Question

How do you divide $\left( {{x}^{4}}+{{x}^{3}}-1 \right)\div \left( x-2 \right)$ using synthetic division?

Answer 1

Hint: Firstly, we have to write the dividend polynomial in the standard from covering all the powers of x from four to zero. Therefore the dividend has to be written as \[{{x}^{4}}+{{x}^{3}}+\left( 0 \right){{x}^{2}}+\left( 0 \right)x-1\]. Then we need to write down all the coefficients of the different powers of x inside a box and the root of the denominator on the outside of the box. Then we will bring down the first coefficient with which the number outside the box will be multiplied and the resultant number will be put under the next coefficient. The next coefficient will then be added to the number below it and the resultant number will be written below the next coefficient. This process will be repeated till the numbers under the box become equal in number to the number of coefficients. The last number under the box will be the remainder and the rest of the numbers will be the coefficients of the quotient.

Complete step by step solution:
According to the question, we need to divide the polynomial $\left( {{x}^{4}}+{{x}^{3}}-1 \right)$ by the polynomial $\left( x-2 \right)$
For using the synthetic division, we need to write the coefficients of all the powers of x in the dividend. Therefore, we write the dividend polynomial $\left( {{x}^{4}}+{{x}^{3}}-1 \right)$ as
$\Rightarrow {{x}^{4}}+{{x}^{3}}+\left( 0 \right){{x}^{2}}+\left( 0 \right)x-1$
We will place these coefficients inside a box in the same order as written above. Also, the root of the divisor $\left( x-2 \right)$, which is equal to $2$, outside the box, as shown below.
\[2\left| \!{\underline {\,
  \begin{matrix}
   1 & 1 & 0 & 0 & -1 \\
\end{matrix} \,}} \right. \]
Now, we will bring down the first coefficient, which is equal to one, as shown below.
\[\begin{align}
  & 2\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & 1 & 0 & 0 & -1 \\
\end{matrix} \\
 & \downarrow \\
\end{align} \,}} \right. \\
 & \text{ 1} \\
\end{align}\]
Now, we will multiply the number down the box, equal to $1$, with the number outside the box, equal to $2$, and write the obtained number, $2\times 1=2$, below the next coefficient, as shown below.
\[\begin{align}
  & 2\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & 1 & 0 & 0 & -1 \\
\end{matrix} \\
 & \downarrow \text{ }2 \\
\end{align} \,}} \right. \\
 & \text{ 1} \\
\end{align}\]
Now, we will add the upper coefficient and the number below it to get
\[\begin{align}
  & 2\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & 1 & 0 & 0 & -1 \\
\end{matrix} \\
 & \downarrow \text{ }2 \\
\end{align} \,}} \right. \\
 & \text{ 1 }3 \\
\end{align}\]
Now, we will multiply the number down the box, equal to $3$, with the number outside the box, equal to $2$, and write the result below the next coefficient as shown.
\[\begin{align}
  & 2\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & 1 & 0 & 0 & -1 \\
\end{matrix} \\
 & \downarrow \text{ }2\text{ }6 \\
\end{align} \,}} \right. \\
 & \text{ 1 }3 \\
\end{align}\]
Now, we will add the coefficient and the number below it.
\[\begin{align}
  & 2\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & 1 & 0 & 0 & -1 \\
\end{matrix} \\
 & \downarrow \text{ }2\text{ }6 \\
\end{align} \,}} \right. \\
 & \text{ 1 }3\text{ }6 \\
\end{align}\]
Following the same procedure, we will get
\[\begin{align}
  & 2\left| \!{\underline {\,
  \begin{align}
  & \begin{matrix}
   1 & 1 & 0 & 0 & -1 \\
\end{matrix} \\
 & \downarrow \text{ }2\text{ }6\text{ }12\text{ 24} \\
\end{align} \,}} \right. \\
 & \text{ 1 }3\text{ }6\text{ }12\text{ }23 \\
\end{align}\]
The last number, equal to \[23\], is the remainder of the division, while the other numbers represent the coefficients of the quotient. Thus, the quotient is ${{x}^{3}}+3{{x}^{2}}+6{{x}^{1}}+12$ and the remainder is equal to \[23\].

Note: The synthetic division method can only be used when the divisor is a linear polynomial. We must keep in mind the fact that the degree of the quotient is one less than that of the dividend. That’s why we kept the highest power of x in the quotient equal to three, which is one less than four, the degree of the dividend $\left( {{x}^{4}}+{{x}^{3}}-1 \right)$.