Answer
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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.
Complete step by step answer:
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,
\[\left( x+1 \right)\overset{{{x}^{10}}-{{x}^{9}}+{{x}^{8}}-{{x}^{7}}+{{x}^{6}}-{{x}^{5}}+{{x}^{4}}-{{x}^{3}}+{{x}^{2}}-x+1}{\overline{\left){\begin{align}
& {{x}^{11}}+1 \\
& \underline{{{x}^{11}}+{{x}^{10}}} \\
& \text{ }0-{{x}^{10}}+1 \\
& \text{ }\underline{-{{x}^{10}}-{{x}^{9}}} \\
& \text{ }0+{{x}^{9}}+1 \\
& \text{ }\underline{{{x}^{9}}+{{x}^{8}}} \\
& \text{ }0-{{x}^{8}}+1 \\
& \text{ }\underline{-{{x}^{8}}-{{x}^{7}}} \\
& \text{ }0+{{x}^{7}}+1 \\
& \text{ }\underline{{{x}^{7}}+{{x}^{6}}} \\
& \text{ }0-{{x}^{6}}+1 \\
& \text{ }\underline{-{{x}^{6}}-{{x}^{5}}} \\
& \text{ }0+{{x}^{5}}+1 \\
& \text{ }\underline{{{x}^{5}}+{{x}^{4}}} \\
& \text{ }0-{{x}^{4}}+1 \\
& \text{ }\underline{-{{x}^{4}}-{{x}^{3}}} \\
& \text{ }0+{{x}^{3}}+1 \\
& \text{ }\underline{{{x}^{3}}+{{x}^{2}}} \\
& \text{ }0-{{x}^{2}}+1 \\
& \text{ }\underline{-{{x}^{2}}-x} \\
& \text{ }0+x+1 \\
& \text{ }\underline{x+1} \\
& \text{ }0 \\
\end{align}}\right.}}\]
So, finally the remainder for this is 0.And if we calculate it by direct method, then,
Remainder $=p\left( -1 \right)={{\left( -1 \right)}^{11}}+1=-1+1=0$
So, in both conditions the remainder is zero and it can be calculated by both the methods, the long division method and the direct method.
So, the correct answer is “Option A”.
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.
Complete step by step answer:
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,
\[\left( x+1 \right)\overset{{{x}^{10}}-{{x}^{9}}+{{x}^{8}}-{{x}^{7}}+{{x}^{6}}-{{x}^{5}}+{{x}^{4}}-{{x}^{3}}+{{x}^{2}}-x+1}{\overline{\left){\begin{align}
& {{x}^{11}}+1 \\
& \underline{{{x}^{11}}+{{x}^{10}}} \\
& \text{ }0-{{x}^{10}}+1 \\
& \text{ }\underline{-{{x}^{10}}-{{x}^{9}}} \\
& \text{ }0+{{x}^{9}}+1 \\
& \text{ }\underline{{{x}^{9}}+{{x}^{8}}} \\
& \text{ }0-{{x}^{8}}+1 \\
& \text{ }\underline{-{{x}^{8}}-{{x}^{7}}} \\
& \text{ }0+{{x}^{7}}+1 \\
& \text{ }\underline{{{x}^{7}}+{{x}^{6}}} \\
& \text{ }0-{{x}^{6}}+1 \\
& \text{ }\underline{-{{x}^{6}}-{{x}^{5}}} \\
& \text{ }0+{{x}^{5}}+1 \\
& \text{ }\underline{{{x}^{5}}+{{x}^{4}}} \\
& \text{ }0-{{x}^{4}}+1 \\
& \text{ }\underline{-{{x}^{4}}-{{x}^{3}}} \\
& \text{ }0+{{x}^{3}}+1 \\
& \text{ }\underline{{{x}^{3}}+{{x}^{2}}} \\
& \text{ }0-{{x}^{2}}+1 \\
& \text{ }\underline{-{{x}^{2}}-x} \\
& \text{ }0+x+1 \\
& \text{ }\underline{x+1} \\
& \text{ }0 \\
\end{align}}\right.}}\]
So, finally the remainder for this is 0.And if we calculate it by direct method, then,
Remainder $=p\left( -1 \right)={{\left( -1 \right)}^{11}}+1=-1+1=0$
So, in both conditions the remainder is zero and it can be calculated by both the methods, the long division method and the direct method.
So, the correct answer is “Option A”.
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.
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