How do you express #(x² + 2x-7 )/( (x+3)(x-1)²)# in partial fractions?

Question

1383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-29T11:43:30

The answer is #==(-1/4)/(x+3)+(5/4)/(x-1)-1/(x-1)^2#

Let's do the decomposition in partial fractions

#(x^2+2x-7)/((x+3)(x-1)^2)=A/(x+3)+B/(x-1)+C/(x-1)^2#

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

Therefore,

#x^2+2x-7=A(x-1)^2+B(x+3)(x-1)+C(x+3)#

Let #x=1#, #=>#, #-4=4C#, #=>#, #C=-1#

Let #x=-3#, #=>#, #-4=16A#, #=>#, #A=-1/4#

Coefficients of #x^2#, #=>#, #1=A+B#

So, #B=1+1/4=5/4#

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