First implies .
Now say is non-constant so . Where are not necessrily distinct. (Note must be a monic polynomial. This is easily to see. Just expand LHS and RHS to the leading term).
Now here is the result that we need: The equation has complex zeros if and only if .
Look at the LHS at the factors . These has complex zeros if . But if this is the case then the RHS cannot have the same zeros because it has only real solutions. A contradiction! Thus, for all . And hence the only possibility is .
And this works because if we have .