Divide $$$x^{2}$$$ by $$$1 - x$$$
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Find $$$\frac{x^{2}}{1 - x}$$$ using long division.
Solution
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: $$$\left(x^{2}\right) - \left(x^{2}- x\right) = x$$$.
$$\begin{array}{r|rrr:c}&{\color{Peru}- x}&&&\\\hline\\{\color{Magenta}- x}+1&{\color{Peru}x^{2}}&+0 x&+0&\frac{{\color{Peru}x^{2}}}{{\color{Magenta}- x}} = {\color{Peru}- x}\\&-\phantom{x^{2}}&&&\\&x^{2}&- x&&{\color{Peru}- x} \left(- x+1\right) = x^{2}- x\\\hline\\&&x&+0&\end{array}$$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: $$$\left(x\right) - \left(x-1\right) = 1$$$.
$$\begin{array}{r|rrr:c}&- x&{\color{GoldenRod}-1}&&\\\hline\\{\color{Magenta}- x}+1&x^{2}&+0 x&+0&\\&-\phantom{x^{2}}&&&\\&x^{2}&- x&&\\\hline\\&&{\color{GoldenRod}x}&+0&\frac{{\color{GoldenRod}x}}{{\color{Magenta}- x}} = {\color{GoldenRod}-1}\\&&-\phantom{x}&&\\&&x&-1&{\color{GoldenRod}-1} \left(- x+1\right) = x-1\\\hline\\&&&1&\end{array}$$Since the degree of the remainder is less than the degree of the divisor, we are done.
The resulting table is shown once more:
$$\begin{array}{r|rrr:c}&{\color{Peru}- x}&{\color{GoldenRod}-1}&&\text{Hints}\\\hline\\{\color{Magenta}- x}+1&{\color{Peru}x^{2}}&+0 x&+0&\frac{{\color{Peru}x^{2}}}{{\color{Magenta}- x}} = {\color{Peru}- x}\\&-\phantom{x^{2}}&&&\\&x^{2}&- x&&{\color{Peru}- x} \left(- x+1\right) = x^{2}- x\\\hline\\&&{\color{GoldenRod}x}&+0&\frac{{\color{GoldenRod}x}}{{\color{Magenta}- x}} = {\color{GoldenRod}-1}\\&&-\phantom{x}&&\\&&x&-1&{\color{GoldenRod}-1} \left(- x+1\right) = x-1\\\hline\\&&&1&\end{array}$$Therefore, $$$\frac{x^{2}}{1 - x} = \left(- x - 1\right) + \frac{1}{1 - x}$$$.
Answer
$$$\frac{x^{2}}{1 - x} = \left(- x - 1\right) + \frac{1}{1 - x}$$$A