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

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Answer 1 · 2016-12-03T10:57:23

The answer is #=-2/(x+2)^2-1/(x+2)+1/(x-2)^2+1/(x-2)#

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)#