Question

Let $r\left( x \right)$ be the remainder when the polynomial ${{x}^{135}}+{{x}^{125}}-{{x}^{115}}+{{x}^{5}}+1$ is divided by ${{x}^{3}}-x$ then, \[\]
A.$r\left( x \right)$ is the zero polynomial\[\]
B. $r\left( x \right)$is a nonzero constant \[\]
C. Degree of $r\left( x \right)$ is one\[\]
D. Degree of $r\left( x \right)$ is two\[\]

Answer 1

Hint: We use the fact that if the degree of divisor polynomial is $m$ degree of remainder polynomial can be at most $m-1$ and assume the remainder polynomial as $r\left( x \right)=a{{x}^{2}}+bx+c$ where $a,b,c\in \mathsf{\mathbb{R}}$ because the degree of given divisor polynomial $d\left( x \right)={{x}^{3}}-1$. We formulate the equation $p\left( x \right)=q\left( x \right)d\left( x \right)+r\left( x \right)$ where $p\left( x \right)$is the given dividend polynomial and $q\left( x \right)$ is the quotient polynomial. We put factorize $d\left( x \right)$ and put $x=a$ obtained from $d\left( x \right)=0$ in the equation to find $a,b,c$ then $r\left( x \right)$.\[\]

Complete step by step answer:
We know Euclidean division of polynomials that when we divide a divided polynomial $p\left( x \right)$ with degree $n$ by some divisor polynomial $d\left( x \right)$ with degree $m\le n$ then we get the quotient polynomial $q\left( x \right)$ of degree $n-m$ and the remainder polynomial as $r\left( x \right)$ of degree equal to $m-1$ when $n\ne 2m$and equal to at most $m-1$.
\[p\left( x \right)=d\left( x \right)q\left( x \right)+r\left( x \right).....\left( 1 \right)\]
Let us denote the given dividend polynomial in the question as $p\left( x \right)$. We have
\[p\left( x \right)={{x}^{135}}+{{x}^{125}}-{{x}^{115}}+{{x}^{5}}+1\]
Let us denote the divisor polynomial in the question as $d\left( x \right)$.So we have,
\[d\left( x \right)={{x}^{3}}-x\]
We see that the degree of $p\left( x \right)$ is $n=135$ and degree of $d\left( x \right)$ is $=3.$ So the degree of remainder polynomial $r\left( x \right)$ will be at most$3-1=2$. Let us assume that polynomial with degree 2 with some real constants $a,b,c$ as
\[r\left( x \right)=a{{x}^{2}}+bx+c\]
We put $p\left( x \right),d\left( x \right),q\left( x \right),r\left( x \right)$ in the relation of Euclidean division relation (1) as,
 \[{{x}^{135}}+{{x}^{125}}-{{x}^{115}}+{{x}^{5}}+1=\left( {{x}^{3}}-x \right)q\left( x \right)+a{{x}^{2}}+bx+c\]
We further factorize $d\left( x \right)$ by taking $x$ common and the using the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for $a=x,b=1$ and have,
\[\begin{align}
  & \Rightarrow {{x}^{135}}+{{x}^{125}}-{{x}^{115}}+{{x}^{5}}+1=x\left( {{x}^{2}}-1 \right)q\left( x \right)+a{{x}^{2}}+bx+c \\
 & \Rightarrow {{x}^{135}}+{{x}^{125}}-{{x}^{115}}+{{x}^{5}}+1=x\left( x+1 \right)\left( x-1 \right)q\left( x \right)+a{{x}^{2}}+bx+c.....\left( 2 \right) \\
\end{align}\]
Let us put $x=0$ in above equation (2) to have,
\[\begin{align}
  & 0+0+0+0+1=0+0+0+c \\
 & \Rightarrow c=1 \\
\end{align}\]
We put $x=1$ in equation (2) to have,
\[\begin{align}
  & 1+1-1+1+1=0+a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c \\
 & \Rightarrow 3=a+b+1\left( \because c=1 \right) \\
 & \Rightarrow a+b=2.....\left( 3 \right) \\
\end{align}\]
We put $x=-1$ in equation (2) to have,
\[\begin{align}
  & {{\left( -1 \right)}^{135}}+{{\left( -1 \right)}^{125}}-{{\left( -1 \right)}^{115}}+{{\left( -1 \right)}^{5}}+1=0+a{{\left( -1 \right)}^{2}}+b\left( -1 \right)+c \\
 & \Rightarrow -1+\left( -1 \right)+1+\left( -1 \right)+1=a-b+1\left( \because c=1 \right) \\
 & \Rightarrow -1=a-b+1 \\
 & \Rightarrow a-b=-2......\left( 4 \right) \\
\end{align}\]
We solve equation (3) and (4) to have,
\[a=0,b=2\]
So the remainder polynomial is
\[r\left( x \right)=a{{x}^{2}}+bx+c=0\cdot {{x}^{2}}+2\cdot x+1=2x+1\]

So, the correct answer is “Option C”.

Note: We alternatively use remainder theorem find $a,b,c$ which states that that if linear polynomial $x-a$ divides the polynomial $p\left( x \right)$ then it will leave the remainder $p\left( a \right)$ and the divisor $pqr$ of number will leave the same remainder as $p,q,r$. We can also use the congruent modulo relation ${{x}^{3}}-x:{{x}^{1+2n}}\equiv x$ and put ${{x}^{3}}=x$ in $p\left( x \right)$ to get $r\left( x \right)$.