Let #f(x) = 1+2x+x^3+4x^5# and note that for every #x#, #x# is a root of the equation if and only if #x# is a zero of #f#.
#f# has at least one real zero (and the equation has at least one real root).
#f# is a polynomial function, so it is continuous at every real number. In particular, #f# is continuous on the closed interval #[-1,0]#.
#f(-1) = 1-2-1-4 = -8# and #f(0)=1#
#0# is between #f(-1)# and #f(0#, so the Intermediate Value Theorem tells us that there is at least one number #c# in #(-1,0)# with #f(c) = 0#.
This #c# is a zero of #f# and a root of the equation.
#f# cannot have two (or more) zeros
Suppose that #f# had two (or more) zeros, call them #a# and #b#. So #f(a)=0=f(b)#
#f# is continuous on the closed interval #[a,b]# (it's still a polynomial) and
#f# is differentiable on the open interval #(a,b)#
(#f'(x) = 2+3x^2+20x^4# exists for all #x# in the interval.)
and #f(a) = f(b)# so by Rolle's Theorem (or by the Mean Value Theorem) there is a #c# in #(a,b)# with #f'(c)=0#
However,
#f'(x) = 2+3x^2+20x^4# can never be #0#.
(Look at each term. Both #20x^4# and #3x^2# are at least #0# and the constant adds #2#. So, #f'(x) >= 2#.)
That means that no #c# with #f'(c)=0# can exist.
Conclusion
If there were two (or more) zeros, then we could make #f'(c)=0#.
But, clearly, we cannot make #f'(c)=0#, so there cannot be two (or more) zeros.