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$ \frac{125x^{2}+x+3}{x^{2}(x-5)} = > \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{(x-5)} | * x^{2}(x-5)$

$125x^{2}+x+3 = Ax(x-5) + B(x-5) + C (x^{2})$

$125x^{2}+x+3 = A x^{2} - 5Ax + Bx -5B +Cx^{2}$

$125x^{2}+x+3 = x^{2}(A+C) -x(A+B)-5B$

$3 = -5B \Rightarrow B = \frac{-3}{5}$

$-1 = A+B \Rightarrow A = -1 - B \Rightarrow A = \frac{-5}{5} - \frac{-3}{5} \Rightarrow A=\frac{-8}{5}$

$125 = A+C$

Where I did wrong in calculating of variable $A$, because correct answer is $A = \frac{-8}{25}$, but I get $A = \frac{-8}{5}$.

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    Are you using `=>` as an arrow, or as "greater than or equal"? For arrows, you can use `\Rightarrow`, that will produce $\Rightarrow$. For "greater than or equal", you should use `\geq`, which produces $\geq$.2011-09-05
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    The coefficient of $x$ is all wrong. You had $-5Ax+Bx$, and this became $-x(A+B)$; it should have been $x(B-5A)$.2011-09-05
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    thanks, i see now that missing -5A. it's time to stop math for today obvieously :)2011-09-05

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