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I was trying to factor a polynomial with Wolfram and I noticed a quadratic form I've never considered.

$$n^2-4+\frac{6}{n}=0$$ The purpose of it is not important but it made me wonder. How do you solve it? A normal quadratic $\quad ax^2+bx+c\quad$ is solved by the quadratic formula:

$$n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

but this equation seems to take the form $\quad ax^2+0x-c+dx^{-1}=0.\quad$ How do you solve it? Do I multiply through by $n$ and then solve it as a cubic?

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    $\begingroup$ It's not a quadratic equation; $n^3-4n+6=0$ is a cubic equation $\endgroup$ – J. W. Tanner Feb 5 at 18:30
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    $\begingroup$ @J. W. Tanner That makes sense. Thanks. BTW, do you have ideas about factoring the numerator of the division I gave to Wolfram Alpha? $\endgroup$ – poetasis Feb 5 at 18:32
  • $\begingroup$ $n=1$ is a factor of that numerator $\endgroup$ – J. W. Tanner Feb 5 at 18:38
  • $\begingroup$ @J. W. Tanner I know that $n=1$. That is the solution I seek in a formula $f(P,m)$ for solving a Bring-Jerrard quintic here and here and here. $\endgroup$ – poetasis Feb 5 at 18:48
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Hint Multiplying by $n$ you get the cubic equation $$n^3-4n+6=0$$

The Rational Root Test tells you that the only potential rational roots are $\pm1, \pm 2, \pm 3, \pm6$ and you can see that none of them works. This means that you cannot factor this equation easily; you need to use the Cubic Formula (or solve it as a cubic).

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  • $\begingroup$ Thanks. I know how hard it is to solve a cubic because I needed help to find a solution here. The final solution for $that$ is here. $\endgroup$ – poetasis Feb 5 at 18:41
  • $\begingroup$ The so called "Cardano method" is very easy to remember. In your case the equation is already in the right form. Since $$(u+v)^3-3uv(u+v)-u^3-v^3=0$$ if you can make $$-3uv=-4 \\-u^3-v^3=6$$ then $x=u+v$ is a solution. Replacing $u=\frac{4}{3v}$ in the second equatiion gives you a quadratic in $v^3$. Solve. $\endgroup$ – N. S. Feb 5 at 18:50
  • $\begingroup$ Can this type of solution apply to a quintic $\quad n^5-m^4n+\frac{P}{2m}=0\quad$ as in this question? The Wolfram test was for one example: $\quad n^5-16n+15=0\quad$ where it's easy to see that $n=1$. I'm looking for a general $n=f(P,m)$. $\endgroup$ – poetasis Feb 5 at 18:54
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    $\begingroup$ @poetasis Galois theory tells us that there is no general formula which solves the quitic. Quitic equations can be solvable exactly when the Galois group is solvable. For that equation, some values of $m$ and $P$ lead to non-solvable Galois groups, and then, Galois Theory tells us that in those scenarios this type of approach simply cannot work. $\endgroup$ – N. S. Feb 5 at 22:10
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    $\begingroup$ @poetasis If don't know Galois theory, and you try exactly the same approach, you are going to see that when you go degree 4 or higher, the approach only works when the coefficients of $X^3,X^2,X$ and $1$ are higly related (you are gonna get a sytem of four I think polynomial equations with 2 variables, you need some consistency conditions to have solutions). So this approach only works extremely extremely rarely (when the coefficients satisfy a particular pattern)... And it never works when the Galois group is unsolvable, which is the case for some values of $m,P$ in the above equation. $\endgroup$ – N. S. Feb 5 at 22:16
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First, you have to multiply by $n$, obtaining: $$n^3-4n+6$$. Now you can use the cubic formula, obtaining the only real solution $x=2.5251022548143$.

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You need to solve for $n$ the cubic equation $$n^3-4n+6=0$$

Just follow the steps given here with $a=1$, $b=0$, $c=-4$ and $d=6$. You will get $$\Delta=-716 \qquad p=-4 \qquad q=6$$ So, only one real root.

To get it, use the hyperbolic method and get $$n=-\frac{4 }{\sqrt{3}}\cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{9 \sqrt{3}}{8}\right)\right)=-2.52510225481432$$

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