Finding the root of a degree $5$ polynomial

Question

$\textbf{Question}$: 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.

Answer 1

Hint:

If a reduced rational number $\,\frac rs\,$ is a root of an integer polynomial $\,a_0+a_1x+...+a_nx^n\,$ , then

$$r\mid a_0\;,\;\;s\mid a_n$$

The above is called the Rational Root Theorem, sometimes

Answer 2

Hint: Use Rational Root Theorem.

Answer 3

Use the rational root theorem, and note that the denominator of one of the options given does not divide $9$...

Answer 4

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