Question

For the following exercises, perform the indicated operations.
$\frac{12p^2+3p}{p+3}÷\frac{p^2+2p-63}{p^2-p-12}.\frac{p-7}{9p^3-9p^2}$

Short answer

The simplified expression is$\frac{\left(4p+1\right)\left(p-4\right)}{3p\left(p+9\right)\left(p-1\right)}.$

Step-by-step solution

Step 1. Given Information

$$\begin{gathered} \frac{12p^2+3p}{p+3}÷\frac{p^2+2p-63}{p^2-p-12}.\frac{p-7}{9p^3-9p^2} \\ =\frac{12p^2+3p}{p+3}.\frac{p^2-p-12}{p^2+2p-63}.\frac{p-7}{9p^3-9p^2} \end{gathered}$$

Step 2. Factor the numerators and denominators

$\frac{3p\left(4p+1\right)}{p+3}.\frac{\left(p-4\right)\left(p+3\right)}{\left(p+9\right)\left(p-7\right)}.\frac{p-7}{9p^2\left(p-1\right)}$

Step 3. Remove the common factors

$$\begin{gathered} =\frac{3p\left(4p+1\right)\left(p-4\right)\left(p+3\right)\left(p-7\right)}{9p\times p(p+3)\left(p+9\right)\left(p-7\right)\left(p-1\right)} \\ =\frac{\left(4p+1\right)\left(p-4\right)}{3p\left(p+9\right)\left(p-1\right)} \end{gathered}$$