Question

In the following exercises, multiply the rational expressions.

$\frac{28-4b}{3b-3}.\frac{b^2+8b-9}{b^2-49}$

Short answer

The multiplication of the expressions is $\frac{-4\left(b+9\right)}{\left(b+7\right)}.$

Step-by-step solution

Step 1. Given information 

The rational expression is$\frac{28-4b}{3b-3}.\frac{b^2+8b-9}{b^2-49}$

Step 2. Factor each numerator and denominator completely.

Factor $28-4b$and $b^2-49.$

$28-4b=4(7-b)$and

$b^2-49=\left(b-7\right)\left(b+7\right)$and

$$\begin{gathered} b^2+8b-9=b^2+9b-b-9 \\ =b(b+9)-1(b+9) \\ =\left(b+9\right)\left(b-1\right) \end{gathered}$$

Step 3. Multiply the numerators and denominators.

$$\begin{gathered} \frac{28-4b}{3b-3}.\frac{b^2+8b-9}{b^2-49}=\frac{4\left(7-b\right)}{3\left(b-1\right)}.\frac{\left(b+9\right)\left(b-1\right)}{\left(b-7\right)\left(b+7\right)} \\ =\frac{-4\left(b-7\right)\left(b+9\right)}{\left(b-7\right)\left(b+7\right)} \\ =\frac{-4\left(b+9\right)}{\left(b+7\right)}. \end{gathered}$$