# What are the real zeros of f(x) = 3x^6 + 1?

Refer to explanation

#### Explanation:

There aren't any. Let assume that there is value $a$ for which
$f \left(a\right) = 0$ hence

$3 \cdot {a}^{6} + 1 = 0 \implies {a}^{6} = - \frac{1}{3}$

which is impossible because the first part is always positive and the second is negative.