# How do you multiply (6p+7)/(p+2)div(36p^2-49)?

Sep 1, 2016

 1/(6p^2 -5p-14  This is a division problem not a multiplication problem.

#### Explanation:

$\frac{6 p + 7}{p + 2}$ is a division problem written as a fraction.

Then divided by a "whole number " fraction  36p^2 - 49)/1

Set the problem up as complex fraction

 (6p +7) / (p+2)/ (36p^2 - 49) / 1 )

To make the second fraction (bottom fraction or denominator) disappear multiply both fractions by the multiplicative inverse.

$\frac{6 p + 7}{p + 2} \times \frac{1}{36 {p}^{2} - 49}$ and

$\frac{36 {p}^{2} - 49}{1} \times \frac{1}{36 {p}^{2} - 49}$ It is necessary to multiply both the nominator and the denominator by the same "number" or quantity. This is the equality property of multiplication.

$\frac{36 {p}^{2} - 49}{1} \times \frac{1}{36 {p}^{2} - 49}$ = 1

The denominator disappears leaving

$\frac{6 p + 7}{p + 2} \times \frac{1}{36 {p}^{2} - 49}$

$\left(36 {p}^{2} - 49\right)$ is the difference between two squares so

$\left(36 {p}^{2} - 49\right)$ = $\left(6 p + 7\right) \times \left(6 p - 7\right)$

Now you have

 ( 6p +7)/{ (p+2) xx (6p +7) xx 6p -7)}

6p +7 on the top divided by the 6p + 7 on the bottom equals 1

leaving

$\frac{1}{\left(p + 2\right) \times \left(6 p - 7\right)}$

Now multiply $\left(p + 2\right) \times \left(6 p - 7\right)$

This gives $6 {p}^{2} + 5 p - 14$ so the answer is

$\frac{1}{6 {p}^{2} + 5 p - 14}$

Sep 1, 2016

$\frac{1}{\left(p + 2\right) \left(6 p - 7\right)}$

#### Explanation:

The key is to recognize that $36 {p}^{2} - 49 = \left(6 p + 7\right) \left(6 p - 7\right)$ and that
the $6 p + 7$ will cancel the one on the numerator. That leaves $\left(p + 2\right) \left(6 p - 7\right)$ in the denominator.

Sep 1, 2016

$\frac{1}{\left(p + 2\right) \left(6 p - 7\right)}$

#### Explanation:

$\frac{6 p + 7}{p + 2} \div \frac{\textcolor{red}{\left(36 {p}^{2} - 49\right)}}{1}$

To divide by a fraction, change $\div \to \times$ and invert the fraction.

=(6p+7)/(p+2) xx1/color(red)((36p^2-49)) larr " factorise " color(red)("diff of two squares")

=$\frac{\cancel{\left(6 p + 7\right)}}{\left(p + 2\right)} \times \frac{1}{\textcolor{red}{\cancel{\left(6 p + 7\right)} \left(6 p - 7\right)}} \leftarrow$ cancel

=$\frac{1}{\left(p + 2\right) \left(6 p - 7\right)} \leftarrow$ can be left in this form.