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

1 Answer
Dec 3, 2016

The answer is =-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^2-2ab+b^2

and a^2-b^2=(a+b)(a-b)

So,

x^4-8x^2+16=(x^2-4)^2=(x+2)^2(x-2)^2

Therefore,

(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

(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)

Therefore,

(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, =>, 16=16C, =>, C=1

Let x=-2, =>, -32=16A, =>, A=-2

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

B-D=-2

Coefficients of x^3

0=B+D

So, B=-1 and D=1

So,

(3x^2+12x-20)/(x^4-8x^2+16)=-2/(x+2)^2-1/(x+2)+1/(x-2)^2+1/(x-2)