Question

How do you divide $\left( {{x}^{4}}-6{{x}^{3}}-40x+33 \right)\div \left( x-7 \right)$ using synthetic division?

Answer 1

Hint: In this question, we have to divide a given polynomial of four degree with a linear polynomial using synthetic division. For this, we will just use the coefficient of dividend and the root exhibited by the divisor to solve. We will get the result in the form of coefficient. These coefficients will be in the order for $x^3, x^2, x\ and\ 1$. The last term will give us a remainder.

Complete step-by-step answer:
Here we are given the polynomial as $\left( {{x}^{4}}-6{{x}^{3}}-40x+33 \right)\text{ and }\left( x-7 \right)$. We need to divide terms using synthetic division. Here the dividend is equal to $\left( {{x}^{4}}-6{{x}^{3}}-40x+33 \right)$ and the divisor is equal to x-7. Let us first find the root of x-7 i.e. x-7 = 0 Therefore x = 7.
So we will divide the coefficient of $\left( {{x}^{4}}-6{{x}^{3}}-40x+33 \right)$ by 7. The division like configuration looks like this, $07\left| \!{\underline {\,
  \begin{matrix}
   1 & -6 & 0 & -40 & 33 \\
\end{matrix} \,}} \right. $.
Let us place the first number in the division as the first position of result. The result is written below the line i.e. we get $\begin{align}
  & 07\left| \!{\underline {\,
  \begin{matrix}
   1 & -6 & 0 & -40 & 33 \\
\end{matrix} \,}} \right. \\
 & \begin{matrix}
   {} & 1 \\
\end{matrix} \\
\end{align}$.
Now let us multiply 7 with first position of result and place the product down the second coefficient of divided we get \[\begin{align}
  & 07\left| \underline{\begin{matrix}
   1 & -6 & 0 & -40 & 33 \\
   {} & 7 & {} & {} & {} \\
\end{matrix}} \right. \\
 & \begin{matrix}
   {} & 1 & {} & {} & {} \\
\end{matrix} \\
\end{align}\].
Now let us add the product 7 to the coefficient -6 and write it on the second position of result we get \[\begin{align}
  & 07\left| \underline{\begin{matrix}
   1 & -6 & 0 & -40 & 33 \\
   {} & 7 & {} & {} & {} \\
\end{matrix}} \right. \\
 & \begin{matrix}
   {} & 1 & 01 & {} & {} \\
\end{matrix} \\
\end{align}\].
Similarly taking the product of 7 with number at second position of result and adding to the third number of coefficient of divided we get \[\begin{align}
  & 07\left| \underline{\begin{matrix}
   1 & -6 & 0 & -40 & 33 \\
   {} & 7 & 7 & {} & {} \\
\end{matrix}} \right. \\
 & \begin{matrix}
   {} & 1 & 01 & 7 & {} \\
\end{matrix} \\
\end{align}\].
Let us multiply the new entry 7 with 7 again to get 49 and then place it under the fourth coefficient of divided. After that let us add the product 49 with fourth coefficient of dividend we get \[\begin{align}
  & 07\left| \underline{\begin{matrix}
   1 & -6 & 0 & -40 & 33 \\
   {} & 7 & 7 & 49 & {} \\
\end{matrix}} \right. \\
 & \begin{matrix}
   {} & 1 & 01 & 07 & 09 \\
\end{matrix} \\
\end{align}\].
Let us multiply the new entry 9 by 7 to get 63 and then place it under the last coefficient of dividend. After that, let us add the product 63 with fifth coefficient of dividend we get \[\begin{align}
  & 07\left| \underline{\begin{matrix}
   1 & -6 & 0 & -40 & 33 \\
   {} & 7 & 7 & 49 & 63 \\
\end{matrix}} \right. \\
 & \begin{matrix}
   {} & 1 & 01 & 07 & 09 & 96 \\
\end{matrix} \\
\end{align}\].
Here all the numbers of results except the last becomes the coefficient of quotient polynomial. The value of the result is the remainder. Therefore, the quotient polynomial becomes $\left( 1 \right){{x}^{3}}-\left( 1 \right){{x}^{2}}-\left( 7 \right){{x}^{2}}+9$.
Simplifying it we get ${{x}^{3}}-{{x}^{2}}-7{{x}^{2}}+9$.
The remainder is 96.
This is our required answer.

Note: Students should take care of the signs while adding the coefficient with the product. Note that, first position of the result is always coefficient for the highest degree variable of quotient polynomials. If the last answer at result would be zero, then we could say that x-7 divides $\left( {{x}^{4}}-6{{x}^{3}}-40x+33 \right)$ completely.