Question

On dividing a function ${{x}^{3}}-3{{x}^{2}}+x+2$ by a polynomial g (x), the quotient and remainder were (x-2) and (-2x+4) respectively. Find g (x).

Answer 1

Hint: In order to solve this question, we should know about the concept of division algorithm, when 2 functions a and b are given then, $a=b\times q+r$, where a represents the dividend, b represents the divisor, q represents the quotient and r represents the remainder. By using these concepts, we will solve this question.

Complete step-by-step solution -
In this question, we have been given a function, ${{x}^{3}}-3{{x}^{2}}+x+2$ which gives (x-2) and (-2x+4) as the quotient and the remainder respectively when divided by g (x). And we have been asked to find the function, g (x). To solve this question, we should know about the division algorithm, when 2 functions a and b are given then, $a=b\times q+r$, where a represents the dividend, b represents the divisor, q represents the quotient and r represents the remainder.
So, for the given question, we can say that $a={{x}^{3}}-3{{x}^{2}}+x+2$ and b = g (x), q = (x-2) and r = (-2x+4). We can write the division algorithm as,
${{x}^{3}}-3{{x}^{2}}+x+2=\left[ g\left( x \right) \right]\left[ x-2 \right]+\left[ -2x+4 \right]$
Now, we will simplify it by taking all polynomials on one side and keeping g (x) on the other side. So, we get,
\[\begin{align}
  & \dfrac{\left( {{x}^{3}}-3{{x}^{2}}+x+2 \right)-\left( -2x+4 \right)}{x-2}=g\left( x \right) \\
 & \dfrac{{{x}^{3}}-3{{x}^{2}}+x+2+2x-4}{x-2}=g\left( x \right) \\
 & g\left( x \right)=\dfrac{{{x}^{3}}-3{{x}^{2}}+3x-2}{x-2} \\
\end{align}\]
Now, we will divide \[{{x}^{3}}-3{{x}^{2}}+3x-2\] by (x-2) using a long division method. So, we can write,
\[x-2\overset{{{x}^{2}}-x+1}{\overline{\left){\begin{align}
  & {{x}^{3}}-3{{x}^{2}}+3x-2 \\
 & \underline{{{x}^{3}}-2{{x}^{2}}} \\
 & 0-{{x}^{2}}+3x-2 \\
 & \underline{0-{{x}^{2}}+2x} \\
 & x-2 \\
 & \underline{x-2} \\
 & 0 \\
\end{align}}\right.}}\]
Hence, we can say that \[\dfrac{{{x}^{3}}-3{{x}^{2}}+3x-2}{x-2}\] gives \[{{x}^{2}}-x+1\]. Therefore, we can say that,
\[g\left( x \right)={{x}^{2}}-x+1\]
Hence, we get \[g\left( x \right)={{x}^{2}}-x+1\], when it is given that the polynomial ${{x}^{3}}-3{{x}^{2}}+x+2$ when divided by g (x), it gives (x-2) and (-2x+4) as quotient and remainder respectively.

Note: While solving this question, we need to be very focused as there are high possibilities of making calculation mistakes. Also, we can solve this question without dividing the polynomial, ${{x}^{3}}-3{{x}^{2}}+3x-2$ by (x-2), by taking out (x-2) as common by writing $-3{{x}^{2}}$ as $-2{{x}^{2}}-{{x}^{2}}$ and 3x as (2x + x) and then by further simplifying we will get the answer.