Question

How do you divide the polynomial \[\dfrac{2{{x}^{4}}+4{{x}^{3}}-5{{x}^{2}}+3x-2}{{{x}^{2}}+2x-3}\]?

Answer 1

Hint: In this problem, we have to divide the given equation and the factor. We can use the polynomial long division method to divide the given problem by dividing the highest order term in the dividend to the highest order term in the divisor step by step until we get the remainder.

Complete step-by-step solution:
We know that the given fraction is,
\[\dfrac{2{{x}^{4}}+4{{x}^{3}}-5{{x}^{2}}+3x-2}{{{x}^{2}}+2x-3}\]
Now we can set up the polynomials to be divided in long division, we get
\[{{x}^{2}}+2x-3\overset{{}}{\overline{\left){2{{x}^{4}}+4{{x}^{3}}-5{{x}^{2}}+3x-2}\right.}}\]
Now we can divide the highest order term in the dividend \[2{{x}^{4}}\] by the highest order term in the divisor \[{{x}^{2}}\], we get
\[{{x}^{2}}+2x-3\overset{2{{x}^{2}}}{\overline{\left){2{{x}^{4}}+4{{x}^{3}}-5{{x}^{2}}+3x-2}\right.}}\]
We can now multiply the quotient term to the divisor, we get
\[{{x}^{2}}+2x-3\overset{2{{x}^{2}}}{\overline{\left){\begin{align}
  & 2{{x}^{4}}+4{{x}^{3}}-5{{x}^{2}}+3x-2 \\
 & 2{{x}^{4}}+4{{x}^{3}}-6{{x}^{2}} \\
\end{align}}\right.}}\]
We know that the expression is to be subtracted in the dividend, so we can change the sign in \[2{{x}^{4}}+4{{x}^{3}}-6{{x}^{2}}\], we get
\[\begin{align}
  & {{x}^{2}}+2x-3\overset{2{{x}^{2}}}{\overline{\left){\begin{align}
  & 2{{x}^{4}}+4{{x}^{3}}-5{{x}^{2}}+3x-2 \\
 & \underline{-2{{x}^{4}}-4{{x}^{3}}+6{{x}^{2}}} \\
\end{align}}\right.}} \\
 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ {{x}^{2}} \\
\end{align}\]
Now we can bring down the next two term from the dividend to the current dividend,
\[\begin{align}
  & {{x}^{2}}+2x-3\overset{2{{x}^{2}}}{\overline{\left){\begin{align}
  & 2{{x}^{4}}+4{{x}^{3}}-5{{x}^{2}}+3x-2 \\
 & \underline{-2{{x}^{4}}-4{{x}^{3}}+6{{x}^{2}}} \\
\end{align}}\right.}} \\
 & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ {{x}^{2}}+3x-2 \\
\end{align}\]
Now we should divide the highest order term in the dividend by the divisor x, we get
\[\begin{align}
  & {{x}^{2}}+2x-3\overset{2{{x}^{2}}+1}{\overline{\left){\begin{align}
  & 2{{x}^{4}}+4{{x}^{3}}-5{{x}^{2}}+3x-2 \\
 & \underline{-2{{x}^{4}}-4{{x}^{3}}+6{{x}^{2}}} \\
\end{align}}\right.}} \\
 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ {{x}^{2}}+3x-2 \\
 & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ {{x}^{2}}+2x-3 \\
\end{align}\]
We can now multiply the new quotient to the divisor, then subtract those expression to get the next dividend value, we get
\[\begin{align}
& {{x}^{2}}+2x-3\overset{2{{x}^{2}}+1}{\overline{\left){\begin{align}
& 2{{x}^{4}}+4{{x}^{3}}-5{{x}^{2}}+3x-2 \\
& \underline{-2{{x}^{4}}-4{{x}^{3}}+6{{x}^{2}}} \\
\end{align}}\right.}} \\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; {{x}^{2}}+3x-2 \\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \underline{-{{x}^{2}}-2x+3} \\
&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x+1 \\
\end{align}\]
We know that dividend is equal to quotient plus the remainder over the divisor.
Therefore, the final answer is \[2{{x}^{2}}+1+\dfrac{x+1}{{{x}^{2}}+2x-3}\].

Note: Students make mistakes while dividing the number to the divisor which we will place it in the quotient. We should always concentrate on each and every step while bringing the terms from the given equation step by step. We should also remember that the formula for the final result is the dividend is equal to quotient plus remainder over the divisor.