How do you split a sum of two rational expressions with unlike denominators in the fraction #\frac { 2x + 3} { x ^ { 2} + 3x + 2}#?

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Answer 1 · 2017-03-13T17:52:29

THe answer is #=1/(x+2)+1/(x+1)#

First, we factorise the denominator

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

We perform a decomposition into partial fractions

#(2x+3)/(x^2+3x+2)=(2x+3)/((x+2)(x+1))#

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

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

The denominators are the same, we compare the numerators

#2x+3=A(x+1)+B(x+2)#

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

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

Therefore,

#(2x+3)/(x^2+3x+2)=1/(x+2)+1/(x+1)#