<s>[INST] How to know which values of $x$ can be used to find the unknowns $A$, $B$ and $C$? I'm having trouble understanding what my maths textbook says, and I've not been able to find help. I would appreciate an explanation in simple (high-school) terms.

Worked example
Question

Given that $5x^2 - 7x + 3 = A(x - 1)(x - 2) + B(x - 1) + C$ for all values of $x$, find the values of $A$, $B$ and $C$.

Answer

Since $5x^2 - 7x + 3 = A(x - 1)(x - 2) + B(x - 1) + C$ for all values of $x$, we can select any value of $x$ to find the unknowns $A$, $B$ and $C$.
Let $x$ = $1$, then $5(1)^2 - 7(1) + 3 = A(1 - 1)(1 - 2) + B(1 - 1) + C$,
i.e. $C = 1$.
Let $x$ = $2$, then $5(2)^2 - 7(2) + 3 = A(2 - 1)(2 - 2) + B(2 - 1) + C$,
i.e. $B = 8$.
Let $x$ = $0$, then $5(0)^2 - 7(0) + 3 = A(0 - 1)(0 - 2) + B(0 - 1) + C$,
i.e. $A = 5$.
$\therefore$ The values of $A$, $B$ and $C$ are $5$, $8$ and $1$ respectively.


Question


*

*Why where the values $x = 0, 1, 2$ chosen?

*Can we use other numbers such as $x = -3, 5$ and $7$ to find the unknowns?



I'm not sure about the answers to the two questions above.
For 1, I think the answer is that since any value of $x$ could be chosen, these were 3 random values.
I think that the answer to question 2 is also yes, but I was unable to find the values of $A$, $B$ and $C$ when $x = -3, 5$ and $7$.
Please could someone tell me if I'm right, or if not, where my reasoning is wrong?
 [/INST] For 1, the values $1$ and $2$ are chosen such that $(x-1)=0$ and $(x-2)=0$ respectively. This simplifies the calculations. $x=0$ simplifies the polynomial on the left hand side.
For 2, you can with $-3,5, 7$, but you need to solve the system:
$$\begin{cases}
69=20A-4B+C\\
93=12A+4B+C\\
199=30A+6B+C
\end{cases}$$
which is harder to solve than
$$\begin{cases}
3=2A-B+C\\
1=C\\
9=B+C
\end{cases}$$
obtained by using the textbook's choice $0,1,2$.

EDIT: To solve the first system, we subtract the first and second equations from the third, obtaining the equations:
$$\begin{cases}
130=10A+10B\\
106=18A+2B
\end{cases}$$
Now with only two unknowns, it is easy to show that $A=5$, $B=8$.
Finally substitute the values back to any equation and we will have $C=1$, as desired.
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