<s>[INST] question about partial fractions why $\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-2)^2}$ can anyone tell me why
$$\frac{6x^2+19x+15}{(x-1)(x-2)^2}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{(x-2)^2}$$
I don't undestand the $\frac{C}{(x-2)^2}$ and also what is wrong according to basic math rules if I write the equation in the following way
$$\frac{6x^2+19x+15}{(x-1)(x-2)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-2}$$
Thanks
 [/INST] An easy way to see why your proposal doesn't work is the following: Multiply both sides of your equation with $(x-1)(x-2)^2$ and extend the formula continuously to $x=1$ and $x=2$. We now have
$$6x^2+19x+15 = A(x-2)^2+B(x-1)(x-2)+C(x-1)(x-2).$$
If we now enter $x=2$, we get $24+38+15 = A\cdot 0^2+B\cdot 1\cdot 0 + C\cdot 1\cdot 0 = 0$, which is obviously false.
On the other hand
$$6x^2+19x+15 = A(x-2)^2 + B(x-1)(x-2)+C(x-1)$$
would have reduced to $24+38+15 = C$, which is perfectly fine.
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