Number of positive integers $n$ The question is to find out the number of positive integers $n$ such that $nx^4+4x+3 \leq 0 $ for some real $x$ (without using a graphic calculator).
My attempt at the solution:
We have $nx^4+4x+3 \leq 0 $ which is $nx^4 \leq -3-4x$ or $n \leq -\frac{(3+4x)}{x^4}$. It can easily be observed that for $x=-1$ the condition of $n$ being a positive integer is satisfied. But I could not get others. Moreover to satisfy that $n$ is positive we must have $x<-3/4$.
Please help me in this regard. Thanks.

Solution WITHOUT calculus.
Note that
$$nx^4+4x+3=(n-1)x^4+x^4+4x+3=(n-1)x^4+((x-1)^2+2)(x+1)^2.$$ 
Hence for $n=1$, we have that the above polynomial is $0$ at $x=-1$.
If $n\geq 2$ then, for any $x\in\mathbb{R}$,
$$nx^4+4x+3\geq x^4+2(x+1)^2>0$$
because the squares $x^4$ and $(x+1)^2$ are zero for different values of $x$ (namely $x=0$ and $x=-1$).
