Question

In the following exercises, divide the rational expressions.

$\frac{{\displaystyle \frac{12c^2-12}{2c^2-3c+1}}}{{\displaystyle \frac{4c+4}{6c^2-13c+5}}}$

Short answer

The simplified expression is$3\left(3c-5\right).$

Step-by-step solution

Step 1. Given Information 

$\frac{{\displaystyle \frac{12c^2-12}{2c^2-3c+1}}}{{\displaystyle \frac{4c+4}{6c^2-13c+5}}}=\frac{12c^2-12}{2c^2-3c+1}÷\frac{4c+4}{6c^2-13c+5}$

Step 2. Rewrite as product of first times reciprocalof second.

$\frac{12c^2-12}{2c^2-3c+1}÷\frac{4c+4}{6c^2-13c+5}=\frac{12c^2-12}{2c^2-3c+1}\times \frac{6c^2-13c+5}{4c+4}$

Step 3. Factor the numerators and thedenominators, and then multiply.

$$\begin{gathered} =\frac{12\left(c^2-1\right)}{\left(c-1\right)\left(2c-1\right)}\times \frac{\left(3c-5\right)\left(2c-1\right)}{4\left(c+1\right)} \\ =\frac{12\left(c-1\right)\left(c+1\right)\left(3c-5\right)\left(2c-1\right)}{4\left(c-1\right)\left(2c-1\right)\left(c+1\right)} \end{gathered}$$

Step 4. Simplify by dividing out common factors. 

$=3\left(3c-5\right)$