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Last updated date: 03rd Dec 2023
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# What is the remainder when $\left( {{x}^{11}}+1 \right)$ is divided by $\left( x+1 \right)$?A. 0B. 2C. 11D. 12

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Hint: For solving this question we should know about the long division method. This method is used for dividing one large multi digit number into another large multi digit number. The formula for checking that our answer is correct or not is, dividend = divisor $\times$ quotient + remainder.

In the given question it is asked to find the remainder when $\left( {{x}^{11}}+1 \right)$ is divide by $\left( x+1 \right)$. So, as we know that, dividend = divisor $\times$ quotient + remainder, so by long division method,
\left( x+1 \right)\overset{{{x}^{10}}-{{x}^{9}}+{{x}^{8}}-{{x}^{7}}+{{x}^{6}}-{{x}^{5}}+{{x}^{4}}-{{x}^{3}}+{{x}^{2}}-x+1}{\overline{\left){\begin{align} & {{x}^{11}}+1 \\ & \underline{{{x}^{11}}+{{x}^{10}}} \\ & \text{ }0-{{x}^{10}}+1 \\ & \text{ }\underline{-{{x}^{10}}-{{x}^{9}}} \\ & \text{ }0+{{x}^{9}}+1 \\ & \text{ }\underline{{{x}^{9}}+{{x}^{8}}} \\ & \text{ }0-{{x}^{8}}+1 \\ & \text{ }\underline{-{{x}^{8}}-{{x}^{7}}} \\ & \text{ }0+{{x}^{7}}+1 \\ & \text{ }\underline{{{x}^{7}}+{{x}^{6}}} \\ & \text{ }0-{{x}^{6}}+1 \\ & \text{ }\underline{-{{x}^{6}}-{{x}^{5}}} \\ & \text{ }0+{{x}^{5}}+1 \\ & \text{ }\underline{{{x}^{5}}+{{x}^{4}}} \\ & \text{ }0-{{x}^{4}}+1 \\ & \text{ }\underline{-{{x}^{4}}-{{x}^{3}}} \\ & \text{ }0+{{x}^{3}}+1 \\ & \text{ }\underline{{{x}^{3}}+{{x}^{2}}} \\ & \text{ }0-{{x}^{2}}+1 \\ & \text{ }\underline{-{{x}^{2}}-x} \\ & \text{ }0+x+1 \\ & \text{ }\underline{x+1} \\ & \text{ }0 \\ \end{align}}\right.}}
Remainder $=p\left( -1 \right)={{\left( -1 \right)}^{11}}+1=-1+1=0$
Note: During the long division method, we should be very much careful while dividing the ${{x}^{n}}$ values and putting the values under the same power of $x$ and after this you must change the sign if the value has positive sign then it will be negative and if value has negative then it will be positive.