Polynomials - Solutions

Question

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.

Answer 1

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.

Answer 2

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.

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

Answer 3

You may enjoy the trigonometric form of the solutions:

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

Answer 4

The method described here is quite simple and neat.

Answer 5

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.