Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$. 
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.

Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no roots hence proving that above polynomial has $4$ real roots.
But using online calculators I get zero real roots. Where am I wrong?

I think the notion that the fourth derivative having no real roots proves that the polynomial itself has four real roots is your problem.  Can you explain your reasoning a bit more?  I mean to say, $x^6+x^4+1$ clearly has no real roots (it is everywhere positive), but its fourth derivative $360x^2+24$ has no real roots either (it is likewise everywhere positive).
The polynomial in your problem does indeed have no real roots.
