<s>[INST] For what $a$ and $b$ is  $9x^4-12x^3+28x^2+ax+b$ a perfect square? If $9x^4-12x^3+28x^2+ax+b$ is a perfect square, find the value of $a$ and $b$. 
This is one of my past year examination's questions, some help on it?
(The answer for this problem is $a=-16$, $b=16$.)
 [/INST] Even though it doesn't affect the final outcome in this problem, I think it's better long-term practice not to assume that the leading term is positive. This emphasizes that there are two square roots to a square polynomial, differing by sign, just like with integers.

Suppose $9x^4-12x^3+28x^3+ax+b=g(x)^2$ for some polynomial $g(x)$. Because the degree of $9x^4-12x^3+28x^3+ax+b$ is 4, we have $4=\deg(g(x)^2)=2\deg(g(x))$, so that $g$ is a degree  2 polynomial. 
Write $g(x)=cx^2+dx+e$. Then
$$9x^4-12x^3+28x^3+ax+b=(cx^2+dx+e)^2=(c^2)x^4+(2cd)x^3+(2ce+d^2)x^2+(2de)x+e^2.$$
Thus $c^2=9$, $2cd=-12$, $2ce+d^2=28$, $2de=a$, and $e^2=b$.
$c^2=9$ implies $c=3$ or $c=-3$. 
Combined with $2cd=-12$, this implies that $d=2$ or $d=-2$.
Combined with $2ce+d^2=28$, this implies that $e=4$ or $e=-4$.
If $c=3$, then $d=-2$ and $e=4$. If $c=-3$, then $d=2$ and $e=-4$. Thus, 
$$g(x)=\pm(3x^2-2x+4),$$
and $a=2de=-16$ and $b=e^2=16$.
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