<s>[INST] How do you solve this quadratic equation? The number of values of a for which
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
is an identity in x is?
Here's how much I was able to solve through:-
$$ (a^2 - 3a + 2)x^2 + (a^2-5a + 6)x + a^2-4 = 0$$
$$ ((a-1)(a-2))x^2 + ((a-3)(a-2))x + (a+2)(a-2) = 0$$
$$ (a-2)[(a-1)x^2 + (a-3)x + (a+2)] = 0$$
$$ so \hspace{5pt} (a-2) = 0 \hspace{5pt} or \hspace{5pt} [(a-1)x^2 + (a-3)x + (a+2)] = 0$$
$$ so \hspace{5pt} a = 2$$
$$ Now \hspace{5pt} [(a-1)x^2 + (a-3)x + (a+2)] = 0 $$
I don't know what to do next. Thanks in advance
 [/INST] HINT:
If $Ax^2+Bx+C=ax^2+bx+c$ is an identity
$A=a,B=b,C=c$ 
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