Question

Divide Rational Expressions:$\left\{\frac{12p^2\hspace{0.17em}+\hspace{0.17em}3p}{p\hspace{0.17em}+\hspace{0.17em}3}÷\frac{p^2+\hspace{0.17em}2p\hspace{0.17em}-\hspace{0.17em}63}{p^2\hspace{0.17em}-\hspace{0.17em}p\hspace{0.17em}-\hspace{0.17em}12}\right\}\hspace{0.17em}\hspace{0.17em}\times \frac{p\hspace{0.17em}-\hspace{0.17em}7}{9p^3\hspace{0.17em}-\hspace{0.17em}9p^2}$

Short answer

The required answer 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. 

The given expression is$\left\{\frac{12p^2\hspace{0.17em}+\hspace{0.17em}3p}{p\hspace{0.17em}+\hspace{0.17em}3}÷\frac{p^2+\hspace{0.17em}2p\hspace{0.17em}-\hspace{0.17em}63}{p^2\hspace{0.17em}-\hspace{0.17em}p\hspace{0.17em}-\hspace{0.17em}12}\right\}\hspace{0.17em}\hspace{0.17em}\times \frac{p\hspace{0.17em}-\hspace{0.17em}7}{9p^3\hspace{0.17em}-\hspace{0.17em}9p^2}$

Step 2.  Simplify the expression. 

$\left\{\frac{12p^2\hspace{0.17em}+\hspace{0.17em}3p}{p\hspace{0.17em}+\hspace{0.17em}3}÷\frac{p^2+\hspace{0.17em}2p\hspace{0.17em}-\hspace{0.17em}63}{p^2\hspace{0.17em}-\hspace{0.17em}p\hspace{0.17em}-\hspace{0.17em}12}\right\}\hspace{0.17em}\hspace{0.17em}\times \frac{p\hspace{0.17em}-\hspace{0.17em}7}{9p^3\hspace{0.17em}-\hspace{0.17em}9p^2}$

Apply the fraction rule.

$$\begin{gathered} \frac{\frac{12p^2+3p}{p+3}}{\frac{p^2+2p-63}{p^2-p-12}}\times \frac{p-7}{9p^3-9p^2} \\ =\frac{\left(12p^2+3p\right)(p^2-p-12)\left(p-7\right)}{(p+3)\left(p^2+2p-63\right)\left(9p^3-9p^2\right)} \\ =\frac{\left(12p^2+3p\right)\left(p-4\right)(p+3)\left(p-7\right)}{(p+3)\left(p^2+2p-63\right)\left(9p^3-9p^2\right)} \\ =\frac{\left(12p^2+3p\right)\left(p-4\right)\left(p-7\right)}{\left(p^2+2p-63\right)\left(9p^3-9p^2\right)} \\ =\frac{\left(12p^2+3p\right)\left(p-4\right)\left(p-7\right)}{\left(p-7\right)\left(p+9\right)\left(9p^3-9p^2\right)} \\ =\frac{\left(12p^2+3p\right)\left(p-4\right)}{\left(p+9\right)\left(9p^3-9p^2\right)} \end{gathered}$$

Step 3. Factor the numerators and denominators and then multiply.  

$$\begin{gathered} =\frac{3p\left(4p+1\right)\left(p-4\right)}{\left(p+9\right)\left(9p^3-9p^2\right)} \\ =\frac{3p\left(4p+1\right)\left(p-4\right)}{\left(p+9\right)\times \hspace{0.17em}9p^2\left(p-1\right)} \\ =\frac{3p\left(4p+1\right)\left(p-4\right)}{\left(p+9\right)\times \hspace{0.17em}3\times \hspace{0.17em}3p^2\left(p-1\right)} \\ =\frac{\left(4p+1\right)\left(p-4\right)}{\left(p+9\right)\times \hspace{0.17em}3p\left(p-1\right)} \\ =\frac{\left(4p+1\right)\left(p-4\right)}{3p\left(p+9\right)\left(p-1\right)} \end{gathered}$$