Show that the polynomial f(x)

Let $t=x^{2}$

So, $f(t)=t^{2}+4 t+6$

Now, to find the zeroes, we will equate $f(t)=0$.

$\Rightarrow t^{2}+4 t+6=0$

Now, $t=\frac{-4 \pm \sqrt{16-24}}{2}$

$=\frac{-4 \pm \sqrt{-8}}{2}$

$=-2 \pm \sqrt{-2}$

i. e., $x^{2}=-2 \pm \sqrt{-2}$

$\Rightarrow x=\sqrt{-2 \pm \sqrt{-2}}$, which is not a real number.

The zeroes of a polynomial should be real number $s$.

$\therefore$ The given $f(x)$ has no zeroes.