How do you write the partial fraction decomposition of the rational expression $\frac{x^{2} + 4}{(x^{2} - 4) (x^{2} + 2)}$ $(x^2 + 4)/ ((x^2 - 4)(x^2 +2))$ ?

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How do you write the partial fraction decomposition of the rational expression $\frac{x^{2} + 4}{(x^{2} - 4) (x^{2} + 2)}$ $(x^2 + 4)/ ((x^2 - 4)(x^2 +2))$ ?

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Answer 1 · 2015-12-14T05:45:19

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

Explanation:

Factor the denominator.

$\frac{x^{2} + 4}{(x + 2) (x - 2) (x^{2} + 2)} = \frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C x + D}{x^{2} + 2}$ $(x^2+4)/((x+2)(x-2)(x^2+2))=A/(x+2)+B/(x-2)+(Cx+D)/(x^2+2)$

$x^{2} + 4 = A (x - 2) (x^{2} + 2) + B (x + 2) (x^{2} + 2) + (C x + D) (x^{2} - 4)$ $x^2+4=A(x-2)(x^2+2)+B(x+2)(x^2+2)+(Cx+D)(x^2-4)$

If $x = - 2$ $x=-2$ :

$8 = - 24 A$ $8=-24A$
$A = - \frac{1}{3}$ $A=-1/3$

If $x = 2$ $x=2$ :

$8 = 24 B$ $8=24B$
$B = \frac{1}{3}$ $B=1/3$

If $x = 0$ $x=0$ :

$4 = - 4 A + 4 B - 4 D$ $4=-4A+4B-4D$
$1 = - A + B - D$ $1=-A+B-D$

Plug in $A$ $A$ and $B$ $B$ to see that $D = - \frac{1}{3}$ $D=-1/3$ .

If $x = 1$ $x=1$ :

$5 = - 3 A + 9 B - 3 C - 3 D$ $5=-3A+9B-3C-3D$

Plug in all the known values to find that $C = 0$ $C=0$ .

Plug these all back in to see that the expression decomposes into

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

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How do you write the partial fraction decomposition of the rational expression $\frac{x^{2} + 4}{(x^{2} - 4) (x^{2} + 2)}$ $(x^2 + 4)/ ((x^2 - 4)(x^2 +2))$ ?

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Explanation:

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How do you write the partial fraction decomposition of the rational expression (x^2 + 4)/ ((x^2 - 4)(x^2 +2))x2+4(x2−4)(x2+2) (x^2 + 4)/ ((x^2 - 4)(x^2 +2))?

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Explanation:

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How do you write the partial fraction decomposition of the rational expression $\frac{x^{2} + 4}{(x^{2} - 4) (x^{2} + 2)}$ $(x^2 + 4)/ ((x^2 - 4)(x^2 +2))$ ?