Divide $$$u^{3}$$$ by $$$u^{2} + 1$$$
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Your Input
Find $$$\frac{u^{3}}{u^{2} + 1}$$$ using long division.
Solution
Write the problem in the special format (missed terms are written with zero coefficients):
$$$\begin{array}{r|r}\hline\\u^{2}+1&u^{3}+0 u^{2}+0 u+0\end{array}$$$
Step 1
Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{u^{3}}{u^{2}} = u$$$.
Write down the calculated result in the upper part of the table.
Multiply it by the divisor: $$$u \left(u^{2}+1\right) = u^{3}+u$$$.
Subtract the dividend from the obtained result: $$$\left(u^{3}\right) - \left(u^{3}+u\right) = - u$$$.
$$\begin{array}{r|rrrr:c}&{\color{Chartreuse}u}&&&&\\\hline\\{\color{Magenta}u^{2}}+1&{\color{Chartreuse}u^{3}}&+0 u^{2}&+0 u&+0&\frac{{\color{Chartreuse}u^{3}}}{{\color{Magenta}u^{2}}} = {\color{Chartreuse}u}\\&-\phantom{u^{3}}&&&&\\&u^{3}&+0 u^{2}&+u&&{\color{Chartreuse}u} \left(u^{2}+1\right) = u^{3}+u\\\hline\\&&&- u&+0&\end{array}$$Since the degree of the remainder is less than the degree of the divisor, we are done.
The resulting table is shown once more:
$$\begin{array}{r|rrrr:c}&{\color{Chartreuse}u}&&&&\text{Hints}\\\hline\\{\color{Magenta}u^{2}}+1&{\color{Chartreuse}u^{3}}&+0 u^{2}&+0 u&+0&\frac{{\color{Chartreuse}u^{3}}}{{\color{Magenta}u^{2}}} = {\color{Chartreuse}u}\\&-\phantom{u^{3}}&&&&\\&u^{3}&+0 u^{2}&+u&&{\color{Chartreuse}u} \left(u^{2}+1\right) = u^{3}+u\\\hline\\&&&- u&+0&\end{array}$$Therefore, $$$\frac{u^{3}}{u^{2} + 1} = u + \frac{- u}{u^{2} + 1}$$$.
Answer
$$$\frac{u^{3}}{u^{2} + 1} = u + \frac{- u}{u^{2} + 1}$$$A