High degree polynomial with complex roots only (no real roots) Suppose a given polynomial $p(x)=a_0+a_1x^1+...+a_{2n}x^{2n}$ for $n\in{N}$.
How can I show that if all roots are complex? note that all coefficients are real. I actually need the solution for multinomials with $m$ variables but for simplicity, I have chosen a simple polynomial.
 A: One of the key ideas here is to express the polynomial as a sum of squares. If there is no value of the variable(s) for which the squares are all zero, then there are no real roots. I have linked below to various statements that such polynomials can be decomposed into sums of squares, but have found limited material on effective methods for doing this.
There is plenty of information and links in the Wikipedia article on positive polynomials
For your single variable case the article states that any non-negative polynomial over $\mathbb R$ can be expressed as the sum of two squares of polynomials over $\mathbb R$. In fact this is easy to see as the roots then have to come in complex conjugate pairs and give a factor of the form $(x-a)^2+b^2$ for each pair of roots $a\pm bi$. Then the identity $$(p^2+q^2)(r^2+s^2)=(pr+qs)^2+(ps-qr)^2$$ shows that the product of such factors is itself a sum of two squares. This, of course, does not help unless there is an effective way of finding a decomposition into a sum of squares in the first place.
The multi-variable case was Hilbert's 17th problem, solved by Artin, who showed that a non-negative polynomial was the sum of squares of rational functions. Again, this excludes real solutions if you avoid the squares being simultaneously zero. Again, I think this is non-constructive.
The question of effective methods seems to be addressed in this Mathoverflow question and answers. See also this, which has some material on polynomials, and this also.
See further Polynomial Sums of Squares and there is a paper (in English) here.
