How many zeros does $f(x)= 3x^4 + x + 2 $ have? 
How many zeros does this function have?
  $$f(x)=  3x^4 + x + 2 $$


$f(x)$ has $4$ zeros by the fundamental theorem of algebra.
Here is an alternative way to find that none of them is real:
$$ \ \ \ \ \ \ \ \ 3x^4+x+2=0 \ \ \ \ \ (1)\\ 3x^4=-x-2,$$ which, as $0\ne-2$, is equivalent to $$3x^3=-1-\frac{2}{x}.$$
Letting $g(x)=3x^3$ and $\displaystyle h(x)=-1-\frac{2}{x}$ we have $$
\left\{ 
\begin{array}{c}
x>0 \\ 
g(x)=h(x)\\
g(x)>0 \\ 
h(x)<-1 \\
\end{array}
\right. 
  \text{and}   \left\{ 
\begin{array}{c}
x<0 \\ 
g(x)=h(x)\\g(x)<0 \\ 
h(x)>-1 \\
\end{array}
\right. .$$
The first has no solutions whereas the second implies $$\left\{ 
\begin{array}{c}
-1<g(x)<0 \\ 
-1<h(x)<0 \\
\end{array}
\right. \leftrightarrow \left\{ 
\begin{array}{c}
-\frac{1}{\sqrt[3]{3}}<x<0 \\ 
x<-2 \\
\end{array}
\right. ,$$ which is an inconsistent system. Thus, $(1)$ is impossible in $\mathbb{R}$.
