Answer
Verified
375.6k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE