# How do you find the quotient of (n&2+7n+12)/(16n^2)div(n+3)/(2n)?

Aug 5, 2017

See a solution process below:

#### Explanation:

First, rewrite the expression as:

$\frac{\frac{{n}^{2} + 7 n + 12}{16 {n}^{2}}}{\frac{n + 3}{2 n}}$

Next, use this rule of dividing fractions to rewrite the fraction as:

$\frac{\frac{\textcolor{red}{a}}{\textcolor{b l u e}{b}}}{\frac{\textcolor{g r e e n}{c}}{\textcolor{p u r p \le}{d}}} = \frac{\textcolor{red}{a} \times \textcolor{p u r p \le}{d}}{\textcolor{b l u e}{b} \times \textcolor{g r e e n}{c}}$

$\frac{\frac{\textcolor{red}{\left({n}^{2} + 7 n + 12\right)}}{\textcolor{b l u e}{\left(16 {n}^{2}\right)}}}{\frac{\textcolor{g r e e n}{\left(n + 3\right)}}{\textcolor{p u r p \le}{\left(2 n\right)}}} = \frac{\textcolor{red}{\left({n}^{2} + 7 n + 12\right)} \times \textcolor{p u r p \le}{\left(2 n\right)}}{\textcolor{b l u e}{\left(16 {n}^{2}\right)} \times \textcolor{g r e e n}{\left(n + 3\right)}}$

Next, factor the quadratic as:

color(red)(((n + 3)(n + 4)) xx color(purple)((2n)))/(color(blue)((16n^2)) xx color(green)((n + 3)

Then cancel common terms in the numerator and denominator:

$\frac{\textcolor{red}{\left(\textcolor{g r e e n}{\cancel{\textcolor{red}{\left(n + 3\right)}}} \left(n + 4\right)\right) \times \textcolor{p u r p \le}{\left(\textcolor{b l u e}{\cancel{\textcolor{p u r p \le}{2}}} \textcolor{b l u e}{\cancel{\textcolor{p u r p \le}{n}}}\right)}}}{\textcolor{b l u e}{\left(\textcolor{p u r p \le}{\cancel{\textcolor{b l u e}{16}}} 8 \textcolor{p u r p \le}{\cancel{\textcolor{b l u e}{{n}^{2}}}} n\right)} \times \textcolor{red}{\cancel{\textcolor{g r e e n}{\left(n + 3\right)}}}} \implies$

$\frac{n + 4}{8 n}$