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.

$\frac{9}{z^2+2z-8},\frac{4z}{z^2-4}$

Short answer

(a) The LCD is$\left(z-2\right)\left(z+2\right)\left(z+4\right)$

(b) The expressions with equal denominators are$\frac{9\left(z+2\right)}{\left(z-2\right)\left(z+2\right)\left(z+4\right)},\frac{4z\left(z+4\right)}{\left(z-2\right)\left(z+2\right)\left(z+4\right)}$

Step-by-step solution

Part (a) Step 1. Given Information

The given expressions are$\frac{9}{z^2+2z-8},\frac{4z}{z^2-4}$

Part (a) Step 2. Find LCD 

• Factor the denominators of both the given rational expressions.

$\begin{array}{rcl}{z}^2+2z-8& =& \left(z+4\right)\left(z-2\right)\\ z^2-4& =& \left(z+2\right)\left(z-2\right)\end{array}$

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

Part (b) Step 1. Given Information

The given expressions are$\frac{9}{z^2+2z-8},\frac{4z}{z^2-4}$.

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.

$\begin{array}{rcl}\frac{9}{z^2+2z-8}& =& \frac{9\left(z+2\right)}{\left(z-2\right)\left(z+2\right)\left(z+4\right)}\\ \frac{4z}{z^2-4}& =& \frac{4z\left(z+4\right)}{\left(z-2\right)\left(z+2\right)\left(z+4\right)}\end{array}$

So, the expressions with equal denominators are$\frac{9\left(z+2\right)}{\left(z-2\right)\left(z+2\right)\left(z+4\right)},\frac{4z\left(z+4\right)}{\left(z-2\right)\left(z+2\right)\left(z+4\right)}$