Partial Fractions 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.
 A: $$A=\frac 12,B-1=C=\frac 12$$
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)
$$A+B+C=0\implies \frac 12-1+C=0\implies C=\frac 12$$
A: 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 $-3A-2B-C$.
A: 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.
