(x^2+2x+8) / (x^3-4x^2+4x)
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Originally Posted by cityismine Express $\displaystyle \frac{x^2+2x+8}{x^3-4x^2+4x}$ as a sum of partial fractions first factor the denominator $\displaystyle \frac{x^2+2x+8}{x(x-2)^2}$ $\displaystyle \frac{x^2+2x+8}{x(x-2)^2} \equiv \frac{A}{x} + \frac{B}{x-2} + \frac{C}{(x-2)^2}$ Now continue doing this with the method your familiar with. Bobak
Originally Posted by cityismine (x^2+2x+8) / (x^3-4x^2+4x) Factor the Denominator first: $\displaystyle x^3-4x^2+4x=x(x^2-4x+4)=x(x-2)^2$ Thus, the PFD is : $\displaystyle \frac{x^2+2x+8}{x(x-2)^2}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$ Try taking it from here.
2/x - 1/(x-2) + 8/(x-2)2 Is that the right answer?
Originally Posted by cityismine 2/x - 1/(x-2) + 8/(x-2)2 Is that the right answer? Yes, Good job Bobak
OMG, the book was showing a different answer, so I thought it was wrong. I guess the book is wrong.
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