How do you write $\frac{x^{4}}{{(x - 1)}^{3}}$ $x^4/(x-1)^3$ as a partial fraction decomposition?

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How do you write $\frac{x^{4}}{{(x - 1)}^{3}}$ $x^4/(x-1)^3$ as a partial fraction decomposition?

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Answer 1 · 2016-10-27T13:07:03

The result is
$\frac{x^{4}}{{(x - 1)}^{3}} = x + 3 + \frac{6}{x - 1} + \frac{4}{{(x - 1)}^{2}} + \frac{1}{{(x - 1)}^{3}}$ $x^4/(x-1)^3=x+3+6/(x-1)+4/(x-1)^2+1/(x-1)^3$

Explanation:

Since the degree of the numerator is greater than the degree of the denominator, we perform a long division

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

$x^{4}$ $x^4$ $a a a a a a a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaaaaaaaa)$ $∣$ $∣$ $x^{3} - 3 x^{2} + 3 x - 1$ $x^3-3x^2+3x-1$

$x^{4} - 3 x^{3} + 3 x^{2} - x$ $x^4-3x^3+3x^2-x$ $a a a a a$ $color(white)(aaaaa)$ $∣$ $∣$ $x + 3$ $x+3$

$0 + 3 x^{3} - 3 x^{2} + x$ $0+3x^3-3x^2+x$

$a a a$ $color(white)(aaa)$ $3 x^{3} - 9 x^{2} + 9 x - 3$ $3x^3-9x^2+9x-3$

$a a a a a$ $color(white)(aaaaa)$ $0 + 6 x^{2} - 8 x + 3$ $0+6x^2-8x+3$

So we get

$\frac{x^{4}}{{(x - 1)}^{3}} = x + 3 + \frac{6 x^{2} - 8 x + 3}{{(x - 1)}^{3}}$ $x^4/(x-1)^3=x+3+(6x^2-8x+3)/(x-1)^3$

now we form the partial fraction decomposition

$\frac{6 x^{2} - 8 x + 3}{{(x - 1)}^{3}} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}}$ $(6x^2-8x+3)/(x-1)^3=A/(x-1)+B/(x-1)^2+C/(x-1)^3$

$\frac{6 x^{2} - 8 x + 3}{{(x - 1)}^{3}} = \frac{A {(x - 1)}^{2} + B (x - 1) + C}{{(x - 1)}^{3}}$ $(6x^2-8x+3)/(x-1)^3=(A(x-1)^2+B(x-1)+C)/(x-1)^3$

So $6 x^{2} - 8 x + 3 = A {(x - 1)}^{2} + B (x - 1) + C$ $6x^2-8x+3=A(x-1)^2+B(x-1)+C$

Let $x = 1$ $x=1$ then $1 = C$ $1=C$

Compare the coefficients of $x^{2}$ $x^2$

$6 = A$ $6=A$

Let $x = 0$ $x=0$ , then $3 = A - B + C$ $3=A-B+C$

$B = 6 + 1 - 3 = 4$ $B=6+1-3=4$

So the final result is

$\frac{x^{4}}{{(x - 1)}^{3}} = x + 3 + \frac{6}{x - 1} + \frac{4}{{(x - 1)}^{2}} + \frac{1}{{(x - 1)}^{3}}$ $x^4/(x-1)^3=x+3+6/(x-1)+4/(x-1)^2+1/(x-1)^3$

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How do you write $\frac{x^{4}}{{(x - 1)}^{3}}$ $x^4/(x-1)^3$ as a partial fraction decomposition?

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How do you write x4(x−1)3x^4/(x-1)^3 as a partial fraction decomposition?

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How do you write $\frac{x^{4}}{{(x - 1)}^{3}}$ $x^4/(x-1)^3$ as a partial fraction decomposition?