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 ->
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)
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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(5830) (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.
