Say I have a quartic polynomial $f(x) = ax^4 + bx^3 + cx^2 + d$. I am told that $f(x)$ is nonnegative iff it can be expressed as a sum of squares as follows. $f(x) = \sum_{i=1}^4 q_i(x)^2$. As an intermediate step, I am trying to prove that if $f(x) >= 0$ for all $x \in R$, then $f(x) = p_0 ((x-a_1)^2 + b_1^2)((x-a_2)^2 + b_2^2)$.
However, I am having trouble with this. Could someone give me a hint in the right drection?
On
Real algebraic geometry is a really hard subject, and the theorem you are referencing is a topic in a real algebrraic geometry. However what you are trying to prove is simpler. The fundamental theorem of algebra says that every polynomial can be factored over the reals as a product of linear and irreducible quadratic factors. Your polynomial, if it is strictly positive, has no linear factors because it is strictly positive. So now you just need to show that an irreducible polynomial over the reals of degree 2 that is positive valued has the form you prescribed. For this, you can use the quadratic formula.
Hint: If the complex roots of $f$ are $z_1$ through $z_4$, then $f(x)=a(x-z_1)(x-z_2)(x-z_3)(x-z_4)$. But since $f$ has no real roots, its complex roots come in conjugate pairs, so the linear factors combine together two and two to form upwards-pointing parabolas.
$$ f(x) = a((x-a_1)^2+s)(x-a_2)^2+t) $$
for some $a_1,a_2\in \mathbb R$ and $s,t> 0$.