Question

Rewrite as equivalent rational expressions with denominator $\left(x+3\right)\left(x-4\right)\left(x+4\right)$.

$\frac{2}{x^2-x-12},\frac{1}{x^2-16}$

Short answer

The simplification of the rational expression is$\frac{2x+8}{\left(x+3\right)\left(x-4\right)\left(x+4\right)},\frac{x+3}{\left(x+3\right)\left(x-4\right)\left(x+4\right)}$

Step-by-step solution

Given information 

The given rational expression is$\frac{2}{x^2-x-12},\frac{1}{x^2-16}$

The simplification of the expression

$$\begin{gathered} \frac{2}{x^2-4x+3x-12},\frac{1}{\left(x-4\right)\left(x+4\right)} \\ \frac{2}{x\left(x-4\right)+3\left(x-4\right)},\frac{1}{\left(x-4\right)\left(x+4\right)} \\ \frac{2}{\left(x+3\right)\left(x-4\right)},\frac{1}{\left(x-4\right)\left(x+4\right)} \end{gathered}$$

The LCD of the expression is $\left(x+4\right)\left(x-4\right)\left(x+3\right)$

Multiply each numerator by the same factor .

$$\begin{gathered} \frac{2\left(x+4\right)}{\left(x+3\right)\left(x-4\right)\left(x+4\right)},\frac{1\left(x+3\right)}{\left(x-4\right)\left(x+4\right)\left(x+3\right)} \\ \frac{2x+8}{\left(x+3\right)\left(x-4\right)\left(x+4\right)},\frac{x+3}{\left(x-4\right)\left(x+4\right)\left(x+3\right)} \end{gathered}$$