Question

In the following exercise, divide.

$\frac{{\displaystyle \frac{c^2-64}{3c^2+26c+16}}}{{\displaystyle \frac{c^2-4c-32}{15c+10}}}$

Short answer

The simplified expression is$\frac{5}{c+4}$

Step-by-step solution

Step 1. Given information

Expression is$\frac{{\displaystyle \frac{c^2-64}{3c^2+26c+16}}}{{\displaystyle \frac{c^2-4c-32}{15c+10}}}$

Step 2. Rewrite the complex fraction as division 

$\left(\frac{c^2-64}{3c^2+26c+16}\right)÷\left(\frac{c^2-4c-32}{15c+10}\right)$

Step 3. Rewrite as the product of first times the reciprocal of the second.

$\left(\frac{c^2-64}{3c^2+26c+16}\right)\times \left(\frac{15c+10}{c^2-4c-32}\right)$

Step 4. Factorize the denominator and numerator 

$$\begin{gathered} \left(\frac{(c-8)(c+8)}{3c^2+24c+2c+16}\right)\times \left(\frac{5(3c+2)}{c^2-8c+4c-32}\right) \\ =\left(\frac{(c-8)(c+8)}{3c(c+8)+2(c+8)}\right)\times \left(\frac{5(3c+2)}{c(c-8)+4(c-8)}\right) \\ =\frac{(c-8)(c+8)}{(3c+2)(c+8)}\times \frac{5(3c+2)}{(c-8)(c+4)} \end{gathered}$$

Step 5.Multiply and remove common factors

$$\begin{gathered} \frac{5(c-8)(c+8)(3c+2)}{(3c+2)(c+8)(c-8)(c+4)} \\ =\frac{5}{c+4} \end{gathered}$$