Question

How do you use synthetic division to divide $\dfrac{6{{x}^{3}}-9{{x}^{2}}-12x-6}{3x+4}$ ?

Answer 1

Hint: Here in this question we have been asked to perform synthetic division between the given polynomial expressions that is $\dfrac{6{{x}^{3}}-9{{x}^{2}}-12x-6}{3x+4}$ . Here $3x+4$ is the divisor and $6{{x}^{3}}-9{{x}^{2}}-12x-6$ is the dividend. Now we need to perform synthetic division between them and derive the quotient.

Complete step by step solution:
Now considering from the question we have been asked to perform synthetic division between the given polynomial expressions that is $\dfrac{6{{x}^{3}}-9{{x}^{2}}-12x-6}{3x+4}$ .
Here $3x+4$ is the divisor and $6{{x}^{3}}-9{{x}^{2}}-12x-6$ is the dividend.
Now we need to perform synthetic division between them and derive the quotient.
For performing synthetic division we need to write only the coefficients of $x$ inside a L-shape.
Then we will have $\left| \!{\underline {\,
  \begin{matrix}
   6 & -9 & -12 & -6 \\
   {} & {} & {} & {} \\
\end{matrix} \,}} \right. $.
The divisor is $3x+4$, so we need to evaluate for $x=\dfrac{-4}{3}$ .
Now we will have $\begin{matrix}
   \dfrac{-4}{3} \\
   {} \\
\end{matrix}\left| \!{\underline {\,
  \begin{matrix}
   6 & -9 & -12 & -6 \\
   {} & {} & {} & {} \\
\end{matrix} \,}} \right. $ .
Now we need to drop the first coefficient of the dividend below the division symbol then we will have $\begin{align}
  & \begin{matrix}
   \dfrac{-4}{3} \\
   {} \\
\end{matrix}\left| \!{\underline {\,
  \begin{matrix}
   6 & -9 & -12 & -6 \\
   {} & {} & {} & {} \\
\end{matrix} \,}} \right. \\
 & \begin{matrix}
   {} & 6 & {} & {} & {} \\
\end{matrix} \\
\end{align}$ .
Now we need to multiply the drop-down by the divisor and put the result in the next column and then drop-down the sum and repeat this process for all coefficients remaining. By doing this we will have
$\begin{align}
  & \begin{matrix}
   \dfrac{-4}{3} \\
   {} \\
\end{matrix}\left| \!{\underline {\,
  \begin{matrix}
   6 & -9 & -12 & -6 \\
   {} & -8 & \dfrac{68}{3} & \dfrac{-128}{9} \\
\end{matrix} \,}} \right. \\
 & \begin{matrix}
   {} & 6 & -17 & \dfrac{32}{3} & \dfrac{-182}{9} \\
\end{matrix} \\
\end{align}$
Therefore we can conclude that the result of the synthetic division $\dfrac{6{{x}^{3}}-9{{x}^{2}}-12x-6}{3x+4}$ will be given as $2{{x}^{2}}-\dfrac{17}{3}x+\dfrac{32}{9}$ and the remainder will be $\dfrac{-182}{9}$ .

Note: While answering questions of this type we should be sure with our concepts that we are going to apply in the process and the calculations that we are going to perform in between the steps. This is a very simple and easy question and can be answered accurately in a short span of time. Very few mistakes are possible in questions of this type. This question can also be answered by performing simple basic polynomial division as follows:
$\begin{align}
  & \text{ }2{{x}^{2}}-\dfrac{17}{3}x+\dfrac{32}{9} \\
 & 3x+4\left| \!{\overline {\,
 \begin{align}
  & 6{{x}^{3}}-9{{x}^{2}}-12x-6 \\
 & 6{{x}^{3}}+8{{x}^{2}} \\
 & \_\_\_\_\_\_\_\_\_\_\_ \\
 & -17{{x}^{2}}-12x-6 \\
 & -17{{x}^{2}}-\dfrac{68x}{3} \\
 & \_\_\_\_\_\_\_\_ \\
 & \dfrac{-32x}{3}-6 \\
 & \dfrac{-32x}{3}+\dfrac{128}{9} \\
 & \_\_\_\_\_\_\_\_\_\_\_ \\
 & \dfrac{-182}{9} \\
\end{align} \,}} \right. \\
\end{align}$