For a given polynomial of $n$th degree

$$P_n(x)=\sum_{i=0}^n a_ix^i = a_nx^n+a_{n-1}x^{n-1}+…+a_1x+a_0$$

you can guess rational solutions $x$ for the corresponding problem $P_n(x)=0$ by applying the following two rules:

1. $$x=\frac{p}{q}\text{, with } p \in \mathbb{Z} \land q \in \mathbb{N}\land p\mid a_0 \land q\mid a_n$$
2. $$\lvert x\rvert\le2\cdot \max\left\lbrace \sqrt[k]{\frac{\lvert a_{n-k}\rvert}{\lvert a_n\rvert}}, k=1, …, n\right\rbrace$$

Example

The equations above might be a little bit confusing so let’s apply them to the following example.

$$f(x)=4x^3+8x^2-2x-4$$

$$f(x)=0\Rightarrow2x^3+4x^2-x-2=0$$

After stating $a_0=-2$ and $a_n=2$ we can determine $p$ and $q$. $p$ is every divisor of $a_0$ that is an integer. Hence possible values are:

$$p=\pm 1\lor p=\pm 2$$

For $q$ all positive integer divisors of $a_n$ are possible:

$$q=1\lor q=2$$

Rule 1. $x=\frac{p}{q}$ now leads to six possible real solutions:

$$x=\pm \frac{1}{2}\lor x=\pm 1\lor x=\pm 2$$

The next step is validating each of them with rule 2. For the given example that doesn’t reduce the number of possible solutions because the following equation is true for every determined value of x.

$$\lvert x\rvert\le2\cdot \max\left\lbrace 2,\frac{\sqrt{2}}{2},1\right\rbrace$$

Finally $f(x)=0$ can be tested for every value of $x$. That will lead to one real solution $x_1=-2$. After conducting e.g. a polynomial division the other two non-real solutions can be determined as well ($x_{2/3}=\pm\frac{\sqrt{2}}{2}$).