Question

How do you divide \[\dfrac{2{{x}^{4}}+7}{{{x}^{2}}-1}\] using polynomial long division?

Answer 1

Hint: In this problem, we have to find the division using the polynomial long division method. We can first set up the polynomials to be divided in long division and we can divide the highest order term in the dividend by the highest order term in the divisor for every term by bringing it down step by step. If there are any missing terms for every exponent, add one with a value of 0.

Complete step-by-step solution:
We know that the given fraction is,
\[\dfrac{2{{x}^{4}}+7}{{{x}^{2}}-1}\]
Now we can set up the polynomials to be divided in long division, if there is any missing terms for every exponent, add one with a value of 0, we get
\[{{x}^{2}}-1\overset{{}}{\overline{\left){2{{x}^{4}}+0{{x}^{2}}+7}\right.}}\]
Now we can divide the highest order term in the dividend \[2{{x}^{2}}\] by the highest order term in the divisor x, we get
\[{{x}^{2}}-1\overset{2{{x}^{2}}}{\overline{\left){2{{x}^{4}}+0{{x}^{2}}+7}\right.}}\]
We can now multiply the quotient term to the divisor, we get
\[{{x}^{2}}-1\overset{2{{x}^{2}}}{\overline{\left){\begin{align}
  & 2{{x}^{4}}+0{{x}^{2}}+7 \\
 & 2{{x}^{4}}-2{{x}^{2}} \\
\end{align}}\right.}}\]
We know that the expression is to be subtracted in the dividend, so we can change the sign in \[2{{x}^{4}}-2{{x}^{2}}\], we get
\[{{x}^{2}}-1\overset{2{{x}^{2}}}{\overline{\left){\begin{align}
  & 2{{x}^{4}}+0{{x}^{2}}+7 \\
 & -2{{x}^{4}}+2{{x}^{2}} \\
\end{align}}\right.}}\]
Now we can bring down the next term from the dividend to the current dividend,
\[\begin{align}
  & {{x}^{2}}-1\overset{2{{x}^{2}}}{\overline{\left){\begin{align}
  & 2{{x}^{4}}+0{{x}^{2}}+7 \\
 & -\underline{2{{x}^{4}}+2{{x}^{2}}} \\
\end{align}}\right.}} \\
 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2{{x}^{2}}+7 \\
\end{align}\]
Now we should divide the highest order term in the dividend 2 by the divisor x, we get
\[\begin{align}
  & {{x}^{2}}-1\overset{2{{x}^{2}}+2}{\overline{\left){\begin{align}
  & 2{{x}^{4}}+0{{x}^{2}}+7 \\
 & -\underline{2{{x}^{4}}+2{{x}^{2}}} \\
\end{align}}\right.}} \\
 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2{{x}^{2}}+7 \\
 &\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;2{{x}^{2}}-2 \\
\end{align}\]
We can now multiply the new quotient to the divisor, then subtract those expression to get the next dividend value, we get
\[\begin{align}
  & {{x}^{2}}-1\overset{2{{x}^{2}}+2}{\overline{\left){\begin{align}
  & 2{{x}^{4}}+0{{x}^{2}}+7 \\
 & -\underline{2{{x}^{4}}+2{{x}^{2}}} \\
\end{align}}\right.}} \\
 & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2{{x}^{2}}+7 \\
 & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \underline{-2{{x}^{2}}+2} \\
 & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 9 \\
\end{align}\]
We know that dividend is equal to quotient plus the remainder over the divisor.
Therefore, the final answer is \[\left( 2{{x}^{2}}+2\right) +\dfrac{9}{{{x}^{2}}-1}\].

Note: Students may get confused on seeing the given problem where we can see some missing terms. In such a case we can add 0 for the missing terms. In this problem, we have added only \[0{{x}^{2}}\] as we will have only even terms in power when multiplied. We should also remember the formula for the final answer that is Dividend = Quotient + Remainder/Divisor.