How do you write the partial fraction decomposition of the rational expression # (x+13)/(x^3+2x^2-5x-6)#?

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Answer 1 · 2017-03-26T12:23:10

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

We need to factorise the denominator

Let #f(x)=x^3+2x^2-5x-6#

Then

#f(2)=8+8-10-6=0#

Therefore,

#(x-2)# is a factor of #f(x)#

We can do a long division

#color(white)(aaaa)##x^3+2x^2-5x-6##color(white)(aaaa)##|##x-2#

#color(white)(aaaa)##x^3-2x^2##color(white)(aaaaaaaaaaaa)##|##x^2+4x+3#

#color(white)(aaaaa)##0+4x^2-5x#

#color(white)(aaaaaaa)##+4x^2-8x#

#color(white)(aaaaaaaa)##+0+3x-6#

#color(white)(aaaaaaaaaaaa)##+3x-6#

#color(white)(aaaaaaaaaaaa)##+0-0#

so,

#x^3+2x^2-5x-6=(x-2)(x^2+4x+3)#

#=(x-2)(x+1)(x+3)#

Now, we perform the decomposition into partial fractions

#(x+13)/(x^3+2x^2-5x-6)=(x+13)/((x-2)(x+1)(x+3))#

#=A/(x-2)+B/(x+1)+C/(x+3)#

#=(A(x+1)(x+3)+B(x-2)(x+3)+C(x-2)(x+1))/((x-2)(x+1)(x+3))#

The denominators are the same, we compare the numerators

#x+13=A(x+1)(x+3)+B(x-2)(x+3)+C(x-2)(x+1)#

Let, #x=2#, #=>#,#15=15A#, #=>#,#A=1#

Let, #x=-1#, #=>#, #12=-6B#, #=>#, #B=-2#

Let, #x=-3#, #=>#, #10=10C#, #=>#, #C=1#

So,

#(x+13)/(x^3+2x^2-5x-6)=1/(x-2)-2/(x+1)+1/(x+3)#