How do I prove a given polynomial is not factor-able using only radicals, basic arithmetic operations, and real, rational numbers? For example, how can I prove that
$$0=x^3-3x+1$$
Has no solution with the given restraints?
Persistently, I want an algebraic (pre-calc. type of algebra) proof.
An idea I've had is to assume it does have one solution expressible with the given and then disproving using contradictions, but it seems be fruitless.
Let $a,b,c$ be the three real roots of $p(x) = x^3-3x+1$.
Consider the interpolating polynomials $f,g$ such that $f(a)=b, f(b)=c, f(c)=a$ and $g(a)=c, g(b)=a, g(c)=b$.
With some Galois theory, one can show that they are defined over $\Bbb Q(\sqrt \Delta)$ (and since all roots are real, the polynomials are real, so $\Delta > 0$). In fact since here $\Delta$ is a square, they have rational coefficients.
It turns out that $f,g$ are the polynomials $x \mapsto x^2-2$ and $x^2 \mapsto -x^2-x+2$. It's a bit hard to find, but easy to check because a straightforward computations shows that if $a$ is a root, then so are $f(a)$ and $g(a)$.
Well this tell us that if $K$ is any subfield of $\Bbb R$ then if $p(x)$ has a root in $K$ then the other two roots are also in $K$. So either $p(x)$ is irreducible or it has three linear factors.
Now, if $K \subset L = K(\sqrt[n] y) \subset \Bbb R$ with $y \in K$ and $n$ prime, then if $p(x)$ splits in $L$ then it was already split in $K$ :
Since $n$ is prime, there is no intermediate subfield between $K$ and $L$, so if $p$ splits in $L$ and is irreducible over $K$ then $L = K(a)$ and $n=3$.
Then $L$ is Galois over $K$ which is impossible because cube root extensions are not Galois :
If $q(a)$ is a root of $Y^3-y$ for some polynomial $q \in K[X]$, then $Y^3-y$ splits into $(Y-q(a))(Y-q(b))(Y-q(c))$ over $K(a)$ (remember that $b=f(a)$ and $c=g(a)$), and we get $(q(a) -q(b))^2 + 3q(c)^2 = 0$, from which we get $q(c)=0$ then $y=0$, and $K=L$, contradiction.
Now if you had a way to obtain $a$ from rationals with the usual operations and taking real roots, then there would be a finite tower of radical extensions $\Bbb Q \subset K_1 \subset \ldots \subset K_n \subset \Bbb R$ with $a \in K_n$. With the above result, if $a \in K_n$ then $a \in K_{n-1}$ and so on, which implies that $a \in \Bbb Q$ in the first place.
Since the roots are not rationals, it is impossible to obtain them from rationals with those operations.
The discriminant $\Delta=(-4\times-27)-27=3\times 27=81.$ Since the discriminant is not a perfect cube, then the polynomial is not factorable because of the formula to the cubic equation.
But I am not quite sure about this. The discriminant is given by the formula $\Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$.
Here's about the discriminant.
A good place to start for rational numbers is Eisenstein's criterion. This following link will tell you more about it:
See "What Is Mathematics?" by Courant and Robbins, ch. 3. If you take $x=2\cos t$ then you'll get an equation $\cos 3t=-1/2$. It correspond to trisection of the angle $120^\circ$, which is impossible (see Courant - Robbins again).
Let us define our cubic as:
$$f(x)=ax^3+bx^2+cx+d=0$$
We consider the polynomial:
$$D(x)=3ax^2+2bx+c$$
We test for when $D(x)=0$, say $x_1,x_2$.
If $x_1,x_2$ don't exist, then $f$ has only one real root.
If $x_1,x_2$ both exist, and $f(x_1),f(x_2)$ have different signs, then $f$ has three real roots.
If either or both of $f(x_1),f(x_2)$ are zero, then $f$ has three real roots, two (or all three if both) being the same.
Otherwise, i.e. $\operatorname{sign}(f(x_1))=\operatorname{sign}(f(x_2))$, there is only one real root.
So we have that $$x^3-3x+1=0 \tag1$$ We can rewrite this as: $$\begin{align} & x^3+3x^2+3x+1=3x^2+6x \\ \implies & (x+1)^3=3x(x+2)=3\{(x+1)-1\}\{(x+1)+1\}=3(x+1)^2+2(x+1)-1 \\ \implies & (x+1)^3-3(x+1)^2-2(x+1)+1=0 \\ \implies & (x+1)^3-3(x+1)^2+3(x+1)-1=5(x+1)-2 \\ \implies & (x+1-1)^3=5(x+1)-2 \\ \implies & x^3=5x+3 \\ \implies & x^3-5x-3=0 \tag2\end{align} $$
Hence, we see that we can re-arrange $(1)$ to get $(2)$.
Now, $(1)-(2) \implies 5x+3-3x+1=0 \implies x=-2$
So $x=-2$ is a common solution of both $(1)$ and $(2)$. But on verification, we see that $x=-2$ does not satisfy either equation.
This is a clear implication that $(1)$ has no solution in $\mathbb{Q}$.
On application of Cardan's Theorem, we get that the equation, being a cubic one has all $3$ solutions in $\mathbb{R-Q}$.
