Question

Divide:$\frac{4b^2+7b-2}{1-b^2}÷\frac{4b^2+15b-4}{b^2-2b+1}$.

Short answer

The required answer is$-\frac{\left(b+2\right)\left(b-1\right)}{\left(1+b\right)\left(b+4\right)}$.

Step-by-step solution

Step 1. Given information.

The given expression is:

$\frac{4b^2+7b-2}{1-b^2}÷\frac{4b^2+15b-4}{b^2-2b+1}$

Step 2. Simplify the expression.

$\frac{4b^2+7b-2}{1-b^2}÷\frac{4b^2+15b-4}{b^2-2b+1}$

Rewrite the division as the product of the first rational expression and the reciprocal of the second.

$\frac{4b^2+7b-2}{1-b^2}·\frac{b^2-2b+1}{4b^2+15b-4}$

Factor the numerators and denominators and then multiply.

localid="1657033116737" $$\begin{gathered} \frac{\left(4b-1\right)\left(b+2\right)}{\left(1+b\right)\left(1-b\right)}·\frac{\left(b-1\right)\left(b-1\right)}{\left(4b-1\right)\left(b+4\right)} \\ \left(-1\right)\frac{\left(4b-1\right)\left(b+2\right)\left(1-b\right)\left(b-1\right)}{\left(1+b\right)\left(1-b\right)\left(4b-1\right)\left(b+4\right)} \end{gathered}$$

Simplify by dividing out common factors.

$-\frac{\left(b+2\right)\left(b-1\right)}{\left(1+b\right)\left(b+4\right)}$