# SOLUTION: Partial fraction decomposition: (5x^2+20x+6)/x^3+2x^2+x Here's what I got but I think it is wrong: (5x^2+20x+6)/x(x^2+2x+1) (5x^2+20x+6)/x(x+1)^2 (A/x)+(B/(x+1))+(C/(x+1)^2)

Algebra ->  Algebra  -> Exponents-negative-and-fractional -> SOLUTION: Partial fraction decomposition: (5x^2+20x+6)/x^3+2x^2+x Here's what I got but I think it is wrong: (5x^2+20x+6)/x(x^2+2x+1) (5x^2+20x+6)/x(x+1)^2 (A/x)+(B/(x+1))+(C/(x+1)^2)      Log On

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 Question 467157: Partial fraction decomposition: (5x^2+20x+6)/x^3+2x^2+x Here's what I got but I think it is wrong: (5x^2+20x+6)/x(x^2+2x+1) (5x^2+20x+6)/x(x+1)^2 (A/x)+(B/(x+1))+(C/(x+1)^2) A(x+1)(x+1)^2 + B(x)(x+1)^2 + C(x)(x+1) (5x^2+20x+6)=Ax^3+Bx^3+3Ax^2+2Bx^2+Cx^2+3Ax+Bx+Cx+A (5x^2+20x+6)=x^3(A+B) + x^2(3A+2B+C) + x(3A+B+C) + A A+B=0 3A+2B+C=5 3A+B+C=20 A=6 A=6 B=-6 C=-1 But this isn't right. Where did I go wrong?Answer by robertb(4012)   (Show Source): You can put this solution on YOUR website! should have been written as , and then equated to . Compare term-by-term to get the system A + B = 5, 2A + B + C = 20, A = 6. ==> B = -1, and C = 9.