
How do you long divide $\left( {{x}^{5}}-1 \right)\div \left( x-1 \right)$ ?
Answer
542.1k+ views
Hint: Here in this question we have been asked to perform a long division and find the value of the expression $\left( {{x}^{5}}-1 \right)\div \left( x-1 \right)$. For answering this question we need to check the first term of dividend polynomial ${{x}^{5}}-1$ is divisible by the divisor $x-1$ and consider the nearest multiple and make it as the first term of the quotient and repeat the same process till the remainder is zero.
Complete step-by-step solution:
Now considering from the question we have been asked to perform long division and find the value of the expression $\left( {{x}^{5}}-1 \right)\div \left( x-1 \right)$.
For answering this question we need to check the first term of dividend polynomial ${{x}^{5}}-1$ is divisible by the divisor $x-1$ and consider the nearest multiple and make it as the first term of the quotient and repeat the same process till the remainder is zero.
So firstly the first term of dividend is ${{x}^{5}}$ therefore we will consider ${{x}^{4}}$ as the first term of quotient.
$\begin{align}
& \text{ }{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1 \\
& x-1\left| \!{\overline {\,
\begin{align}
& {{x}^{5}}-1 \\
& {{x}^{5}}-{{x}^{4}} \\
& - \; + \\
& \_\_\_\_\_\_ \\
& {{x}^{4}}-1 \\
& {{x}^{4}}-{{x}^{3}} \\
& - \; + \\
& \_\_\_\_\_\_ \\
& {{x}^{3}}-1 \\
& {{x}^{3}}-{{x}^{2}} \\
& - \; + \\
& \_\_\_\_\_\_ \\
& {{x}^{2}}-1 \\
& {{x}^{2}}-x \\
& - \; + \\
& \_\_\_\_\_ \\
& x-1 \\
& x-1 \\
& - \; + \\
& \_\_\_\_\_ \\
& 0 \\
\end{align} \,}} \right. \\
\end{align}$
Therefore we can conclude that the quotient of the long division $\left( {{x}^{5}}-1 \right)\div \left( x-1 \right)$ will be ${{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1$.
Note: While answering questions of this type we should be sure with 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. This is similar to basic arithmetic division for example $24\div 2$ the quotient will be given by
$\begin{align}
& \text{ }12 \\
& 2\left| \!{\overline {\,
\begin{align}
& 24 \\
& 2 \\
& - \\
& \_\_\_\_\_\_ \\
& \text{ 0}4 \\
& \text{ 0}4 \\
& - \\
& \_\_\_\_\_\_ \\
& 000 \\
\end{align} \,}} \right. \\
\end{align}$.
Complete step-by-step solution:
Now considering from the question we have been asked to perform long division and find the value of the expression $\left( {{x}^{5}}-1 \right)\div \left( x-1 \right)$.
For answering this question we need to check the first term of dividend polynomial ${{x}^{5}}-1$ is divisible by the divisor $x-1$ and consider the nearest multiple and make it as the first term of the quotient and repeat the same process till the remainder is zero.
So firstly the first term of dividend is ${{x}^{5}}$ therefore we will consider ${{x}^{4}}$ as the first term of quotient.
$\begin{align}
& \text{ }{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1 \\
& x-1\left| \!{\overline {\,
\begin{align}
& {{x}^{5}}-1 \\
& {{x}^{5}}-{{x}^{4}} \\
& - \; + \\
& \_\_\_\_\_\_ \\
& {{x}^{4}}-1 \\
& {{x}^{4}}-{{x}^{3}} \\
& - \; + \\
& \_\_\_\_\_\_ \\
& {{x}^{3}}-1 \\
& {{x}^{3}}-{{x}^{2}} \\
& - \; + \\
& \_\_\_\_\_\_ \\
& {{x}^{2}}-1 \\
& {{x}^{2}}-x \\
& - \; + \\
& \_\_\_\_\_ \\
& x-1 \\
& x-1 \\
& - \; + \\
& \_\_\_\_\_ \\
& 0 \\
\end{align} \,}} \right. \\
\end{align}$
Therefore we can conclude that the quotient of the long division $\left( {{x}^{5}}-1 \right)\div \left( x-1 \right)$ will be ${{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x+1$.
Note: While answering questions of this type we should be sure with 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. This is similar to basic arithmetic division for example $24\div 2$ the quotient will be given by
$\begin{align}
& \text{ }12 \\
& 2\left| \!{\overline {\,
\begin{align}
& 24 \\
& 2 \\
& - \\
& \_\_\_\_\_\_ \\
& \text{ 0}4 \\
& \text{ 0}4 \\
& - \\
& \_\_\_\_\_\_ \\
& 000 \\
\end{align} \,}} \right. \\
\end{align}$.
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