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

Hint: We are given two rational expressions, $\dfrac{{{p}^{2}}-11p+28}{2p}$ and $\dfrac{p-7}{2p}$ and are asked to divide them. To do so, we will first simplify our equation using the division rule, $\dfrac{a}{b}\div \dfrac{c}{d}=\dfrac{a}{b}\times \dfrac{d}{c}$, then we will learn how to divide polynomials using the long division method.

Complete step by step answer:
We are asked to solve $\dfrac{{{p}^{2}}-11p+28}{2p}\div \dfrac{p-7}{2p}$. We can see that there are two rational expressions involving the polynomials. In order to divide them, we will first simplify it. As we know that when we divide two fractions like $\dfrac{a}{b}\div \dfrac{c}{d}$, we can write them as $\dfrac{a}{b}\times \dfrac{d}{c}$.
So, considering $\dfrac{a}{b}=\dfrac{{{p}^{2}}-11p+28}{2p}$ and $\dfrac{c}{d}=\dfrac{p-7}{2p}$, we can write,
$\Rightarrow \dfrac{{{p}^{2}}-11p+28}{2p}\div \dfrac{p-7}{2p}=\dfrac{{{p}^{2}}-11p+28}{2p}\times \dfrac{2p}{p-7}$
Now, we can see that 2p is can be cancelled, so we will get,
$\Rightarrow \dfrac{{{p}^{2}}-11p+28}{p-7}$
Now, we are left with two polynomials, ${{p}^{2}}-11p+28$ and $p-7$ and we have to divide them. So, we will learn about how we can divide polynomials using the long division method.
In order to divide the polynomial, we should see whether the polynomials in the numerator and denominator have been arranged in decreasing order of their power or not.
For example, let us consider $\dfrac{8+{{x}^{2}}+3{{x}^{3}}}{{{x}^{2}}+2x}$. Here we can see that the numerator is not well defined, so we will arrange it in a descending order based on their power. So, we will get the numerator after arranging it as $3{{x}^{3}}+{{x}^{2}}+8$.
Now, let us consider our question, so we have $\dfrac{{{p}^{2}}-11p+28}{p-7}$. In our polynomial both the numerator and the denominator are in the correct order. Now, we will learn how to perform the long division of polynomials with the help of the following steps.
Step 1: Divide the first term of the numerator by the first term of the denominator and then put that in the quotient.
Step 2: Multiply the denominator by that value and then put that below the numerator.
Step 3: Now, subtract them to create the next polynomial.
Step 4: Repeat these steps again with the new polynomial.
Now, we have $\dfrac{{{p}^{2}}-11p+28}{p-7}$. The first term of the numerator is ${{p}^{2}}$ and that of the denominator is p. So, on dividing them, we will get $\dfrac{{{p}^{2}}}{p}=p$. So, we will multiply the denominator by it and put it below the numerator and subtract.
\[p-7\overline{\left){\begin{align}
  & {{p}^{2}}-11p+28 \\
 & {{p}^{2}}-7p \\
 & \overline{-4p+28} \\
\end{align}}\right.}\left( p \right.\]
The highest power is -4p in the numerator while in p-7, we have p. So, on dividing -4p by p, we get \[\dfrac{-4p}{p}=-4\]. So, now multiplying -4 with p-7 and then subtracting, we get,
\[p-7\overline{\left){\begin{align}
  & {{p}^{2}}-11p+28 \\
 & {{p}^{2}}-7p \\
 & \overline{\begin{align}
  & -4p+28 \\
 & -4p+28 \\
 & \overline{\text{ 0 }} \\
\end{align}} \\
\end{align}}\right.}\left( p-4 \right.\]
Hence, when we divide ${{p}^{2}}-11p+28$ by p-7, we the solution as p-4.
Therefore, p-4 is the answer.

Note:
We can also solve this question using an alternative method in which we will expand ${{p}^{2}}-11p+28$ using a middle term splitting method. As we know that 4+7=11 and $4\times 7=28$, we can write, ${{p}^{2}}-11p+28={{p}^{2}}-7p-4p+28$. On simplifying it we will get,
$\Rightarrow \left( p-7 \right)\left( p-4 \right)$
Now, on solving our problem, we will get,
 $\begin{align}
  & \dfrac{\left( p-7 \right)\left( p-4 \right)}{2p}\times \dfrac{2p}{p-7} \\
 & \Rightarrow p-4 \\
\end{align}$
Hence, we get the answer as p-4.