partial fraction question

Question

$ \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 $A = \frac{-8}{5}$.

Answer 1

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 $\rm\: 3133 = 25\:c\:.$

Answer 2

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.

Answer 3

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 $$\frac{125x^{2}+x+3}{x^{2}(x-5)}=-\frac{8}{25x}-\frac{3}{5x^{2}}+\frac{3133}{25\left( x-5\right) }.$$

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