Question

For the following exercises, perform the indicated operations.

$\frac{4n^2+32n}{3n+2}.\frac{3n^2-n-2}{n^2+n-30}÷\frac{108n^2-24n}{n+6}$

Short answer

The simplified expression is$\frac{\left(n+8\right)\left(n-1\right)}{3n\left(n-5\right)\left(9n-2\right)}.$

Step-by-step solution

Step 1. Given Information

$$\begin{gathered} \frac{4n^2+32n}{3n+2}.\frac{3n^2-n-2}{n^2+n-30}÷\frac{108n^2-24n}{n+6} \\ =\frac{4n^2+32n}{3n+2}.\frac{3n^2-n-2}{n^2+n-30}.\frac{n+6}{108n^2-24n} \end{gathered}$$

Step 2. Factor numerators and denominators 

$$\begin{gathered} \frac{4n^2+32n}{3n+2}.\frac{3n^2-n-2}{n^2+n-30}.\frac{n+6}{108n^2-24n} \\ =\frac{4n\left(n+8\right)}{3n+2}.\frac{\left(n-1\right)\left(3n+2\right)}{\left(n+6\right)\left(n-5\right)}.\frac{n+6}{12n\left(9n-2\right)} \end{gathered}$$

Step 3. Multiply numerators and denominators 

$=\frac{4n\left(n+8\right)\left(n-1\right)\left(3n+2\right)\left(n+6\right)}{12n\left(n+6\right)\left(n-5\right)(3n+2)\left(9n-2\right)}.$

Step 4. Remove common factors 

$=\frac{\left(n+8\right)\left(n-1\right)}{3n\left(n-5\right)\left(9n-2\right)}.$