How do you express $\frac{x^{3} + 2}{x^{4} - 16}$ $(x^3 + 2) / (x^4 -16)$ in partial fractions?

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How do you express $\frac{x^{3} + 2}{x^{4} - 16}$ $(x^3 + 2) / (x^4 -16)$ in partial fractions?

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Answer 1 · 2016-11-20T07:20:49

The answer is $= \frac{\frac{3}{16}}{x + 2} + \frac{\frac{5}{16}}{x - 2} + \frac{\frac{x}{2} - \frac{1}{4}}{x^{2} + 4}$ $=(3/16)/(x+2)+(5/16)/(x-2)+(x/2-1/4)/(x^2+4)$

Explanation:

Let's factorise the denominator

$x^{4} - 16 = (x^{2} - 4) (x^{2} + 4) = (x + 2) (x - 2) (x^{2} + 4)$ $x^4-16=(x^2-4)(x^2+4)=(x+2)(x-2)(x^2+4)$

Therefore,

$\frac{x^{3} + 2}{x^{4} - 16} = \frac{x^{3} + 2}{(x + 2) (x - 2) (x^{2} + 4)}$ $(x^3+2)/(x^4-16)=(x^3+2)/((x+2)(x-2)(x^2+4))$

The decomposition in partial fractions is

$\frac{x^{3} + 2}{x^{4} - 16} = \frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C x + D}{x^{2} + 4}$ $(x^3+2)/(x^4-16)=A/(x+2)+B/(x-2)+(Cx+D)/(x^2+4)$

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

Therefore,

$(x^{3} + 2) = (A (x - 2) (x^{2} + 4) + B (x + 2) (x^{2} + 4) + (C x + D) (x - 2) (x + 2))$ $(x^3+2)=(A(x-2)(x^2+4)+B(x+2)(x^2+4)+(Cx+D)(x-2)(x+2))$

let $x = 2$ $x=2$
$10 = 32 B$ $10=32B$ ; $\Rightarrow$ $=>$ $B = \frac{5}{16}$ $B=5/16$

Let $x = - 2$ $x=-2$
$- 6 = - 32 A$ $-6=-32A$ ; $\Rightarrow$ $=>$ $A = \frac{3}{16}$ $A=3/16$

Coefficients of $x^{3}$ $x^3$
$1 = A + B + C$ $1=A+B+C$ , $\Rightarrow$ $=>$ $C = 1 - \frac{5}{16} - \frac{3}{16} = \frac{8}{16} = \frac{1}{2}$ $C=1-5/16-3/16=8/16=1/2$

Let $x = 0$ $x=0$
$2 = - 8 A + 8 B - 4 D$ $2=-8A+8B-4D$ : $\Rightarrow$ $=>$ $4 D = - 2 - \frac{3}{2} + \frac{5}{2} = - 1$ $4D=-2-3/2+5/2=-1$
$D = - \frac{1}{4}$ $D=-1/4$

$\frac{x^{3} + 2}{x^{4} - 16} = \frac{\frac{3}{16}}{x + 2} + \frac{\frac{5}{16}}{x - 2} + \frac{\frac{x}{2} - \frac{1}{4}}{x^{2} + 4}$ $(x^3+2)/(x^4-16)=(3/16)/(x+2)+(5/16)/(x-2)+(x/2-1/4)/(x^2+4)$

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How do you express $\frac{x^{3} + 2}{x^{4} - 16}$ $(x^3 + 2) / (x^4 -16)$ in partial fractions?

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