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]}