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

1 Answer
Dec 22, 2016

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

Explanation:

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

color(white)((x^2+2x)/(x^3-x^2+x-1)) = A/(x-1)+(Bx+C)/(x^2+1)

color(white)((x^2+2x)/(x^3-x^2+x-1)) = (A(x^2+1)+(Bx+C)(x-1))/(x^3-x^2+x-1)

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

Equating coefficients, we find:

{ (A+B=1), (-B+C=2), (A-C=0) :}

Adding all three of these equations, the B and C terms cancel out, leaving us with:

2A = 3

Hence A=3/2

Then from the first equation, we find:

B=1-A = 1-3/2 = -1/2

From the third equation, we find:

C = A = 3/2

So:

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