How do I find an irrational root of a quartic function?

Question

For example, the equation $2x^4 - 3x^3 - 1 = 0$ has no rational roots, but it does have an irrational root between x = 1 and x = 2. If it had a rational root, I would be able to find it, but I am a bit confused on how to find an irrational root.

Answer 1

In this case consider: \begin{align} 2 \, x^4 - 3 \, x^3 -1 &= (a \, x^2 + b \, x + 1)(x^2 + c \, x -1) \\ &= a \, x^4 + (ac + b) \, x^3 + (-a + bc + 1) \, x^2 + (-b + c) \, x - 1 \end{align} This leads to $b = c$, $a=2$, $1 - a + bc = -1 + b^2 = 0$ or $b = \pm 1$, and $ac + b = 3b = -3$ which narrows $b$ to $b = -1$. Now, $$2 \, x^4 - 3 \, x^3 -1 = (a \, x^2 + b \, x + 1)(x^2 + c \, x -1) = (2 \, x^2 - x +1)(x^2 - x - 1).$$ The polynomial $2 \, x^2 - x + 1$ can be factored as $2(x - a)(x - b)$, where $4a = 1 + i \, \sqrt{7}$ and $4b = 1 - i \sqrt{7}$, and $x^2 - x - 1$ can be factored as $(x - \alpha)(x - \beta)$ where $2 \, \alpha = 1 + \sqrt{5}$ and $2 \, \beta = 1 - \sqrt{5}$. This brings the factorization into the form: $$2 \, x^4 - 3 \, x^3 -1 = 2(x - a)(x - b)(x - \alpha)(x - \beta).$$ The four roots would then be $$x \in \{ a, b, \alpha, \beta \} = \left\{ \frac{1 + i \, \sqrt{7}}{4}, \frac{1 - i \, \sqrt{7}}{4}, \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2} \right\}.$$

Answer 2

$$ -1 - 3x^3 + 2x^4 = (-1-x+x^2)(1-x+2x^2) $$ Can you continue from here?