Divide $$$x^{2}$$$ by $$$x + 1$$$

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

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

$$$\begin{array}{r|r}\hline\\x+1&x^{2}+0 x+0\end{array}$$$

Step 1

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

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

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

Subtract the dividend from the obtained result: $$$\left(x^{2}\right) - \left(x^{2}+x\right) = - x$$$.

Step 2

Divide the leading term of the obtained remainder by the leading term of the divisor: $$$\frac{- x}{x} = -1$$$.

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

Multiply it by the divisor: $$$- \left(x+1\right) = - x-1$$$.

Subtract the remainder from the obtained result: $$$\left(- x\right) - \left(- x-1\right) = 1$$$.

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

The resulting table is shown once more:

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

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