Question

Expand \[\dfrac{1}{{{x}^{4}}-5{{x}^{3}}+7{{x}^{2}}+x-8}\] in descending powers of x to four terms, and find the remainder.

Answer 1

Hint: We start solving the problem by finding the dividend and divisor for performing the long division method to find the quotient and remainder. We then perform a step by step process in the long division to get the four terms of x with different powers on it. We then make the calculations required and stop the division after getting four terms in the quotient. We find the resulting remainder from the leftover of long division.

Complete step-by-step solution
According to the problem, we need to expand the given \[\dfrac{1}{{{x}^{4}}-5{{x}^{3}}+7{{x}^{2}}+x-8}\] in descending powers of x to four terms and the remainder.
We solve this problem by using the long division method of expansion.
We have a dividend ${{x}^{4}}-5{{x}^{3}}+7{{x}^{2}}+x-8$ and divisor 1. After the long division, whatever the quotient we obtained here will be the expansion we just needed.
According to the problem, we need to expand \[\dfrac{1}{{{x}^{4}}-5{{x}^{3}}+7{{x}^{2}}+x-8}\] up to four terms. This means that the quotient should have four terms in order to get the required remainder.
Let us do the long division now in order to get a clear view about expansion and remainder.
\[\begin{align}
  & \text{ }\underline{{{x}^{-4}}+5{{x}^{-5}}+18{{x}^{-6}}+54{{x}^{-7}}} \\
 & \left. {{x}^{4}}-5{{x}^{3}}+7{{x}^{2}}+x-8 \right)1 \\
 & \text{ }1-5{{x}^{-1}}+7{{x}^{-2}}+{{x}^{-3}}-8{{x}^{-4}} \\
 & \text{ }\underline{(-)\text{ }(+)\text{ }(-)\text{ }(-)\text{ }\left( + \right)\text{ }} \\
 & \text{ }5{{x}^{-1}}-7{{x}^{-2}}-{{x}^{-3}}+8{{x}^{-4}} \\
 & \text{ }5{{x}^{-1}}-25{{x}^{-2}}+35{{x}^{-3}}+5{{x}^{-4}}-40{{x}^{-5}} \\
 & \text{ }\underline{\text{ }\left( - \right)\text{ }\left( + \right)\text{ }\left( - \right)\text{ }\left( - \right)\text{ }\left( + \right)\text{ }} \\
 & \text{ }18{{x}^{-2}}-36{{x}^{-3}}+3{{x}^{-4}}+40{{x}^{-5}} \\
 & \text{ }18{{x}^{-2}}-90{{x}^{-3}}+126{{x}^{-4}}+18{{x}^{-5}}-144{{x}^{-6}} \\
 & \text{ }\underline{\text{ }\left( - \right)\text{ }\left( + \right)\text{ }\left( - \right)\text{ }\left( - \right)\text{ }\left( + \right)\text{ }} \\
 & \text{ }54{{x}^{-3}}-123{{x}^{-4}}+22{{x}^{-5}}+144{{x}^{-6}} \\
 & \text{ }54{{x}^{-3}}-270{{x}^{-4}}+378{{x}^{-5}}+54{{x}^{-6}}-432{{x}^{-7}} \\
 & \text{ }\underline{\text{ }\left( - \right)\text{ }\left( + \right)\text{ }\left( - \right)\text{ }\left( - \right)\text{ }\left( + \right)\text{ }} \\
 & \text{ }147{{x}^{-4}}-356{{x}^{-5}}+90{{x}^{-6}}+432{{x}^{-7}} \\
\end{align}\].
From the long division method, we have found the quotient as ${{x}^{-4}}+5{{x}^{-5}}+18{{x}^{-6}}+54{{x}^{-7}}$ and the remainder $147{{x}^{-4}}-356{{x}^{-5}}+90{{x}^{-6}}+432{{x}^{-7}}$.
$\therefore$ The remainder we get when we expand \[\dfrac{1}{{{x}^{4}}-5{{x}^{3}}+7{{x}^{2}}+x-8}\] in descending powers of x to four terms is $147{{x}^{-4}}-356{{x}^{-5}}+90{{x}^{-6}}+432{{x}^{-7}}$.

Note: We can see that the quotient of the given division process is not terminating. So, we need to stop the division process by finding up to the number of terms we need. From the problem, we can see that the division process has to be continued till we get the four terms in quotient and remainder. Whenever we get this type of problem, we need to remember to continue the division process up to the number of terms in the quotient is equal to the number of terms mentioned in the problem.