How do you express #(4x^2 - 7x - 12) / ((x) (x+2) (x-3))# in partial fractions?

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Answer 1 · 2016-12-16T06:52:35

The answer is #=2/(x)+(9/5)/(x+2)+(1/5)/(x-3)#

Let's do the decomposition into partial fractions

#(4x^2-7x-12)/(x(x+2)(x-3))=A/(x)+B/(x+2)+C/(x-3)#

#=(A(x+2)(x-3)+B(x)(x-3)+C(x)(x+2))/(x(x+2)(x-3))#

Therefore,

#4x^2-7x-12=A(x+2)(x-3)+B(x)(x-3)+C(x)(x+2)#

Let #x=0#, #=>#, #-12=-6A#, #=>#, #A=2#

Let #x=-2#,#=>#, #18=10B#, #=>#, #B=9/5#

Let #x=3#, #=>#, #3=15C#, #=>#, #C=1/5#

So,

#(4x^2-7x-12)/(x(x+2)(x-3))=2/(x)+(9/5)/(x+2)+(1/5)/(x-3)#

Answer 2 · 2016-12-16T06:53:04

#(4x^2-7x-12)/(x(x+2)(x-3))=2/x+9/(5(x+2))+1/(5(x-3))#

Let #(4x^2-7x-12)/(x(x+2)(x-3))hArrA/x+B/(x+2)+C/(x-3)#

Hence #(4x^2-7x-12)/(x(x+2)(x-3))hArr(A(x+2)(x-3)+Bx(x-3)+Cx(x+2))/(x(x+2)(x-3))#

i.e. #4x^2-7x-12hArrA(x+2)(x-3)+Bx(x-3)+Cx(x+2)#

Now putting #x=0#, we get #-6A=-12# or #A=2#

putting #x=3#, we get #15C=3# or #C=1/5#

and putting #x=-2#, we get #10B=16+14-12=18# or #B=9/5#

Hence #(4x^2-7x-12)/(x(x+2)(x-3))=2/x+9/(5(x+2))+1/(5(x-3))#