Given some degree 6 polynomial $f(x) \in \mathbb{Q}[x]$, is there any invariant of the polynomial (depending on the coefficents) that will tell you if this polynomial has 6 complex roots or just 2 complex roots.
Edit: By complex root I mean a root in $\mathbb{C}$ but not $\mathbb{R}$.
In other words, is it possible to tell between the two cases without explicitly finding the roots.
Thank you
Best Answer
Yes, there is a closed algebraic expression involving only the coefficients of $f$ such that $f$ has exactly 6 non-real roots if and only if this expression is true thanks to the Tarski–Seidenberg quantifier elimination theorem. In short this says that:
``any system of polynomial equations and inequalities over the reals involving or, and, not, for all and there exists is equivalent to a quantifier free one.''
As having 6 non-real roots is equivalent to having no real roots which is equivalent to the statement: $$ \forall x \in \mathbb{R},\; f(x) \neq 0 $$ we can transform this into a quantifier free statement using the cylindrical decomposition algorithm (this is implemented as Mathematicas Resolve function. However this can quickly become unwieldy, for example when $$ f(x) = x^4 + d x^3 + cx^2 + bx + a $$ the command:
results in:
$$ \left(c<\frac{3 d^2}{8}\land \left(\left(b<\frac{1}{8} \left(4 c d-d^3\right)-\frac{\sqrt{-512 c^3+576 c^2 d^2-216 c d^4+27 d^6}}{24 \sqrt{3}}\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,1\right]\right)\lor \left(\frac{1}{8} \left(4 c d-d^3\right)-\frac{\sqrt{-512 c^3+576 c^2 d^2-216 c d^4+27 d^6}}{24 \sqrt{3}}\leq b<\frac{1}{8} \left(4 c d-d^3\right)\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,3\right]\right)\lor \left(b=\frac{1}{8} \left(4 c d-d^3\right)\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,2\right]\right)\lor \left(\frac{1}{8} \left(4 c d-d^3\right)<b\leq \frac{\sqrt{-512 c^3+576 c^2 d^2-216 c d^4+27 d^6}}{24 \sqrt{3}}+\frac{1}{8} \left(4 c d-d^3\right)\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,3\right]\right)\lor \left(b>\frac{\sqrt{-512 c^3+576 c^2 d^2-216 c d^4+27 d^6}}{24 \sqrt{3}}+\frac{1}{8} \left(4 c d-d^3\right)\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,1\right]\right)\right)\right)\lor \left(c\geq \frac{3 d^2}{8}\land a>\text{Root}\left[256 \text{$\#$1}^3+\text{$\#$1}^2 \left(-192 b d-128 c^2+144 c d^2-27 d^4\right)+\text{$\#$1} \left(144 b^2 c-6 b^2 d^2-80 b c^2 d+18 b c d^3+16 c^4-4 c^3 d^2\right)-27 b^4+18 b^3 c d-4 b^3 d^3-4 b^2 c^3+b^2 c^2 d^2\&,1\right]\right) $$