Question

Determine which of the following is the remainder when the polynomial ${{x}^{31}}+31$ is divided by $x+1$.
(A) 30
(B) 31
(C) -1
(d) 0

Answer 1

In this we can divide the polynomial ${{x}^{31}}+31$ in variable $x$ by $x+1$ until the degree of the remainder in less than the degree of $x+1$. i.e., the degree of the remainder function is 0. Or we can also use the remainder theorem which states that –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 $f\left( a \right)=0$. So in order to find the remainder when ${{x}^{31}}+31$ is divided by $x+1$, find the value of $f\left( -1 \right)$.

Complete step-by-step answer:
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

& f\left( -1 \right)={{\left( -1 \right)}^{31}}+31 \\\ & =-1+31 \\\ & =30 \\\ \end{aligned}$$ Thus using the remainder theorem, we get that if $${{x}^{31}}+31$$ is divided by a linear polynomial $$x+1$$, the remainder comes out to be 30. **So, the correct answer is “Option A”.** **Note:** In this problem, while dividing the polynomial $${{x}^{31}}+31$$ by $$x+1$$, make sure the degree of the remainder is not more the degree of the divisor which is 1. Otherwise the answer would be wrong. Also choosing the quotient while dividing $${{x}^{31}}+31$$ by $$x+1$$ should be done in such a way that at least one of the terms of the dividend gets cancelled if you are doing proper division without using the remainder theorem. But in order to make the calculate short and precise remainder theorem is widely used when the divisor is a linear polynomial.