# How do you divide \frac { p ^ { 2} - 11p + 28} { 2p } \div \frac { p - 7} { 2p }?

May 24, 2018

$p - 4$

#### Explanation:

$\frac{\left(p - 7\right) \left(p - 4\right)}{2 p} \cdot \frac{2 p}{p - 7}$

May 24, 2018

$p - 4$

#### Explanation:

First of all, remember that dividing by a fraction is the same as multiplying by the inverse fraction:

$\setminus \frac{{p}^{2} - 11 p + 28}{2 p} \div \setminus \frac{p - 7}{2 p} = \setminus \frac{{p}^{2} - 11 p + 28}{2 p} \setminus \cdot \setminus \frac{2 p}{p - 7}$

Cross simplify common terms:

$\setminus \frac{{p}^{2} - 11 p + 28}{\cancel{2 p}} \setminus \cdot \setminus \frac{\cancel{2 p}}{p - 7} = \setminus \frac{{p}^{2} - 11 p + 28}{p - 7}$

You can also observe that ${p}^{2} - 11 p + 28$ is a quadratic equation. If we find the two roots ${p}_{1}$ and ${p}_{2}$, we may write ${p}^{2} - 11 p + 28 = \left(p - {p}_{1}\right) \left(p - {p}_{2}\right)$.

We can avoid the quadratic formula in this case: if the quadratic coefficient is $1$ (which is our case), the equation can be read as

${p}^{2} - s x + p$

where $s$ is the sum of the solutions and $p$ is their product. So, we're looking for two numbers which sum up to $11$ and give $28$ when multiplied. These numbers are clearly $4$ and $7$. So, you have

${p}^{2} - 11 p + 28 = \left(p - 4\right) \left(p - 7\right)$

The fraction becomes

$\setminus \frac{{p}^{2} - 11 p + 28}{p - 7} = \setminus \frac{\left(p - 4\right) \cancel{\left(p - 7\right)}}{\cancel{p - 7}} = p - 4$