Question

If f(x) = $2{{x}^{4}}-5{{x}^{3}}+{{x}^{2}}+3x-2$ is divided by g(x), the quotient is q(x) = $2{{x}^{2}}-5x+3$ and r(x) = -2x +1 , find g(x):

Answer 1

Hint: We will first write f(x) in the form of g(x), q(x), r(x) and then after doing that we will take all the terms to other side of equality except g(x) and then by performing the required algebraic operations like first subtraction and then we will try to convert the numerator such that it is directly divisible by denominator and then we will cancel the common terms from numerator and denominator to find the value of g(x).

Complete step-by-step answer:
Let’s start the solution by writing f(x) in the form of g(x), q(x), r(x)
$f(x)=g(x)q(x)+r(x)$
Now we will substitute the values of f(x), q(x), r(x)
$g(x)=\dfrac{f(x)-r(x)}{q(x)}$
$\begin{align}
  & g(x)=\dfrac{2{{x}^{4}}-5{{x}^{3}}+{{x}^{2}}+3x-2-\left( -2x+1 \right)}{2{{x}^{2}}-5x+3} \\
 & g(x)=\dfrac{2{{x}^{4}}-5{{x}^{3}}+{{x}^{2}}+3x-2+2x-1}{2{{x}^{2}}-5x+3} \\
 & g(x)=\dfrac{2{{x}^{4}}-5{{x}^{3}}+{{x}^{2}}+5x-3}{2{{x}^{2}}-5x+3} \\
\end{align}$
Now we will add $3{{x}^{2}}-3{{x}^{2}}$ in the numerator we get,
$\begin{align}
  & g(x)=\dfrac{2{{x}^{4}}-5{{x}^{3}}+3{{x}^{2}}-3{{x}^{2}}+{{x}^{2}}+5x-3}{2{{x}^{2}}-5x+3} \\
 & g(x)=\dfrac{2{{x}^{4}}-5{{x}^{3}}+3{{x}^{2}}-\left( 2{{x}^{2}}-5x+3 \right)}{2{{x}^{2}}-5x+3} \\
 & g(x)=\dfrac{{{x}^{2}}\left( 2{{x}^{2}}-5x+3 \right)-\left( 2{{x}^{2}}-5x+3 \right)}{2{{x}^{2}}-5x+3} \\
 & g(x)=\dfrac{\left( {{x}^{2}}-1 \right)\left( 2{{x}^{2}}-5x+3 \right)}{2{{x}^{2}}-5x+3} \\
 & g(x)={{x}^{2}}-1 \\
\end{align}$
Hence after performing subtraction and then by converting the numerator in terms of denominator to cancel the common terms and find the value of g(x), we get,
$g(x)={{x}^{2}}-1$
Hence this is the correct answer to this question.

Note: In this question we have converted the numerator such that it is directly divisible by denominator but it might be difficult for some students so we can also use the division method of two equations and then we can find the value of g(x), as this does not involve that much thinking as in the solution given but it takes some more time and calculation.