Given f(x) = x^4+2x+2
By the rational root theorem, any rational factor p/q in lowest terms must have p is a divisor of the constant term 2 and q is a divisor of the multiplier (1) of the highest order term.
That means that any rational root of f(x) = 0 must be +-1 or +-2
f(-2) = 16-4+2 = 14
f(-1) = 1-2+2 = 1
f(1) = 1+2+2 = 5
f(2) = 16+4+2 = 22
So none of +-1 or +-2 are roots of f(x) = 0.
So it has no rational roots.
f'(x) = 4x^3+2
This is zero when x = -1/root(3)(2) and for no other real values of x. So f(x) has one turning point.
f(-1/root(3)(2)) = (-1/root(3)(2))^4+2(-1/root(3)(2))+2
=1/(2root(3)(2))-2/root(3)(2)+2
=-3/2*1/root(3)(2)+2
> -3/2+2 = 1/2 > 0
So f(x) > 0 for all x in RR
That is f(x) = 0 has no real roots. It has 4 complex roots.
graph{x^4+2x+2 [-11.5, 8.5, -0.8, 9.2]}