Answer
$LCM=(2 x+5)(x-7)(x+3)$
Work Step by Step
Given \begin{equation}
\frac{x-5}{2 x^2-9 x-35}, \quad \frac{x+6}{2 x^2+11 x+15}.
\end{equation} First factor each denominator as shown below.
$$\textbf{Denominator 1}$$ \begin{equation}
\begin{aligned}
2x^2-9x-35&=2x^2+5x-14 x-35 \\
&= \left(2 x^2+5 x\right)+(-14 x-35)\\
&= x(2 x+5)-7(2 x+5)\\
&=(2 x+5)(x-7).
\end{aligned}
\end{equation} $$\textbf{Denominator 2}$$ \begin{equation}
\begin{aligned}
2x^2+11x+15&= 2x^2+5x+6x+15\\
&= \left(2 x^2+5 x\right)+(6 x+15)\\
&=x(2 x+5)+3(2 x+5)\\
&= (2 x+5)(x+3).
\end{aligned}
\end{equation} Rewrite the fractions.
\begin{equation}
\begin{aligned}
\frac{x-5}{2 x^2-9 x-35} &= \frac{(x-5)(x+3)}{(2 x+5)(x-7)(x+3)}\\
\frac{x+6}{2 x^2+11 x+15}&=\frac{(x+6)(x-7)}{ (2 x+5)(x-7)(x+3)}.
\end{aligned}
\end{equation} Hence, the $\mathbf{LCM}=(2 x+5)(x-7)(x+3)$.
