How do you write the partial fraction decomposition of the rational expression $\frac{3 x^{2} + 12 x - 20}{x^{4} - 8 x^{2} + 16}$ $(3x^2 + 12x - 20)/(x^4 - 8x^2 + 16)$ ?

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How do you write the partial fraction decomposition of the rational expression $\frac{3 x^{2} + 12 x - 20}{x^{4} - 8 x^{2} + 16}$ $(3x^2 + 12x - 20)/(x^4 - 8x^2 + 16)$ ?

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Answer 1 · 2016-12-03T10:57:23

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

Explanation:

To factorise the denominator,

We use ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a-b)^2=a^2-2ab+b^2$

and $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$

So,

$x^{4} - 8 x^{2} + 16 = {(x^{2} - 4)}^{2} = {(x + 2)}^{2} {(x - 2)}^{2}$ $x^4-8x^2+16=(x^2-4)^2=(x+2)^2(x-2)^2$

Therefore,

$\frac{3 x^{2} + 12 x - 20}{x^{4} - 8 x^{2} + 16} = \frac{3 x^{2} + 12 x - 20}{{(x + 2)}^{2} {(x - 2)}^{2}}$ $(3x^2+12x-20)/(x^4-8x^2+16)=(3x^2+12x-20)/((x+2)^2(x-2)^2)$

We can now do the decomposition in partial fractions

$\frac{3 x^{2} + 12 x - 20}{x^{4} - 8 x^{2} + 16} = \frac{A}{{(x + 2)}^{2}} + \frac{B}{x + 2} + \frac{C}{{(x - 2)}^{2}} + \frac{D}{x - 2}$ $(3x^2+12x-20)/(x^4-8x^2+16)=A/(x+2)^2+B/(x+2)+C/(x-2)^2+D/(x-2)$

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

Therefore,

$(3 x^{2} + 12 x - 20) = A {(x - 2)}^{2} + B {(x - 2)}^{2} (x + 2) + C {(x + 2)}^{2} + D {(x + 2)}^{2} (x - 2)$ $(3x^2+12x-20)=A(x-2)^2+B(x-2)^2(x+2)+C(x+2)^2+D(x+2)^2(x-2)$

Let $x = 2$ $x=2$ , $\Rightarrow$ $=>$ , $16 = 16 C$ $16=16C$ , $\Rightarrow$ $=>$ , $C = 1$ $C=1$

Let $x = - 2$ $x=-2$ , $\Rightarrow$ $=>$ , $- 32 = 16 A$ $-32=16A$ , $\Rightarrow$ $=>$ , $A = - 2$ $A=-2$

Let $x = 0$ $x=0$ , $\Rightarrow$ $=>$ , $- 20 = 4 A + 8 B + 4 C - 8 D$ $-20=4A+8B+4C-8D$

$B - D = - 2$ $B-D=-2$

Coefficients of $x^{3}$ $x^3$

$0 = B + D$ $0=B+D$

So, $B = - 1$ $B=-1$ and $D = 1$ $D=1$

So,

$\frac{3 x^{2} + 12 x - 20}{x^{4} - 8 x^{2} + 16} = - \frac{2}{{(x + 2)}^{2}} - \frac{1}{x + 2} + \frac{1}{{(x - 2)}^{2}} + \frac{1}{x - 2}$ $(3x^2+12x-20)/(x^4-8x^2+16)=-2/(x+2)^2-1/(x+2)+1/(x-2)^2+1/(x-2)$

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How do you write the partial fraction decomposition of the rational expression $\frac{3 x^{2} + 12 x - 20}{x^{4} - 8 x^{2} + 16}$ $(3x^2 + 12x - 20)/(x^4 - 8x^2 + 16)$ ?

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Explanation:

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How do you write the partial fraction decomposition of the rational expression (3x^2 + 12x - 20)/(x^4 - 8x^2 + 16)3x2+12x−20x4−8x2+16 (3x^2 + 12x - 20)/(x^4 - 8x^2 + 16)?

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Explanation:

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How do you write the partial fraction decomposition of the rational expression $\frac{3 x^{2} + 12 x - 20}{x^{4} - 8 x^{2} + 16}$ $(3x^2 + 12x - 20)/(x^4 - 8x^2 + 16)$ ?