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I here would like to clear my doubt on the question below:

$$\frac{1}{x(x-1)(x-2)}\;,$$ that is, we want to bring it into the form: $$\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}\;,$$ in which the unknown parameters are $A,B$, and $C$. Multiplying these formulas by $x(x − 1)(x − 2)$ turns both into polynomials, which we equate: $$A(x-1)(x-2) + Bx(x-2) + Cx(x-1) = 1\;,$$ or, after expansion and collecting terms with equal powers of $x$: $$(A+B+C)x^2 - (3A+2B+C)x + 2A = 1\;.$$ At this point it is essential to realize that the polynomial $1$ is in fact equal to the polynomial $0x^2 + 0x + 1$, having zero coefficients for the positive powers of $x$. Equating the corresponding coefficients now results in this system of linear equations:

$$\left\{\begin{align*} &A+B+C = 0\\ &3A+2B+C = 0\\ &2A = 1\;. \end{align*}\right.$$

Solving it results in: $$A = \frac{1}{2},\, B = -1,\, C = \frac{1}{2}\;.$$

So from my solving I had different values of $A,B$, and $C$ which gave me:

$$\left\{\begin{align*} &A=\frac12\\ &B= 2\\ &C= -\frac52\;. \end{align*}\right.$$

Can someone please tell me if these answers are correct because when I substitute these values into equation $A+B+C= 0$, it still gave me a zero.

But this time for the $2$nd equation, instead of $3A+2B+C= 0$, I used $-3A+2B+C= 0$, which then by substituting the values of $A, B$, and $C$ I had, also gave me a zero. Only $A= \frac12$ was the same as obtained from $2A= 1$.

Does this mean that the values that I have obtained for $A, B$, and $C$ are also correct? Kindly can someone please give a clear explanation to this?

Many thanks.

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  • $\begingroup$ How did you get the other values? $\endgroup$ – Stefan Jul 30 '13 at 7:46
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    $\begingroup$ Just a quick remark, you can find $A$, $B$ and $C$ directly by multiplying the equation $$ \frac{1}{x(x-1)(x-2)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}$$ by (x-1) and then take the limit when $x$ goes to $1$. We get $$\lim_{x\to 1} \frac{1}{x(x-2)} = \lim_{x\to 1} \frac{A(x-1)}{x} + B + \frac{C(x-1)}{x-2}=B$$ and do the same thing for $A$ and $C$. Another trick to get the equation $A+B+C=0$ you multiply by $x$ the equation and then take the limit when $x$ goes to $+\infty$ $\endgroup$ – user37238 Jul 30 '13 at 9:26
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$$A=\frac 12,B-1=C=\frac 12$$ these valuse are correct

from the step:

$$A(x-1)(x-2)+Bx(x-2)+Cx(x-1)=1$$

put $x=1,x=2,x=0$ you will get right values

even from this equations you also get same values: $$\left\{\begin{align*} &A+B+C = 0\\ &3A+2B+C = 0\\ &2A = 1\;. \end{align*}\right.$$

from 3rd equation $A=\dfrac 12$

after perform (2)-(1)

$$2A+B=0\implies B=-1$$

and in eqn (1)

$$A+B+C=0\implies \frac 12-1+C=0\implies C=\frac 12$$

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No, your values are not correct: they do not satisfy the equation $3A+2B+C=0$. The equation $-3A+B+C=0$ has nothing to do with the problem, so numbers obtained by using it simply aren’t relevant. I suspect that you got the equation $-3A+B+C=0$ by making a sign error in collecting the coefficients of $x$ in $A(x-1)(x-2)+Bx(x-2)+Cx(x-1)$, e.g., by converting $-(3A+2B+C)$ to $-3A+2B+C$ instead of to $-3A-2B-C$.

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Examine the form $$A(x−1)(x−2)+Bx(x−2)+Cx(x−1)=1$$

This is supposed to be true for every value of $x$. If we try $x=0$ we get $2A=1$, $x=1$ gives $=B=1$, and $x=2$ gives $2C=1$.

This short-cut method (known as the "cover-up" rule) can be used to find the numerators for partial fractions where all the denominators are linear, and is very useful in other cases too.

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