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. 

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 $a, b \in \mathbb{Z}$.
