<s>[INST] Finding the root of a degree $5$ polynomial $\textbf{Question}$: which of the following $\textbf{cannot}$ be a root of a polynomial in $x$ of the form $9x^5+ax^3+b$, where $a$ and $b$ are integers? 
A) $-9$
B) $-5$
C) $\dfrac{1}{4}$
D) $\dfrac{1}{3}$
E) $9$ 

I thought about this question for a bit now and can anyone provide any hints because I have no clue how to begin to eliminate the choices? 

Thank you very much in advance. 
 [/INST] If you want to try process of elimination, an easy place to start is by assuming that either $a = 0$ or $b = 0$.  (Hey, zero is an integer!)
If you set $a = 0$, then you get $b = -9x^5$.  Because $\mathbb{Z}$ is closed under multiplication, if $x$ is an integer, then so is $b$, and so you have a valid $(a, b)$ pair with $x$ as a root.  Thus, (A), (B), and (E) cannot be the right answer.  However, for $x = \frac{1}{4}$ or $x = \frac{1}{3}$, you'd get $b = \frac{-9}{1024}$ or $b = \frac{-1}{27}$, respectively, so $b \notin \mathbb{Z}$, and these aren't valid solutions.  So far, (C) and (D) are still possible answer choices.
If you set $b = 0$, then you get $9x^5 + ax^3 = 0$, which factors to $x^3(9x^2+a) = 0$, so either $x = 0$ (which isn't one of the answer choices) or $a = -9x^2$.  If $x = \frac{1}{3}$, then $a = -1$, which is an integer, so that rules out answer choice (D).  But if $x = \frac{1}{4}$, then $a = \frac{-9}{16}$, a non-integer, so (C) is still in the running.
To double-check that (C) is the correct answer, plug in $x = \frac{1}{4}$ into the original equation, to get $\frac{9}{1024} + \frac{1}{64}a + b = 0$.  Moving the constant to the right and multiplying by 64 gives $a + 64b = \frac{-9}{16}$.  Regardless of the specific values of $a$ and $b$, the left-hand side is an integer but the right-hand side is not.  This is a contradiction, so $\frac{1}{4}$ cannot be a root of the polynomial if $a, b \in \mathbb{Z}$.
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