How do you express $\frac{x^{2} - 2 x + 1}{{(x - 2)}^{3}}$ $(x^2-2x+1)/(x-2)^3$ in partial fractions?

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How do you express $\frac{x^{2} - 2 x + 1}{{(x - 2)}^{3}}$ $(x^2-2x+1)/(x-2)^3$ in partial fractions?

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Answer 1 · 2018-08-07T07:30:03

The answer is $= \frac{1}{{(x - 2)}^{3}} + \frac{2}{{(x - 2)}^{2}} + \frac{1}{x - 2}$ $=1/(x-2)^3+2/(x-2)^2+1/(x-2)$

Explanation:

Perform the decomposition into partial fractions as follows

$\frac{x^{2} - 2 x + 1}{{(x - 2)}^{3}} = \frac{A}{{(x - 2)}^{3}} + \frac{B}{{(x - 2)}^{2}} + \frac{C}{x - 2}$ $(x^2-2x+1)/(x-2)^3=A/(x-2)^3+B/(x-2)^2+C/(x-2)$

$= \frac{A + B (x - 2) + C {(x - 2)}^{2}}{{(x - 2)}^{3}}$ $=(A+B(x-2)+C(x-2)^2)/((x-2)^3)$

The denominators are the same, compare the numerators

$x^{2} - 2 x + 1 = A + B (x - 2) + C {(x - 2)}^{2}$ $x^2-2x+1=A+B(x-2)+C(x-2)^2$

Let $x = 2$ $x=2$ , $\Rightarrow$ $=>$ , $4 - 4 + 1 = A$ $4-4+1=A$ , $\Rightarrow$ $=>$ , $A = 1$ $A=1$

Coefficients of $x^{2}$ $x^2$

$1 = C$ $1=C$

Coefficients of $x$ $x$

$- 2 = B - 4 C$ $-2=B-4C$

$B = 4 C - 2 = 2$ $B=4C-2=2$

Finally,

$\frac{x^{2} - 2 x + 1}{{(x - 2)}^{3}} = \frac{1}{{(x - 2)}^{3}} + \frac{2}{{(x - 2)}^{2}} + \frac{1}{x - 2}$ $(x^2-2x+1)/(x-2)^3=1/(x-2)^3+2/(x-2)^2+1/(x-2)$

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How do you express $\frac{x^{2} - 2 x + 1}{{(x - 2)}^{3}}$ $(x^2-2x+1)/(x-2)^3$ in partial fractions?

1 Answer

Explanation:

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How do you express $\frac{x^{2} - 2 x + 1}{{(x - 2)}^{3}}$ $(x^2-2x+1)/(x-2)^3$ in partial fractions?