The idea is basically:
Any monic polynomial can be factored as $f(x) = \prod (x - a_i)$, where $a_{1,\dots,n}$ are the roots of the polynomial.
Now if we expand such a product:
$(x - a_1)(x - a_2) = x^2 - (a_1 + a_2)x + a_1a_2$
$(x - a_1)(x - a_2)(x - a_3) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_1a_3 + a_2a_3)x - a_1a_2a_3$
And so on. The pattern should be clear.
This means that finding the roots of a polynomial is in fact equivalent to solving systems like the following:
For a quadratic polynomial $x^2 - px + q$, find $a_1,a_2$, such that
$p = a_1 + a_2$
$q = a_1a_2$
For a cubic polynomial $x^3 - px^2 + qx - r$, find $a_1,a_2,a_3$, such that
$p = a_1 + a_2 + a_3$
$q = a_1a_2 + a_1a_3 + a_2a_3$
$r = a_1 a_2 a_3$
And similarly for higher degree polynomials.
Not surprisingly, the amount of "unfolding" that needs to be done to solve the quadratic system is much less than the amount of "unfolding" needed for the cubic system.
The reason why polynomials of degree 5 or higher are not solvable by radicals, can be thought of as: The structure (symmetries) of the system for such a polynomial just doesn't match any of the structures that can be obtained by combining the structures of the elementary operations (adding subtracting, multiplication, division, and taking roots).
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:
Resolve[ForAll[x, x^4 + d x^3 + c x^2 + b x + a != 0], Reals]
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) $$
Best Answer
The conjugate factor theorem states that for a polynomial $p(x)$ with real coefficients, the complex roots come in conjugate pairs, or if $a+bi$ is a root, then $a-bi$ is also a root. From this we see that your polynomial has the roots $$\frac{2}{3}, -1, 3+\sqrt{2}i, 3-\sqrt{2}i.$$ Therefore, $$(x-\frac{2}{3})(x+1)(x-3-\sqrt{2}i)(x-3+\sqrt{2}i)=0.$$ $$\therefore (x^2+\frac{x}{3}-\frac{2}{3})(x-3-\sqrt{2}i)(x-3+\sqrt{2}i)=0$$ $$\therefore (x^2+\frac{x}{3}-\frac{2}{3})(x^2-6x+11)=0.$$ Continuing the expansion results in the polynomial $$p(x)=\frac{1}{3}(3x^4-17x^3+25x^2+23x-22).$$