EDITED for correct function.
$f(x) = 1+x^2+x^3+x^4+x^5$
is positive for x>0.
So consider $x < 0$:
One can write $f(x) = (1+x^2)(1+x^3) + x^4$ which is positive at least for $(1+x^3) >0$, i.e. $x > -1$.
$f'(x) = 2x+3x^2+4x^3+5x^4 = x(2+3x)+x^3(4+5x)$
so this is positive at least for $x < -4/5 $ since then all terms are positive.
This means that $f(x) = 0$ can only happen at $x < -1$ and since $f(x)$ is rising there, we have that there can only be one real root, which is at $x < -1$.
Now this root can be further locked in. Observing e.g. that $f(-1.5) = -85/32 <0$, we can give an interval for the root at (-1.5 , -1). This can obviously be improved, knowing that there is only one real root and that f(x) is monotonously increasing in this interval.