Okay, for example, I have 1 value which I can sub x into. And the value is 2. So now I sub x=2 and I have to find unknowns A, B and C.
After that I found A=5 and have unknown B. What should I sub x now?
Checklist:
1. The numerator has a degree less than the denominator. (Tick)
2. The denominator can be factorised. (Tick)
Now, we require a partial fraction of the form
$\displaystyle \frac{A}{3x + 4} + \frac{Bx + C}{x^2 + 4} = \frac{17x^2 + 23x + 12}{(3x + 4)(x^2 + 4)}$.
The numerator is always of one less degree than the denominator.
Getting the common denominator:
$\displaystyle \frac{A(x^2 + 4) + (Bx + C)(3x + 4)}{(3x + 4)(x^2 + 4)} = \frac{17x^2 + 23x + 12}{(3x + 4)(x^2 + 4)}$.
Can you see that since the denominators are equal, so must the numerators...
So $\displaystyle A(x^2 + 4) + (Bx + C)(3x + 4) = 17x^2 + 23x + 12$.
In this case, there does not exist an $\displaystyle x$ that you can use to eliminate the $\displaystyle A$.
So we will use another method. Expand all the brackets...
$\displaystyle Ax^2 + 4A + 3Bx^2 + 4Bx + 3Cx + 4C = 17x^2 + 23x + 12$
Equating like powers of $\displaystyle x$, we get
$\displaystyle (A + 3B)x^2 + (4B + 3C)x + 4A + 4C = 17x^2 + 23x + 12$.
Can you see that for this equation to be true, then
$\displaystyle A + 3B = 17$
$\displaystyle 4B + 3C = 23$
$\displaystyle 4A + 4C = 12$.
Now you have three equations in three unknowns that you can solve simultaneously. Can you go from here?
Yes, if you substitute $\displaystyle -\frac{4}{3}$ you can eliminate $\displaystyle B$ and $\displaystyle C$ to find $\displaystyle A$.
But there does not exist any number that will be able to eliminate $\displaystyle A$ to find $\displaystyle B$ and $\displaystyle C$.
Using your method...
$\displaystyle A(x^2 + 4) + (Bx + C)(3x + 4) = 17x^2 + 23x + 12$
$\displaystyle A\left[\left(-\frac{4}{3}\right)^2 + 4\right] + (Bx + C)\left[3\left(-\frac{4}{3}\right) + 4\right] = 17\left(-\frac{4}{3}\right)^2 + 23\left(-\frac{4}{3}\right) + 12$
$\displaystyle \frac{52}{9}A = \frac{272}{9} - \frac{92}{3} + 12$
$\displaystyle \frac{52}{9}A = \frac{272}{9} - \frac{276}{9} + \frac{108}{9}$
$\displaystyle \frac{52}{9}A = \frac{104}{9}$
$\displaystyle A = 2$.
Now that you have $\displaystyle A$ you should be able to solve the simultaneous equations I gave you easier.
But like I said, in this case, the ONLY method that will get you all three unknowns is to equate like powers of $\displaystyle x$.
You substitute whatever will make the coefficients of A, B, C, etc = 0.
So in your case, we wanted the coefficient of $\displaystyle Bx + C$ to become 0.
So we let $\displaystyle 3x + 4 = 0$
$\displaystyle 3x = -4$
$\displaystyle x = -\frac{4}{3}$.
Notice that the coefficient of A is $\displaystyle x^2 + 4$. To try to eliminate A, we would need to let $\displaystyle x^2 + 4 = 0$
$\displaystyle x^2 = -4$.
Is it possible to square a number and end up with something negative?
You would only substitute $\displaystyle x = 0$ if one of your coefficients happened to be $\displaystyle x$.
You would not be able to find C by substituting $\displaystyle x = 0$ or $\displaystyle x = 1$. You will still have A in the equation.
Edit: Actually, if you have already found A, then yes, you can substitute $\displaystyle x = 0$.
You have
$\displaystyle A(x^2 + 4) + (Bx + C)(3x + 4) = 17x^2 + 23x + 12$.
Substituting $\displaystyle A = 2$ and $\displaystyle x = 0$ we find
$\displaystyle 2(0^2 + 4) + [B(0) + C][3(0) + 4] = 17(0)^2 + 23(0) + 12$
$\displaystyle 8 + 4C = 12$
$\displaystyle 4C = 4$
$\displaystyle C = 1$.