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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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    thanks, i see now that missing -5A. it's time to stop math for today obvieously :)2011-09-05

3 Answers 3

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HINT $\ $ It's simpler to use the Heaviside cover-up method. First, evaluating your $\rm\:E_2 = 2$nd equation at $\rm\:x = 0\:$ yields $\rm\:3 = -5\:b\:.\:$ Next, differentiating $\rm\:E_2\:$ and evaluating at $\rm\:x = 0\:$ yields $\rm\: 1 = b - 5\:a\:.$ Solve those for $\rm\:a,b\:$. Finally evaluating $\rm\:E_2\:$ at $\rm\:x=5\:$ yields $\rm\: 3133 = 25\:c\:.$

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From

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

it should be

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

instead of

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

Hence

$\left\{ \begin{array}{c} 3=-5B \\ 1=B-5A \\ 125=A+C \end{array} \Leftrightarrow \right. \left\{ \begin{array}{c} B=-\frac{3}{5} \\ 1=-\frac{3}{5}-5A \\ 125=A+C \end{array} \Leftrightarrow \right. \left\{ \begin{array}{c} B=-\frac{3}{5} \\ A=-\frac{8}{25} \\ C=\frac{3133}{25} \end{array} \right. $

and the expansion into partial fractions is $\frac{125x^{2}+x+3}{x^{2}(x-5)}=-\frac{8}{25x}-\frac{3}{5x^{2}}+\frac{3133}{25\left( x-5\right) }.$


Second method. Multiply

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

by $x-5$

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

and let $x\rightarrow 5$

$\lim_{x\rightarrow 5}\frac{125x^{2}+5+3}{x^{2}}=\frac{3133}{25}=C.$

Multiply $(\ast )$ by $x^{2}$

$\frac{125x^{2}+x+3}{x-5}=Ax+B+\frac{x^{2}}{x-5}C$

and let $x\rightarrow 0$

$\lim_{x\rightarrow 0}\frac{125x^{2}+x+3}{x-5}=-\frac{3}{5}=B.$

Substitute $C$ and $B$ in $(\ast )$

$\frac{125x^{2}+x+3}{x^{2}(x-5)}=\frac{A}{x}-\frac{3}{5}\frac{1}{x^{2}}+ \frac{3133}{25}\frac{1}{x-5}$

and set, say, $x=1$

$-\frac{125+1+3}{4}=A-\frac{3}{5}-\frac{3133}{25}\frac{1}{4},$

to find $A=-\dfrac{8}{25}$.

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The expression is of the form

$\frac{Ax + b}{ax^2 + bx + c} + \frac{C}{x + d}$ i.e Product of quadratic and Linear Factors.

Keeping the above form in mind,this is what i did.

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

Look for a common denominator in RHS

$\frac{(Ax + B)(x - 5)}{x^2(x-5)} + \frac{c(x^2)}{(x-5)(x^2)}$ Adding the above fraction

$\frac{(Ax + B)(x - 5)+c(x^2)}{x^2(x - 5)}$

Now equating the numerator from LHS with the numerter obtained above {RHS} as the denominator of both LHS and RHS contain the same expression,you obtain,

$125x^2 + x + 3 = Ax^2 - 5Ax + Bx - 5B + cx^2$

Let A = 5,find out the value of C,I'am pretty sure even B would get lost.Obtain the value of B and C by choosing a suitable value for x.

I'am having difficuilty in determining the value of B and C,Let me know if you could.

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    $A$ cannot be $5$.2011-09-08