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.

$\frac{4}{b^2+6b+9},\frac{2b}{b^2-2b-15}$

Short answer

(a) The LCD is $(b+3)(b+3)(b-5)$.

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

Step-by-step solution

Part (a) Step 1. Given Information

The given expressions are$\frac{4}{b^2+6b+9},\frac{2b}{b^2-2b-15}$.

Part (a) Step 2. Find LCD 

• Factor the denominators of both the given rational expressions.

$$\begin{gathered} b^2+6b+9=(b+3)(b+3) \\ {b}^2-2b-15=(b-5)(b+3) \end{gathered}$$

• So, the LCD is localid="1645520419522" $(b+3)(b+3)(b-5)$.

Part (b) Step 1. Given Information 

The given expressions are$\frac{4}{b^2+6b+9},\frac{2b}{b^2-2b-15}$.

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.

$$\begin{gathered} \frac{4}{b^2+6b+9}=\frac{4(b-5)}{(b+3)(b+3)(b-5)} \\ \frac{2b}{b^2-2b-15}=\frac{2b(b+3)}{(b-5)(b+3)(b+3)} \end{gathered}$$

• So, the expressions with equal denominators are $\frac{4(b-5)}{(b+3)(b+3)(b-5)},\frac{2b(b+3)}{(b-5)(b+3)(b+3)}$.