Find the real root of the almost symmetric polynomial $x^7+7x^5+14x^3+7x-1$

Question

Find the real root of following almost symmetric polynomial by radicals $$p(x)=x^7+7x^5+14x^3+7x-1$$

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Here are my attempts.

The coefficients of $p(x)$ are : $1,7,14,7,-1$.

I wanted to try possible factorizations. But Wolfram Alpha can not factorise this polynomial. This can be a reason of our case, so factorisation over $\Bbb R$ seems impossible.

The Rational root theorem also failed.

Again I tried

$$\begin{align} x^7+7x^5+14x^3+7x-1 &=x^7+7x^5+7x^3+7x^3+7x-1 \\ &=x^7+7x^3(x^2+1)+7x(x^2+1)-1 \\ &=x^7+7x(x^2+1)^2-1 \end{align}$$

But, this manipulation also didn't work.

Answer 1

Since $$x^7+7x^5+14x^3+7x=2i\,\text T_7\left(\dfrac{ix}2\right)= 2i \cos\left(7\arccos\,\dfrac{ix}2\right)=1,\tag1$$ where $\;\text T_n(x)\;$ is the Chebyshev polynomial of the first kind (see also WA test),

then $$x_k=-2i\cos\left(\dfrac17\arccos\,\left(-\dfrac i2\right)+\dfrac{2\pi k}7\right),\qquad(k=0,1,2,3,4,5,6)\tag2$$ (see also WA test), wherein $$\arccos\left(-\dfrac i2\right)=-i\ln\left(-\dfrac i2+i\sqrt{1-\left(-\dfrac i2\right)^2}\,\right)=-i\left(\ln e^{^{\Large\frac\pi2i}} + \ln\dfrac1\varphi\right),$$

$$\arccos\left(-\dfrac i2\right)=\dfrac\pi2+i\ln\varphi,\tag3$$ and $\;\varphi=\dfrac{\sqrt5+1}2\;$ is the golden ratio.

From $(2)-(3)$ should $$x_k=-2i\cos\left(\dfrac{4\pi k+\pi+2i\ln\varphi}{14}\right), \qquad(k=0,1,2,3,4,5,6),\tag4$$ (see also WA test).

If $\,k=5,\,$ then $\dfrac{4k+1}{14}\pi=\dfrac32\pi,$ and we have result in radicals: $$x_5= 2i\sin\,\left(\frac i7\ln\varphi\right) =2i\cdot\sin\left(i \ln\sqrt[\large7]\varphi\right) =\sqrt[7]\varphi-\frac1{\sqrt[7]\varphi}\approx0.13759740974800.$$

Answer 2

Remarks: For a cubic equation $x^3 + px + q = 0$, we use the identity $(u + v)^3 \equiv 3uv(u + v) + u^3 + v^3$ and let $x = u + v$.

Similarly, we have the identities $$(a+b)^4 - 4ab(a + b)^2 - (a^4 + b^4 - 2a^2b^2) \equiv 0,$$ $$(a+b)^5 - 5ab(a+b)^3 + 5a^2b^2(a+b) - (a^5+b^5) \equiv 0.$$

In general, $a^n + b^n$ ($n\in \mathbb{Z}_{>0}$) can be expressed in terms of $ab$ and $a + b$. See: 1, and 2.

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We use the identity $$(u + v)^7 - 7uv(u+v)^5 + 14u^2v^2(u + v)^3 - 7u^3v^3(u+v) - (u^7+v^7) \equiv 0. \tag{1}$$

Let $x = u + v$. From (1), we have $$x^7 - 7uvx^5 + 14u^2v^2x^3 - 7u^3v^3x - (u^7+v^7) = 0. \tag{2}$$ If $uv = -1$ and $u^7 + v^7 = 1$, then (2) gives the equation $x^7+7x^5+14x^3+7x-1 = 0$. Since $u^7, v^7$ are roots of $y^2 - y - 1 = 0$, we have $$u = \sqrt[7]{\frac{\sqrt 5 + 1}{2}}, \quad v = - \sqrt[7]{\frac{\sqrt 5 - 1}{2}}.$$ Thus, one root of the equation is given by $$x = u + v = \sqrt[7]{\frac{\sqrt 5 + 1}{2}} - \sqrt[7]{\frac{\sqrt 5 - 1}{2}} \approx 0.1375974100.$$

One can prove that $$x_k = \mathrm{e}^{\mathrm{i}2\pi k/7} u + \mathrm{e}^{- \mathrm{i}2\pi k/7} v$$ are roots of the equation for $k = 0, 1, \cdots, 6$, which are all roots of the equation.

We are done.