Question

Find the remainder when \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] is divided by \[g\left( x \right)=3x-1\] .

Answer 1

Hint: We know the remainder theorem that when a polynomial \[f\left( x \right)\] is divided by a linear expression \[\left( x-a \right)\] , the remainder will be \[f\left( a \right)\] . Here, \[f\left( x \right)\] is replaced by \[p\left( x \right)\] . Now, make \[g\left( x \right)=3x-1\] equal to zero and get the value of x, and this value of x is equal to a. Now, using the value of a get the value of \[p\left( a \right)\] . The value of \[p\left( a \right)\] is the remainder when the polynomial \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] is divided by the linear expression \[g\left( x \right)=3x-1\] .

Complete step-by-step answer:
According to the question, it is given that we have to find the remainder the polynomial \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] is divided by the linear expression, \[g\left( x \right)=3x-1\] .
\[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] …………………………………….(1)
\[g\left( x \right)=3x-1\] ………………………………(2)
We know the remainder theorem that when a polynomial \[f\left( x \right)\] is divided by a linear expression \[\left( x-a \right)\] , the remainder will be \[f\left( a \right)\] ………………………………………(3)
Here, \[f\left( x \right)\] is replaced by \[p\left( x \right)\] .
From equation (2), we have a linear expression.
Now, on simplifying equation (2) to obtain the zero of \[g\left( x \right)\] , we get
\[\begin{align}
  & \Rightarrow g\left( x \right)=3x-1=0 \\
 & \Rightarrow 3\left( x-\dfrac{1}{3} \right)=0 \\
\end{align}\]
\[\Rightarrow \left( x-\dfrac{1}{3} \right)=0\] ………………………………….(4)
Now, on comparing \[\left( x-\dfrac{1}{3} \right)\] and \[\left( x-a \right)\] , we get
\[a=\dfrac{1}{3}\] ……………………………….(5)
Here, we have to get the value of \[p\left( a \right)\] .
Now, using the remainder theorem shown in equation (3), we can say that to get the remainder when the polynomial \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] is divided by \[g\left( x \right)=3x-1\] , we have to replace x by \[\dfrac{1}{3}\] in the polynomial, \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] .
From equation (1), we have the polynomial \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] .
From equation (5), we have the value of \[a\] .
We have to get the value of \[p\left( a \right)\] and for that we have to replace x by \[a\] in the polynomial \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] .
Now, on putting \[x=\dfrac{1}{3}\] in the polynomial \[p\left( x \right)\] shown in equation (1), we get
\[p\left( \dfrac{1}{3} \right)={{\left( \dfrac{1}{3} \right)}^{3}}-6{{\left( \dfrac{1}{3} \right)}^{2}}+2\left( \dfrac{1}{3} \right)-4=\dfrac{1}{27}-\dfrac{6}{9}+\dfrac{2}{3}-4=\dfrac{-107}{27}\] .
Therefore, the remainder when the polynomial \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] is divided by the linear expression \[g\left( x \right)=3x-1\] is \[\dfrac{-107}{27}\] .

Note: In this question, one might think to divide the polynomial \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] directly by the linear expression \[g\left( x \right)=3x-1\] . If we do so then our calculation will be complex which might lead to calculation mistakes. Also, it will take time to directly divide the polynomial \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+2x-4\] by the linear expression \[g\left( x \right)=3x-1\] .