Divide $$$- 2 x^{2} + 5 x - 2$$$ by $$$\left(x - 1\right)^{2}$$$
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Your Input
Find $$$\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}}$$$ using long division.
Solution
Rewrite the divisor: $$$\left(x - 1\right)^{2} = x^{2} - 2 x + 1$$$.
Write the problem in the special format:
$$$\begin{array}{r|r}\hline\\x^{2}- 2 x+1&- 2 x^{2}+5 x-2\end{array}$$$
Step 1
Divide the leading term of the dividend by the leading term of the divisor: $$$\frac{- 2 x^{2}}{x^{2}} = -2$$$.
Write down the calculated result in the upper part of the table.
Multiply it by the divisor: $$$- 2 \left(x^{2}- 2 x+1\right) = - 2 x^{2}+4 x-2$$$.
Subtract the dividend from the obtained result: $$$\left(- 2 x^{2}+5 x-2\right) - \left(- 2 x^{2}+4 x-2\right) = x$$$.
$$\begin{array}{r|rrr:c}&{\color{Red}-2}&&&\\\hline\\{\color{Magenta}x^{2}}- 2 x+1&{\color{Red}- 2 x^{2}}&+5 x&-2&\frac{{\color{Red}- 2 x^{2}}}{{\color{Magenta}x^{2}}} = {\color{Red}-2}\\&-\phantom{- 2 x^{2}}&&&\\&- 2 x^{2}&+4 x&-2&{\color{Red}-2} \left(x^{2}- 2 x+1\right) = - 2 x^{2}+4 x-2\\\hline\\&&x&+0&\end{array}$$Since the degree of the remainder is less than the degree of the divisor, we are done.
The resulting table is shown once more:
$$\begin{array}{r|rrr:c}&{\color{Red}-2}&&&\text{Hints}\\\hline\\{\color{Magenta}x^{2}}- 2 x+1&{\color{Red}- 2 x^{2}}&+5 x&-2&\frac{{\color{Red}- 2 x^{2}}}{{\color{Magenta}x^{2}}} = {\color{Red}-2}\\&-\phantom{- 2 x^{2}}&&&\\&- 2 x^{2}&+4 x&-2&{\color{Red}-2} \left(x^{2}- 2 x+1\right) = - 2 x^{2}+4 x-2\\\hline\\&&x&+0&\end{array}$$Therefore, $$$\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}} = -2 + \frac{x}{\left(x - 1\right)^{2}}$$$.
Answer
$$$\frac{- 2 x^{2} + 5 x - 2}{\left(x - 1\right)^{2}} = -2 + \frac{x}{\left(x - 1\right)^{2}}$$$A