# What is the remainder when \[\left( {{x}^{11}}+1 \right)\] is divided by \[\left( x+1 \right)\]?

A.0

B.2

C.11

D.12

Answer

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**Hint:**In this problem, we have to find the remainder, when \[\left( {{x}^{11}}+1 \right)\] is divided by \[\left( x+1 \right)\]. Here we can see that the polynomial expression has the highest power, so we can use the method remainder theorem concept and find the remainder. If any polynomial \[f\left( x \right)\] is divided by \[f\left( x-h \right)\], then the remainder will be \[f\left( h \right)\], we can now use this theorem to find the remainder.

**Complete step by step answer:**

Here we have to find the remainder, when \[\left( {{x}^{11}}+1 \right)\] is divided by \[\left( x+1 \right)\].

Here we have high power polynomial expression, so we can use the remainder theorem concept.

We know that, If any polynomial \[f\left( x \right)\] is divided by \[f\left( x-h \right)\], then the remainder will be \[f\left( h \right)\].

So, by using the remainder theorem we can say that

Since, \[\left( x+1 \right)\] is the divisor,

Then the remainder is

\[\Rightarrow \operatorname{R}=f\left( -1 \right)\]

We can now write as,

\[\begin{align}

& \Rightarrow f\left( x \right)={{x}^{11}}+1 \\

& \Rightarrow f\left( -1 \right)={{\left( -1 \right)}^{11}}+1=-1+1=0 \\

\end{align}\]

Hence, the remainder is 0 when \[\left( {{x}^{11}}+1 \right)\] is divided by \[\left( x+1 \right)\].

**So, the correct answer is “Option A”.**

**Note:**We should always focus on the divisor of the polynomial which helps to give the remainder value. We should also know that the remainder theorem only works when a function is divided by a linear polynomial, which is of the form x+ number or x- number. Here we have not used the polynomial long division or synthetic division methods as the given polynomial expression has the highest power raised to it.

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