Question

On dividing \[{{x}^{3}}-3{{x}^{2}}+x+2\] by a polynomial \[g\left( x \right)\], the quotient and remainder were
\[x-2\] and \[4-2x\] respectively, then \[g\left( x \right)\] is
(a) \[{{x}^{2}}+x+1\]
(b) \[{{x}^{2}}+x-1\]
(c) \[{{x}^{2}}-x-1\]
(d) \[{{x}^{2}}-x+1\]

Answer 1

Hint: For solving this problem first we use general definition of division that is \[\text{dividend = divisor}\times \text{quotient + remainder}\]. Here, we are given that the remainder as a linear function so the divisor will be a quadratic function. We assume \[g\left( x \right)\] as general function of quadratic that is \[g\left( x \right)=a{{x}^{2}}+bx+c\] and then by taking any three values of \['x'\] we can find values of \[a,b,c\] to get \[g\left( x \right)\].

Complete step-by-step solution
Let us assume that given function as
\[f\left( x \right)={{x}^{3}}-3{{x}^{2}}+x+2\]
Let us assume that quotient and remainder as
\[\begin{align}
  & q\left( x \right)=x-2 \\
 & r\left( x \right)=4-2x \\
\end{align}\]
Here, since the remainder is a linear function then \[g\left( x \right)\] is quadratic.
Let us assume that
\[g\left( x \right)=a{{x}^{2}}+bx+c\].
Now by using definition of division we will get
\[\Rightarrow f\left( x \right)=g\left( x \right)q\left( x \right)+r\left( x \right)\]
By substituting the functions in above equation we will get
\[\begin{align}
  & \Rightarrow {{x}^{3}}-3{{x}^{2}}+x+2=\left( a{{x}^{2}}+bx+c \right)\left( x-2 \right)+4-2x \\
 & \Rightarrow {{x}^{3}}-3{{x}^{2}}+x+2-\left( 4-2x \right)=\left( a{{x}^{2}}+bx+c \right)\left( x-2 \right) \\
 & \Rightarrow {{x}^{3}}-3{{x}^{2}}+3x-2=\left( a{{x}^{2}}+bx+c \right)\left( x-2 \right)........equation(i) \\
\end{align}\]
Now for finding the values of \[a,b,c\] let us assume \[x=0,1,3\]
By substituting \[x=0\] in equation (i) we get
\[\begin{align}
  & \Rightarrow -2=\left( c \right)\left( -2 \right) \\
 & \Rightarrow c=1 \\
\end{align}\]
Now by substituting \[x=1\] in equation (i) we get
\[\begin{align}
  & \Rightarrow \left( 1-3+3-2 \right)=\left( a+b+c \right)\left( 1-2 \right) \\
 & \Rightarrow \left( a+b+1 \right)\left( -1 \right)=-1 \\
 & \Rightarrow a+b=0........equation(ii) \\
\end{align}\]
Bow by substituting \[x=3\] in equation (i) we get
\[\begin{align}
  & \Rightarrow 27-27+9-2=\left( 9a+3b+1 \right)\left( 3-2 \right) \\
 & \Rightarrow 9a+3b+1=7 \\
 & \Rightarrow 9a+3b=6 \\
 & \Rightarrow 3a+b=2........equation(iii) \\
\end{align}\]
Now, by subtracting equation (ii) from equation (iii) we get
\[\begin{align}
  & \Rightarrow \left( 3a+b \right)-\left( a+b \right)=2-0 \\
 & \Rightarrow 2a=2 \\
 & \Rightarrow a=1 \\
\end{align}\]
Let us substitute value of \[a\] in equation (ii) we get
\[\begin{align}
  & \Rightarrow 1+b=0 \\
 & \Rightarrow b=-1 \\
\end{align}\]
Now we got all the values of \[a,b,c\].
Let us substitute the values of \[a,b,c\] in \[g\left( x \right)=a{{x}^{2}}+bx+c\] to get \[g\left( x \right)\]
\[\Rightarrow g\left( x \right)={{x}^{2}}-x+1\]
Therefore, option (d) is the correct answer.

Note: This question can be solved in another method.
 Let us assume that given function as
\[f\left( x \right)={{x}^{3}}-3{{x}^{2}}+x+2\]
Let us assume that quotient and remainder as
\[\begin{align}
  & q\left( x \right)=x-2 \\
 & r\left( x \right)=4-2x \\
\end{align}\]
Here, since the remainder is a linear function then \[g\left( x \right)\] is quadratic.
Now by using definition of division we will get
\[\Rightarrow f\left( x \right)=g\left( x \right)q\left( x \right)+r\left( x \right)\]
By substituting the functions in above equation we will get
\[\begin{align}
  & \Rightarrow {{x}^{3}}-3{{x}^{2}}+x+2=g\left( x \right)\left( x-2 \right)+4-2x \\
 & \Rightarrow {{x}^{3}}-3{{x}^{2}}+x+2-\left( 4-2x \right)=g\left( x \right)\left( x-2 \right) \\
 & \Rightarrow {{x}^{3}}-3{{x}^{2}}+3x-2=g\left( x \right)\left( x-2 \right) \\
 & \Rightarrow g\left( x \right)=\dfrac{{{x}^{3}}-3{{x}^{2}}+3x-2}{x-2} \\
\end{align}\]
By evaluating the above division we will get
\[\Rightarrow g\left( x \right)={{x}^{2}}-x+1\]
So, option (d) is the correct answer.