left( x+2 \right)$ and $\left( x+5 \right)\left( x+3 \right)$.
On solving the LCD is $\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)$.
The rational expression is written as a fraction in such a way that the LCD is the denominator of the fraction. Consider, the individual terms of the provided rational terms;
$\begin{align}
& \frac{-1}{\left( x+5 \right)\left( x+2 \right)}=\frac{-1\left( x+3 \right)}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \\
& =\frac{-x-3}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)}
\end{align}$
And,
$\begin{align}
& \frac{3}{\left( x+5 \right)\left( x+3 \right)}=\frac{3\left( x+2 \right)}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \\
& =\frac{3x+6}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)}
\end{align}$
Since the fractions have the common denominator, add the numerators and the LCD is the denominator.
$\begin{align}
& \frac{-1}{{{x}^{2}}+7x+10}-\frac{3}{{{x}^{2}}+8x+15}=\frac{-x-3}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)}-\frac{3x+6}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \\
& =\frac{-x-3-3x-6}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)} \\
& =\frac{-4x-9}{\left( x+5 \right)\left( x+2 \right)\left( x+3 \right)}
\end{align}$
