Question

In the following exercises, (a) find the LCD for the given rational expressions (b) rewrite them as equivalent rational expressions with the lowest common denominator.

$\frac{5}{x^2-2x-8},\frac{2x}{x^2-x-12}$

Short answer

(a) The LCD is $(x+2)(x+3)(x-4)$.

(b) The expressions with LCD are$\frac{5\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x-4\right)},\frac{2x\left(x+2\right)}{\left(x+2\right)\left(x+3\right)\left(x-4\right)}$.

Step-by-step solution

Part (a) Step 1. Given Information

The given expressions are$\frac{5}{x^2-2x-8},\frac{2x}{x^2-x-12}$

Part (a) Step 2. Find LCD

• Factor the denominators of both the given rational expressions.

$$\begin{gathered} x^2-2x-8=\left(x-4\right)\left(x+2\right) \\ {x}^2-x-12=\left(x-4\right)\left(x+3\right) \end{gathered}$$

• So, the LCD is$\left(x+2\right)\left(x+3\right)\left(x-4\right)$.

Part (b) Step 1. Given Information 

The given expressions are$\frac{5}{x^2-2x-8},\frac{2x}{x^2-x-12}$.

Part (b) Step 2. Express the expressions with LCD

• Multiply and divide the expressions by the missing factor to make the denominator equal to LCD.

$$\begin{gathered} \frac{5}{x^2-2x-8}=\frac{5\left(x+3\right)}{\left(x+2\right)\left(x+3\right)\left(x-4\right)} \\ \frac{2x}{x^2-x-12}=\frac{2x\left(x+2\right)}{\left(x+2\right)\left(x+3\right)\left(x-4\right)} \end{gathered}$$

• So, the expressions with equal denominators arelocalid="1645165348019">5x+3x+2x+3x-4,2xx+2x+2x+3x-4.