Divide $$$u^{5}$$$ by $$$u^{2} + 1$$$

The calculator will divide $$$u^{5}$$$ by $$$u^{2} + 1$$$ using long division, with steps shown.

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Your Input

Find $$$\frac{u^{5}}{u^{2} + 1}$$$ using long division.

Solution

Write the problem in the special format (missed terms are written with zero coefficients):

$$$\begin{array}{r|r}\hline\\u^{2}+1&u^{5}+0 u^{4}+0 u^{3}+0 u^{2}+0 u+0\end{array}$$$

Step 1

Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{u^{5}}{u^{2}} = u^{3}$$$.

Write down the calculated result in the upper part of the table.

Multiply it by the divisor: $$$u^{3} \left(u^{2}+1\right) = u^{5}+u^{3}$$$.

Subtract the dividend from the obtained result: $$$\left(u^{5}\right) - \left(u^{5}+u^{3}\right) = - u^{3}$$$.

$$\begin{array}{r|rrrrrr:c}&{\color{DeepPink}u^{3}}&&&&&&\\\hline\\{\color{Magenta}u^{2}}+1&{\color{DeepPink}u^{5}}&+0 u^{4}&+0 u^{3}&+0 u^{2}&+0 u&+0&\frac{{\color{DeepPink}u^{5}}}{{\color{Magenta}u^{2}}} = {\color{DeepPink}u^{3}}\\&-\phantom{u^{5}}&&&&&&\\&u^{5}&+0 u^{4}&+u^{3}&&&&{\color{DeepPink}u^{3}} \left(u^{2}+1\right) = u^{5}+u^{3}\\\hline\\&&&- u^{3}&+0 u^{2}&+0 u&+0&\end{array}$$

Step 2

Divide the leading term of the obtained remainder 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 remainder from the obtained result: $$$\left(- u^{3}\right) - \left(- u^{3}- u\right) = u$$$.

$$\begin{array}{r|rrrrrr:c}&u^{3}&{\color{Fuchsia}- u}&&&&&\\\hline\\{\color{Magenta}u^{2}}+1&u^{5}&+0 u^{4}&+0 u^{3}&+0 u^{2}&+0 u&+0&\\&-\phantom{u^{5}}&&&&&&\\&u^{5}&+0 u^{4}&+u^{3}&&&&\\\hline\\&&&{\color{Fuchsia}- u^{3}}&+0 u^{2}&+0 u&+0&\frac{{\color{Fuchsia}- u^{3}}}{{\color{Magenta}u^{2}}} = {\color{Fuchsia}- u}\\&&&-\phantom{- u^{3}}&&&&\\&&&- u^{3}&+0 u^{2}&- u&&{\color{Fuchsia}- u} \left(u^{2}+1\right) = - u^{3}- u\\\hline\\&&&&&u&+0&\end{array}$$

Since the degree of the remainder is less than the degree of the divisor, we are done.

The resulting table is shown once more:

$$\begin{array}{r|rrrrrr:c}&{\color{DeepPink}u^{3}}&{\color{Fuchsia}- u}&&&&&\text{Hints}\\\hline\\{\color{Magenta}u^{2}}+1&{\color{DeepPink}u^{5}}&+0 u^{4}&+0 u^{3}&+0 u^{2}&+0 u&+0&\frac{{\color{DeepPink}u^{5}}}{{\color{Magenta}u^{2}}} = {\color{DeepPink}u^{3}}\\&-\phantom{u^{5}}&&&&&&\\&u^{5}&+0 u^{4}&+u^{3}&&&&{\color{DeepPink}u^{3}} \left(u^{2}+1\right) = u^{5}+u^{3}\\\hline\\&&&{\color{Fuchsia}- u^{3}}&+0 u^{2}&+0 u&+0&\frac{{\color{Fuchsia}- u^{3}}}{{\color{Magenta}u^{2}}} = {\color{Fuchsia}- u}\\&&&-\phantom{- u^{3}}&&&&\\&&&- u^{3}&+0 u^{2}&- u&&{\color{Fuchsia}- u} \left(u^{2}+1\right) = - u^{3}- u\\\hline\\&&&&&u&+0&\end{array}$$

Therefore, $$$\frac{u^{5}}{u^{2} + 1} = \left(u^{3} - u\right) + \frac{u}{u^{2} + 1}$$$.

Answer

$$$\frac{u^{5}}{u^{2} + 1} = \left(u^{3} - u\right) + \frac{u}{u^{2} + 1}$$$A


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