Question

How do you divide $\left( {{m}^{3}}-13{{m}^{2}}+24m+18 \right)\div \left( m-3 \right)$ using synthetic division?

Answer 1

Hint: In synthetic division we will consider only the coefficients of the dividend and the constant term in the divisor. So, here, the coefficients will be the dividend and the constant term of the divisor will be the divisor.

Complete step by step answer:
Let us consider the given problem where we are asked to find the quotient of two polynomials using synthetic division, $\left( {{m}^{3}}-13{{m}^{2}}+24m+18 \right)\div \left( m-3 \right).$
So, in our problem, $\left( {{m}^{3}}-13{{m}^{2}}+24m+18 \right)$ is the dividend and $\left( m-3 \right)$ is the divisor.
Since we are asked to find the quotient using synthetic division, we need to consider only the coefficients of the variables in the dividend and the constant term in the divisor.
So, we will write the coefficients of the variables in the dividend which will be the new dividend, $1 -13 24 18$
The constant term that will act as the divisor is $3.$
Now, the division will take place as follows:
In the first step, we will write the coefficients in the same order in which they are located in a standard equation. Since our polynomial is already in the standard form, we do not need to change the order.
So, we will get

 $
   \left| \!{\underline {
  \begin{matrix}
   1 & -13 & 24 & 18 \\
   {} & {} & {} & {} \\
\end{matrix}\\ }} \right. \\ \\
$

We will write $3$ in the left side as


 $
   3\left| \!{\underline {
  \begin{matrix}
   1 & -13 & 24 & 18 \\
   {} & {} & {} & {} \\
\end{matrix}\\ }} \right. \\ \\
$

We will directly write $1$ below the line,

 $
   3\left| \!{\underline {
  \begin{matrix}
   1 & -13 & 24 & 18 \\
   {} & {} & {} & {} \\
\end{matrix} }} \right. \\
\begin{matrix}
  & & 1 & {} & {} & {} \\
  \end{matrix} \\
$

Now we will multiply $1$ with $3$ and write the product beneath $-13$ and above the line,

$
   3\left| \!{\underline {
  \begin{matrix}
   1 & -13 & 24 & 18 \\
   {} & 3 & {} & {} \\
\end{matrix} }} \right. \\
   \begin{matrix}
  & & 1 & {} & {} & {} \\
  \end{matrix} \\
$
Now, we will add $-13$ to $3$ and write the sum below the line next to $1$ and then multiply it with $3$ to get the product which is to be written below $24$ above the line. Similarly, we add the values and the sum is multiplied by $3$ to get a product which is to be written in the position below $18.$ Then we will get the required quotient. So, we will get

 $
   3\left| \!{\underline {
  \begin{matrix}
   1 & -13 & 24 & 18 \\
   {} & 3 & -30 & -18 \\
\end{matrix} }} \right. \\
    \begin{matrix}
   & &1 & -10 & -6 & 0 \\
\end{matrix} \\
$

Hence the required second-degree polynomial is ${{m}^{2}}-10m-6.$

Note: The divisor is the term with which we divide, the dividend is the term to be divided, the quotient is the term we obtain as we divide and the remainder is the term that remains.
Here, the quotient is the second-degree polynomial we have obtained and the remainder is $0.$