Polynomials - Solutions How I can find the exact solutions of this polynomial?
I can not get to the exact roots of the polynomial ... what methods occupy for this "problem"?
$$x^3+3x^2-7x+1=0$$
Thanks for your help.
 A: 
How I can find the exact solutions of this polynomial?
I can not get to the exact roots of the polynomial

You have a real polynomial of the third degree with real coefficients. There are exact formulas to find the roots of any polynomial of this kind.
A general cubic equation of the form 
$$
\begin{equation*}
ax^{3}+bx^{2}+cx+d=0,\tag{1}
\end{equation*}
$$
can be transformed by the substitution 
$$
x=t-\frac{b}{a}\tag{2}
$$ 
into the reduced cubic equation 
$$
\begin{equation*}
t^{3}+pt+q=0.\tag{3}
\end{equation*}
$$
In the present case, we have
$$
\begin{equation*}
x^{3}+3x^{2}-7x+1=0,\quad a=1,b=3,c=-7,d=1.\tag{$\mathrm{A}$}
\end{equation*}
$$
For $x=t-1$, we get the reduced equation
$$
\begin{equation*}
t^{3}-10t+10=0,\qquad p=-10,q=10.\tag{$\mathrm{B}$}
\end{equation*}
$$
It is known from the classical theory of the cubic equation that when  the discriminant 
$$
\Delta =q^{2}+\frac{4p^{3}}{27}=10^{2}+\frac{4\left(-10\right) ^{3}}{27}<0,\tag{$\mathrm{C}$}
$$
 the three roots $t_k$ of $(3)$, with $k\in\{1,2,3\}$, are real and can be written in the following trigonometric form $^1$ 
$$
\begin{eqnarray*}
t_{k} &=&2\sqrt{-p/3}\cos \left( \frac{1}{3}\arccos \left( -\frac{q}{2}\sqrt{-27/p^3}\right) +\frac{(k-1)2\pi }{3}\right).
\end{eqnarray*}\tag{4}
$$ 
The roots of $(1)$ are thus 
$$x_k=t_k-\frac{b}{a}.\tag{5}$$ 
Consequently,
$$
\begin{eqnarray*}
x_{1} &=&2\sqrt{10/3}\cos \left( \frac{1}{3}\arccos \left( -5\sqrt{27/10^3}\right) \right) -1 \approx 1.4236, \\
x_{2} &=&2\sqrt{10/3}\cos \left( \frac{1}{3}\arccos \left( -5\sqrt{27/10^3}\right) +\frac{2\pi }{3}\right) -1 \approx -4.5771, \\
x_{3} &=&2\sqrt{10/3}\cos \left( \frac{1}{3}\arccos \left( -5\sqrt{27/10^3}\right) +\frac{4\pi }{3}\right) -1 \approx 0.15347.
\end{eqnarray*}\tag{$\mathrm{D}$}
$$
--
$^1$ A deduction can be found in this post of mine in Portuguese.
A: There are ways to find the roots of cubic equations by algebraic means (which is what I take your question to mean). See this Wikipedia page for a thorough explanation which there is not much point in repeating here.
Interestingly enough, when the degree of the equation (i.e. the highest exponent of $x$ which in your case is $3$) is greater than 4, finding the roots algebraically is generally impossible.
A: What about subdivision? It isn't particularly fast but it burns lots of hours...
$$f(x)=x^3+3x^2-7x+1=0\;\;;\;\;f(0)f\left(\frac12\right)<0\implies\;\text{there exists a root}\;\;\alpha\in\left(0,\frac12\right)$$
BTW, Descartes' Rule of signs tells us then that there are *exactly*two positive roots.
$$f(0)f\left(\frac14\right)<1\implies\;\text{a root is actually in}\;\;\left(0,\frac14\right)\;,\;\;etc.$$ 
A: You may enjoy the trigonometric form of the solutions:
$$x = -1+\frac{2}{3} \sqrt{30} \sin\left(\frac{\pi}{6} + \frac{2 \pi j}{3} + \frac{1}{3}\arcsin\left(\sqrt{\frac{13}{40}}\right)\right),\ j = 0,1,2$$  
A: The method described here is quite simple and neat.
