How do you express #(2x^3-x^2+x+5)/(x^2+3x+2) # in partial fractions?

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Answer 1 · 2016-12-20T19:12:18

The answer is #=2x-7+1/(x+1)+17/(x+2)#

As the degree of the numerator is #># than the degree of the denominator, we do a long division

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

#color(white)(aaaa)##2x^3+6x^2+4x##color(white)(aaaaaa)##∣##2x-7#

#color(white)(aaaaa)##0-7x^2-3x+5#

#color(white)(aaaaaaa)##-7x^2-21x-14#

#color(white)(aaaaaaaaaaaaaa)##18x+19#

Therefore,

#(2x^3-x^2+x+5)/(x^2+3x+2)=2x-7+(18x+19)/(x^2+3x+2)#

We can now do the decomposition into partial fractions

#(18x+19)/(x^2+3x+2)=(18x+19)/((x+1)(x+2))=A/(x+1)+B/(x+2)#

#=(A(x+2)+B(x+1))/((x+1)(x+2))#

So,

#18x+19=A(x+2)+B(x+1)#

Let #x=-1#, #=>#, #1=A#

Let #x=-2#, #=>, #-17=-B#

#(2x^3-x^2+x+5)/(x^2+3x+2)=2x-7+1/(x+1)+17/(x+2)#