Find the all real roots of the polynomial $x^6+3 x^5+3 x-1=0$ in closed form

Question

Find the all real roots of the polynomial $$x^6+3 x^5+3 x-1=0$$ in exact form.

WolframAlpha gives only numerical results. I've asked a few similar questions before. The source of the problem comes from the algebra precalculus workbook (but not homework workbook).

The rational root theorem doesn't work. Because, $x=1$ and $x=-1$ are not roots.

Wolfram says , we have only $2$ real roots. What kind of factorization should I try?Can a factorization of the form $P(x)=(x^3+ax^2+bx+c)(x^3+dx^2+ex+f)$ work?

I also tried the trick $\frac {P(x)}{x^n}$, but I failed.

Answer 1

There is a way to use $P(x)/x^3$ here, but because the coefficients are only symmetric up to a sign, we have to make a slight modification. Let $Q(x)=P(x)/x^3$, i.e.,

$$(x^3-x^{-3})+3(x^2+x^{-2}).$$

The goal is to express both the terms in parentheses in terms of $w=x-x^{-1}$.

We can compute $w^2=x^2+x^{-2}-2$, so $x^2+x^{-2}=w^2+2$, and $w^3=x^{3}-x^{-3}-3(x-x^{-1})$, so $x^{3}-x^{-3}=w^3+3w$.

Plugging in, we get $Q(w)=w^3+3w+3(w^2+2)=w^3+3w^2+3w+6$. Completing the cube, we are trying to solve $(w+1)^3+5=0$. We can get 3 solutions for $w$ easily from this, and from each solution we can solve for $x$.

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What is going on in this problem is that the roots of the equation satisfy a symmetry. If $Q(z)=0$, then $Q(-1/z)=0$, which manifests in the coefficients being almost palindromic: in $Q(x)$, if $a_n$ is the coefficient of $x^n$, then $a_{-n}=(-1)^na_n$.

Because of this, except for the roots where $\alpha=-1/\alpha$ which need not pair up, we will get that $P(x)$ will factor into pairs of roots $(x-\alpha)(x+1/\alpha)$ in the same way the complex conjugate roots of a real polynomial pair up to give quadratic factors, and multiplying the factors out and then factoring out an $x$, we get $x(x-1/x -(\alpha-1/\alpha))$, and so as long as $0, \pm i$ are not roots of $P(x)$, then we can express $Q(x)$ as a polynomial in $w$.

Answer 2

This problem is not easy, but you can simplify the work a bit by observing that you can divide both sides by $x^3$. It can be shown that $x=0$ is not a solution to the original polynomial, hence this operation is allowed. We get

$$x^3-\frac 1{x^3}+3\left(x^2+\frac 1{x^2}\right)=0$$

Now, use the two identities below

\begin{align*} x^3-\frac 1{x^3} & =\left(x-\frac 1x\right)^3+3\left(x-\frac 1x\right)\\ x^2+\frac 1{x^2} & =\left(x-\frac 1x\right)^2+2 \end{align*}

Our polynomial now becomes

$$\left(x-\frac 1x\right)^3+3\left(x-\frac 1x\right)+3\left(x-\frac 1x\right)^2+6=0$$

Make the substitution $x-\frac 1x=z$ and the equation is now a cubic in $z$.

$$z^3+3z^2+3z+6=0$$

Use Cardano's method to solve this cubic for $z$ and calculate $x$ by using

$$x-\frac 1x=z$$

The resulting six solutions are the roots of your original polynomial.

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EDIT: Here's an actual attempt at solving the problem. Starting with the cubic in $z$, then we get

$$(z+1)^3=-5\qquad\implies\qquad z_0=-1-\sqrt[3]{5}$$

Since the OP is only interested in the real roots, we can neglect the two other complex roots. However, for thoroughness, the two other complex roots would be

\begin{align*} z_1 & =-1-\omega\sqrt[3]{5}\\ z_2 & =-1-\omega^2\sqrt[3]{5} \end{align*}

Where $\omega$ is the cube root of unity

$$\omega=\frac {-1+i\sqrt3}2\qquad\qquad\qquad\omega^2=-\frac {1+i\sqrt3}2$$

Returning to the substitution in $x$, we have that

$$x^2-z_0x-1=0$$

A direct application of the quadratic formula gives the two real roots as

\begin{align*} x_+ & \color{blue}{=\frac 12\left(-1-\sqrt[3]{5}+\sqrt{5+2\sqrt[3]{5}+\sqrt[3]{25}}\right)}\\ x_- & \color{blue}{=\frac 12\left(-1-\sqrt[3]{5}-\sqrt{5+2\sqrt[3]{5}+\sqrt[3]{25}}\right)} \end{align*}

Answer 3

This is such a cursed question.

The only elementary method I can think of is to assume the existence of a factorization of the form $$P(x) = x^6 + 3x^5 + 3x - 1 = (x^2 + kx \pm 1)(x^4 + ax^3 + bx^2 + cx \mp 1) \tag{1}$$ and then after tediously expanding and equating coefficients, we require

$$(a,b,c) = (3-k, 1-3k+k^2, 3-2k+3k^2-k^3) \tag{2}$$ for the case where the quadratic factor is $x^2 + kx - 1$. For the second case where the quadratic is $x^2 + kx + 1$, the solution for $(a,b,c)$ leads to a contradiction $c = k$ and the linear term becomes $0$; thus is inadmissible.

After substituting $(2)$ into $(1)$ we find $$P(x) = x^6 + 3x^5 - k g(k) x^2 + (g(k) + 3)x - 1, \tag{3}$$ where $$g(k) = k^3 - 3k^2 + 3k - 6. \tag{4}$$ Then if $g(k) = 0$, we have a valid factorization. This tells us we are guaranteed a solution in radicals. We note $$\begin{align} g(k) &= k^3 - 3k^2 + 3k - 6 \\ &= (k^3 - 3k^2 + 3k - 1) - 5 \\ &= (k - 1)^3 - 5. \tag{5} \end{align}$$ Therefore, $g(k) = 0$ has the unique real root $k = 1 + \sqrt[3]{5}$, from which we can refer to $(1)$ to recover the two real roots $$x = \frac{-k \pm \sqrt{k^2 + 4}}{2} = \frac{-1 - 5^{1/3} \pm \sqrt{5^{2/3} + 2 \cdot 5^{1/3} + 5}}{2}. \tag{6}$$ The quartic does not have any real solutions, but by this point I'm already exhausted by all the tedious math and don't want to prove it by hand. The problem with this approach is that it is highly speculative--there is no guarantee of it working, and we basically got lucky: the coefficients in $(3)$ turn out "just right" in which the solution of the cubic $(4)$ makes both the linear and quadratic terms correct. Moreover, the solution to $(4)$ is "easy" and does not require the cubic formula.