How do you use the rational roots theorem to find all possible zeros of f(x)=x^4-x-4?

Mar 23, 2016

The rational root theorem helps us determine that this $f \left(x\right)$ has no rational zeros, only irrational and/or Complex ones.

Explanation:

$f \left(x\right) = {x}^{4} - x - 4$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ must be expressible in the form $\frac{p}{q}$ for integers $p$, $q$ with $p$ a divisor of the constant term $- 4$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1$, $\pm 2$, $\pm 4$

Trying each in turn, we find:

$f \left(1\right) = 1 - 1 - 4 = - 4$

$f \left(- 1\right) = 1 + 1 - 4 = - 2$

$f \left(2\right) = 16 - 2 - 4 = 10$

$f \left(- 2\right) = 16 + 2 - 4 = 14$

$f \left(4\right) = 256 - 4 - 4 = 248$

$f \left(- 4\right) = 256 + 4 - 4 = 256$

So there are no rational zeros, but $f \left(x\right)$ changes sign in $\left(- 2 , - 1\right)$ and $\left(1 , 2\right)$, so there are irrational zeros in those intervals.

That's as much as we can learn about this $f \left(x\right)$ from the rational root theorem.