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Question

Answers

(A) 30

(B) 31

(C) -1

(d) 0

Answer
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The given polynomial \[f\left( x \right)\] is \[{{x}^{31}}+31\] which is a polynomial of degree 31.

We can write \[g\left( x \right)\] as \[g\left( x \right)=x+1\] which is a polynomial of degree 1 , hence a linear polynomial.

Then we know the Remainder Theorem –Let \[f\left( x \right)\] be the polynomial of degree \[n\]. Then on dividing \[f\left( x \right)\] by a linear polynomial \[x-a\], the remainder is equal to \[f\left( a \right)\]. That is \[x-a\] is a divisor of \[f\left( x \right)\] if and only if \[f\left( a \right)=0\].

Now comparing the linear polynomial \[g\left( x \right)=x+1\] with \[x-a\] , we get

\[a=-1\]

Now in order find the remainder when the function \[f\left( x \right)={{x}^{31}}+31\] is divided by \[g\left( x \right)=x+1\], we will find \[f\left( -1 \right)\].

Using \[{{\left( -1 \right)}^{2n+1}}=-1\] where \[2n+1\] is an odd natural number, we have

\[\begin{align}

& f\left( -1 \right)={{\left( -1 \right)}^{31}}+31 \\

& =-1+31 \\

& =30 \\

\end{align}\]

Thus using the remainder theorem, we get that if \[{{x}^{31}}+31\] is divided by a linear polynomial \[x+1\], the remainder comes out to be 30.