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Find in each case, the remainder when:
(i) ${{x}^{4}}-3{{x}^{2}}+2x+1$ is divided by $\left( x-1 \right)$
(ii) ${{x}^{3}}+3{{x}^{2}}-12x+4$ is divided by $\left( x-2 \right)$
(iii) ${{x}^{4}}+1$ is divided by $\left( x+1 \right)$

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Last updated date: 19th Jul 2024
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Answer
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Hint: In this type of question we have to use the concept of remainder theorem. We know that remainder theorem states that, If a polynomial \[P\left( x \right)\] is divided by the binomial \[\left( x-a \right)\], then the remainder obtained is equal to \[P\left( a \right)\]. Here, first we find the value of \[x\] by equating the second binomial to zero and then we find the remainder of the first polynomial by using the remainder theorem.

Complete step by step answer:
Now, we have to find the remainder in each case
(i) Let $f\left( x \right)={{x}^{4}}-3{{x}^{2}}+2x+1$ and $g\left( x \right)=\left( x-1 \right)$
Now, let us consider,
$\begin{align}
  & \Rightarrow g\left( x \right)=0 \\
 & \Rightarrow \left( x-1 \right)=0 \\
 & \Rightarrow x=1 \\
\end{align}$
Now, as remainder theorem states that, If a polynomial \[P\left( x \right)\] is divided by the binomial \[\left( x-a \right)\], then the remainder obtained is equal to \[P\left( a \right)\]. Hence, the remainder of $f\left( x \right)={{x}^{4}}-3{{x}^{2}}+2x+1$ when it is divided by $g\left( x \right)=\left( x-1 \right)$ is $f\left( 1 \right)$.
Hence, now we have to find the value of $f\left( 1 \right)$
As $f\left( x \right)={{x}^{4}}-3{{x}^{2}}+2x+1$
$\begin{align}
  & \Rightarrow f\left( 1 \right)={{1}^{4}}-3{{\left( 1 \right)}^{2}}+2\left( 1 \right)+1 \\
 & \Rightarrow f\left( 1 \right)=1-3+2+1 \\
 & \Rightarrow f\left( 1 \right)=1 \\
\end{align}$
Hence, the remainder of $f\left( x \right)={{x}^{4}}-3{{x}^{2}}+2x+1$ when it is divided by $g\left( x \right)=\left( x-1 \right)$ is $f\left( 1 \right)=1$
(ii) Let $f\left( x \right)={{x}^{3}}+3{{x}^{2}}-12x+4$ and $g\left( x \right)=\left( x-2 \right)$
Now, let us consider,
$\begin{align}
  & \Rightarrow g\left( x \right)=0 \\
 & \Rightarrow \left( x-2 \right)=0 \\
 & \Rightarrow x=2 \\
\end{align}$
Now, as remainder theorem states that, If a polynomial \[P\left( x \right)\] is divided by the binomial \[\left( x-a \right)\], then the remainder obtained is equal to \[P\left( a \right)\]. Hence, the remainder of $f\left( x \right)={{x}^{3}}+3{{x}^{2}}-12x+4$ when it is divided by $g\left( x \right)=\left( x-2 \right)$ is $f\left( 2 \right)$.
Hence, now we have to find the value of $f\left( 2 \right)$
As $f\left( x \right)={{x}^{3}}+3{{x}^{2}}-12x+4$
$\begin{align}
  & \Rightarrow f\left( 2 \right)={{2}^{3}}+3{{\left( 2 \right)}^{2}}-12\left( 2 \right)+4 \\
 & \Rightarrow f\left( 2 \right)=8+12-24+4 \\
 & \Rightarrow f\left( 2 \right)=0 \\
\end{align}$
Hence, the remainder of $f\left( x \right)={{x}^{3}}+3{{x}^{2}}-12x+4$ when it is divided by $g\left( x \right)=\left( x-2 \right)$ is $f\left( 2 \right)=0$
(iii) Let $f\left( x \right)={{x}^{4}}+1$ and $g\left( x \right)=\left( x+1 \right)$
Now, let us consider,
$\begin{align}
  & \Rightarrow g\left( x \right)=0 \\
 & \Rightarrow \left( x+1 \right)=0 \\
 & \Rightarrow x=-1 \\
\end{align}$
Now, as remainder theorem states that, If a polynomial \[P\left( x \right)\] is divided by the binomial \[\left( x-a \right)\], then the remainder obtained is equal to \[P\left( a \right)\]. Hence, the remainder of $f\left( x \right)={{x}^{4}}+1$ when it is divided by $g\left( x \right)=\left( x+1 \right)$ is $f\left( -1 \right)$.
Hence, now we have to find the value of $f\left( -1 \right)$
As $f\left( x \right)={{x}^{4}}+1$
$\begin{align}
  & \Rightarrow f\left( -1 \right)={{\left( -1 \right)}^{4}}+1 \\
 & \Rightarrow f\left( -1 \right)=1+1 \\
 & \Rightarrow f\left( -1 \right)=2 \\
\end{align}$
Hence, the remainder of $f\left( x \right)={{x}^{4}}+1$ when it is divided by $g\left( x \right)=\left( x+1 \right)$ is $f\left( -1 \right)=2$

Note: In this type of question students have to remember the remainder theorem. Also students have to note that we can apply the remainder theorem only if the divisor is in the form \[\left( x-a \right)\]. Also students have to take care during the calculation.