Question

If $ {{x}^{3}}+m{{x}^{2}}-x+6 $ has $ \left( x-2 \right) $ as a factor and leaves remainder $ n $ when divided by $ \left( x-3 \right) $ . Find the value of $ m $ and $ n $ ?

Answer 1

Hint: For answering this question we will use the statement for any polynomial $ P\left( x \right) $ when it is divided by $ \left( x-a \right) $ then the remainder will be $ P\left( a \right) $ . Here from the question we have $ P\left( x \right)={{x}^{3}}+m{{x}^{2}}-x+6 $ and it is divided by $ \left( x-3 \right) $ so the remainder will be $ P\left( 3 \right) $ which is given as $ n $ in the question. And in the question it is also given that $ \left( x-2 \right) $ is the factor of $ P\left( x \right) $ which implies that when $ P\left( x \right) $ is divided by $ \left( x-2 \right) $ the remainder will be zero. So we can say that $ P\left( 2 \right)=0 $ .

Complete step by step answer:
Now considering from the question we have a polynomial $ {{x}^{3}}+m{{x}^{2}}-x+6 $
Let us assume $ P\left( x \right)={{x}^{3}}+m{{x}^{2}}-x+6 $ .
From the question, we have been given that $ \left( x-2 \right) $ is a factor of the given polynomial.
Therefore $ 2 $ is one of the zero of the polynomial.
This implies that when $ P\left( x \right) $ is divided by $ \left( x-2 \right) $ the remainder will be zero. So we can say that $ P\left( 2 \right)=0 $ .
By using this we can get the value of $ m $
There we will substitute the values in $ P\left( 2 \right)=0 $ then we will have

$ \begin{align}
  & \Rightarrow {{\left( 2 \right)}^{3}}+m{{\left( 2 \right)}^{2}}-2+6=0 \\
 & \Rightarrow 8+4m+4=0 \\
 & \Rightarrow 12+4m=0 \\
 & \Rightarrow 12=-4m \\
 & \Rightarrow m=-3 \\
\end{align} $
Now, by substituting the $ m $ value in the given polynomial $ P\left( x \right) $ it becomes $ P\left( x \right)={{x}^{3}}-3{{x}^{2}}-x+6 $ .
From the question we have been given that when the polynomial $ P\left( x \right) $ is divided by $ \left( x-3 \right) $ we will have the remainder given as $ n $ .
By using the statement for any polynomial $ P\left( x \right) $ when it is divided by $ \left( x-a \right) $ then the remainder will be $ P\left( a \right) $ we can say that $ P\left( 3 \right)=n $ .
By simplifying this we will have
 $ \begin{align}
  & {{\left( 3 \right)}^{3}}-3{{\left( 3 \right)}^{2}}-\left( 3 \right)+6=n \\
 & \Rightarrow 27-27+3=n \\
 & \Rightarrow n=3 \\
\end{align} $
Therefore we can conclude that If $ {{x}^{3}}+m{{x}^{2}}-x+6 $ has $ \left( x-2 \right) $ as a factor and leaves remainder n when divided by $ \left( x-3 \right) $ then the values of $ m $ and $ n $ are $ -3 $ and $ 3 $ .

Note:
While answering questions of this type we should be sure with calculations. From the basic concept we can say that for any polynomial $ P\left( x \right) $ when it is divided by $ \left( x-a \right) $ then the remainder will be $ P\left( a \right) $ . The proof for this statement can be done using $ \text{Divider = Dividend }\times \text{ quotient + remainder} $ by substituting these values here we will have $ P\left( x \right)=\left( x-a \right)\times \text{ quotient + remainder} $ . When we substitute $ x=a $ we will have $ P\left( a \right)=remainder $ hence it is proved.