Answer
$\frac{h^2-h-14}{(h+3)(h+2)}$
Work Step by Step
The student used the wrong LCD. The correct LCD can be found by first factoring the denominator of the second fraction.
Given \begin{equation}
\frac{h-3}{h+2}-\frac{h+5}{h^2+5 h+6}.
\end{equation} First factor the denominator of the second fraction.
\begin{equation}
\begin{aligned}
h^2+5h+6 & =h^2+2 h+3h+6 \\
& =\left(h^2+2 h\right)+(3 h+6) \\
&=h(h+2)+3(h+2)\\
& =(h+3)(h+2).
\end{aligned}
\end{equation}
Find the LCD of the fractions which is: $\mathbf{LCD}=(h+3)(h+2)$. Perform subtraction:
\begin{equation}
\begin{aligned}
\frac{h-3}{h+2}-\frac{h+5}{h^2+5 h+6}&=\frac{h-3}{h+2}-\frac{h+5}{(h+3)(h+2)} \\
&= \frac{(h+3)}{(h+3)}\cdot \frac{(h-3)}{(h+2)}-\frac{h+5}{(h+3)(h+2)}\\
&= \frac{h^2-9-h-5}{(h+3)(h+2)}\\
& = \frac{h^2-h-14}{(h+3)(h+2)}.
\end{aligned}
\end{equation}
