How do you divide $\left( 2{{n}^{3}}+62n-26{{n}^{2}}+4 \right)\div \left( 2n-6 \right)$ using synthetic division?

Hint: To divide $\left( 2{{n}^{3}}+62n-26{{n}^{2}}+4 \right)\div \left( 2n-6 \right)$ using synthetic division, we will first simplify the dividend and divisor by cancelling their common factors. We will then write down the coefficients of the dividend to the right part. The left section will be the value of x when the divisor is equated to 0 (that is, $x-3=0\Rightarrow x=3$ ). We will bring the leading coefficient to the bottom part. We have to multiply the value of c with the value written in the bottom row. The resultant value must be written below the next coefficient. Now, we will add the second column created. This resultant is written in the bottom row. We have to repeat this. The last value of the bottom row will be the remainder and the remaining values of the bottom row are coefficients of the quotient.

Complete step-by-step solution:
We need to divide $\left( 2{{n}^{3}}+62n-26{{n}^{2}}+4 \right)\div \left( 2n-6 \right)$ using synthetic division. We have $\left( 2{{n}^{3}}+62n-26{{n}^{2}}+4 \right)$ as the dividend and $\left( 2n-6 \right)$ as the divisor.
We can see that the dividend and divisor have a common factor of 2. Hence, we can write
$\begin{align}
  & \dfrac{2\left( {{n}^{3}}+31n-13{{n}^{2}}+2 \right)}{2\left( n-3 \right)} \\
 & \Rightarrow \dfrac{\left( {{n}^{3}}+31n-13{{n}^{2}}+2 \right)}{\left( n-3 \right)} \\
\end{align}$
Thus, the dividend will be $\left( {{n}^{3}}+31n-13{{n}^{2}}+2 \right)$ and divisor will be $\left( n-3 \right)$ .
We have to write the coefficients of the dividend to the right side. The left side will be the c value of the divisor $x-c$ . We can also equate $x-c$ to 0 and find the value of x, which will be c.
$\begin{align}
  & x-c=0 \\
 & \Rightarrow x=c \\
\end{align}$
Now, we can do the same for the divisor $\left( n-3 \right)$ .
$\begin{align}
  & n-3=0 \\
 & \Rightarrow n=3 \\
\end{align}$
Let us write the dividend in standard form as $\left( {{n}^{3}}-13{{n}^{2}}+31n+2 \right)$ .
Now, let us do the synthetic division as shown below.
$3\left| \!{\underline {\,
  \begin{matrix}
   1 & -13 & 31 & 2 \\
\end{matrix} \,}} \right. $
Now, we will bring the leading coefficient, that is, the coefficient of highest degree term, to the bottom part.
\[\begin{align}
  & 3\text{ }\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & -13 & 31 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   \downarrow & \text{ }\text{ }\text{ }\text{ }3 & {} \\
\end{matrix} \\
\end{align} \,}} \right. \\
 & \begin{matrix}
   {} &\text{ }\text{ } 1 & {} \\
\end{matrix} \\
\end{align}\]
Let us multiply the value of c (here its 3) to the value written in the bottom row. We have to write the resultant value below the next coefficient, that is, -13.
\[\begin{align}
  & 3\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & -13 & 31 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   {} & \text{ }\text{ }\text{ }\text{ }3 & {} \\
\end{matrix} \\
\end{align} \,}} \right. \\
 &\text{ }\text{ }\text{ }\text{ }\text{ } \text{ 1} \\
\end{align}\]
Now, we will add the created second column, that is, we will be adding -13 and 3. This resultant value must be written in the bottom row.
\[\begin{align}
  & \text{ }3\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & -13 & 31 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   {} & \text{ }\text{ }\text{ }\text{ }3 & {} \\
\end{matrix} \\
\end{align} \,}} \right. \\
 & \begin{matrix}
   {} & \text{ }\text{ }1 & -10 \\
\end{matrix} \\
\end{align}\]
We will repeat the steps until all the columns get filled. We will multiply 3 and -10 and write the product in the third column.
\[\begin{align}
  & \text{ }3\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & -13 & 31 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   {} & \begin{matrix}
   \text{ }\text{ }\text{ }\text{ }3 & -30 & {} \\
\end{matrix} & {} \\
\end{matrix} \\
\end{align} \,}} \right. \\
 & \begin{matrix}
   {} & \text{ }\text{ }1 & -10 \\
\end{matrix} \\
\end{align}\]
Now, we will sum the third column.
\[\begin{align}
  & \text{ }3\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & -13 & 31 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   {} & \begin{matrix}
   \text{ }\text{ }\text{ }\text{ 3} & -30 & {} \\
\end{matrix} & {} \\
\end{matrix} \\
\end{align} \,}} \right. \\
 & \begin{matrix}
   {} & \text{ }\text{ }1 & \begin{matrix}
   -10 & 1 & {} \\
\end{matrix} \\
\end{matrix} \\
\end{align}\]
We can now multiply 3 with 2 and do the similar operation as above.
\[\begin{align}
  & \text{ }3\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & -13 & 31 & 2 \\
\end{matrix} \\
 & \begin{matrix}
   {} & \begin{matrix}
   \text{ }\text{ }\text{ }\text{ }3 & -30\text{ } \text{ } \text{ 3} & {} \\
\end{matrix} & {} \\
\end{matrix} \\
\end{align} \,}} \right. \\
 & \begin{matrix}
   {} & \text{ }\text{ }1 & \begin{matrix}
   -10 & \text{ }1 & \text{ }5 \\
\end{matrix} \\
\end{matrix} \\
\end{align}\]
In synthetic division, we know that the last value of the bottom row will be the remainder. The remaining values of the bottom row are coefficients of the quotient.
Hence, we can write the quotient as ${{n}^{2}}-10x+1$ and the remainder is 5
Hence, $\left( {{n}^{3}}+31n-13{{n}^{2}}+2 \right)\div \left( n-3 \right)={{n}^{2}}-10n+1+\dfrac{5}{\left( n-3 \right)}$.

Note: Students must simplify the dividend and divisor at the beginning so that the division process becomes easy. They must be careful when writing the coefficients as the coefficients must be in the order of the degree. They must note that when we write the quotient and remainder, it must be of the form $\text{Quotient}+\dfrac{\text{Remainder}}{\text{Divisor}}$ and not like the mixed fraction form of division of the numbers.