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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.

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  • $\begingroup$ en.wikipedia.org/wiki/Gauss's_lemma_(polynomial) Gauss's lemma states that, if you can factorise a polynomial (with integer coefficients) into a product of two polynomials with rational coefficients, then you can make those two polynomials have integer coefficients. In particular, if this polynomial has a rational root, then it has an integer root, and that root must divide +1 (the constant term), so it must be +1 or -1. Neither of these works, so there are no rational solutions to this polynomial. So you're going to have to use a cubic formula or similar, like in the above comment. $\endgroup$ – Billy Jun 24 '13 at 18:18
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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,

$$ \begin{eqnarray*} x_{1} &=&2\sqrt{10/3}\cos \left( \frac{1}{3}\arccos \left( -5\sqrt{27/10^3}\right) \right) -1 \approx 1.4236, \\ x_{2} &=&2\sqrt{10/3}\cos \left( \frac{1}{3}\arccos \left( -5\sqrt{27/10^3}\right) +\frac{2\pi }{3}\right) -1 \approx -4.5771, \\ x_{3} &=&2\sqrt{10/3}\cos \left( \frac{1}{3}\arccos \left( -5\sqrt{27/10^3}\right) +\frac{4\pi }{3}\right) -1 \approx 0.15347. \end{eqnarray*}\tag{$\mathrm{D}$} $$

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$^1$ A deduction can be found in this post of mine in Portuguese.

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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.

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    $\begingroup$ It would be better to say "it is generally impossible to find the roots algebraically" for a polynomial with degree greater than four. There are a number of special cases that can be managed using algebra... $\endgroup$ – colormegone Jun 24 '13 at 18:24
  • $\begingroup$ I agree and recall having that wording in my head at some point during writing that. :) $\endgroup$ – fuglede Jun 24 '13 at 18:25
  • $\begingroup$ I bring it up because, historically, there was a certain amount of wrestling with the problem of which quintic, sextic, etc. polynomial equations could be solved using radicals, which finally led to the proofs that no general method could be produced. $\endgroup$ – colormegone Jun 24 '13 at 18:33
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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.

$$f(0)f\left(\frac14\right)<1\implies\;\text{a root is actually in}\;\;\left(0,\frac14\right)\;,\;\;etc.$$

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You may enjoy the trigonometric form of the solutions:

$$x = -1+\frac{2}{3} \sqrt{30} \sin\left(\frac{\pi}{6} + \frac{2 \pi j}{3} + \frac{1}{3}\arcsin\left(\sqrt{\frac{13}{40}}\right)\right),\ j = 0,1,2$$

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The method described here is quite simple and neat.

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