How do you express $\frac{x^{2} - 8 x + 44}{(x + 2) {(x - 2)}^{2}}$ $(x^2 - 8x + 44) / ((x + 2) (x - 2)^2)$ in partial fractions?

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How do you express $\frac{x^{2} - 8 x + 44}{(x + 2) {(x - 2)}^{2}}$ $(x^2 - 8x + 44) / ((x + 2) (x - 2)^2)$ in partial fractions?

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Answer 1 · 2018-01-02T20:31:33

$\frac{x^{2} - 8 x + 44}{(x + 2) \cdot {(x - 2)}^{2}} = \frac{4}{x + 2} - \frac{3}{x - 2} + \frac{8}{{(x - 2)}^{2}}$ $(x^2-8x+44)/[(x+2)*(x-2)^2]=4/(x+2)-3/(x-2)+8/(x-2)^2$

Explanation:

$\frac{x^{2} - 8 x + 44}{(x + 2) \cdot {(x - 2)}^{2}}$ $(x^2-8x+44)/[(x+2)*(x-2)^2]$

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

After expanding denominator,

$A \cdot {(x - 2)}^{2} + B \cdot (x^{2} - 4) + C \cdot (x + 2) = x^{2} - 8 x + 44$ $A*(x-2)^2+B*(x^2-4)+C*(x+2)=x^2-8x+44$

Set $x = - 2$ $x=-2$ , $16 A = 64$ $16A=64$ , so $A = 4$ $A=4$

Set $x = 2$ $x=2$ , $4 C = 32$ $4C=32$ , so $C = 8$ $C=8$

Set $x = 0$ $x=0$ , $4 A - 4 B + 2 C = 44$ $4A-4B+2C=44$ , so $B = - 3$ $B=-3$

Thus,

$\frac{x^{2} - 8 x + 44}{(x + 2) \cdot {(x - 2)}^{2}} = \frac{4}{x + 2} - \frac{3}{x - 2} + \frac{8}{{(x - 2)}^{2}}$ $(x^2-8x+44)/[(x+2)*(x-2)^2]=4/(x+2)-3/(x-2)+8/(x-2)^2$

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How do you express $\frac{x^{2} - 8 x + 44}{(x + 2) {(x - 2)}^{2}}$ $(x^2 - 8x + 44) / ((x + 2) (x - 2)^2)$ in partial fractions?

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Explanation:

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