I came across what seems to be a very difficult "solve for $x$" type of problem, primarily because there should be $6$ real roots of this problem: $$(x^2 - 3x - 4)(x^2 - 5x + 6)(x^2 + 2x) + 30 = 0.$$ My first step was to (tediously) expand this into the degree $6$ polynomial that it is: $$x^6-6 x^5+x^4+36 x^3-20 x^2-48 x+30 = 0.$$ I would imagine that the first step would be to simplify this into a set of smaller roots, i.e. $$(ax + b)(cx^5 + dx^4 + ex^3 + fx^2 + gx + h) = 0,$$ and then perform a similar factorization on the polynomial of degree $5$. However, after trying hard to find the constant terms for this factorization, I can never quite get the correct factorization. Any recommendations on problems like these? I would really like to solve this without using Wolfram.

  • $\begingroup$ if there is any chance of this being solved by hand, it means that the revised polynomial factors nicely into into linear terms and quadratics. $\endgroup$ – Will Jagy Sep 22 '16 at 0:45
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    $\begingroup$ This factors as $(x^2-2x-5)(x^4-4x^3-2x^2+12x-6)=0$, and no further over the integers. Four of the six roots come from the quartic, and aren't particularly nice. $\endgroup$ – vadim123 Sep 22 '16 at 0:46
  • $\begingroup$ @vadim123 What's not particularly nice about $1 \pm \sqrt{6}/2 \pm \sqrt{10}/2$? $\endgroup$ – Robert Israel Sep 22 '16 at 0:56
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    $\begingroup$ @vadim123 The quartic can be rewritten as the biquadratic $(x-1)^4 - 8 (x-1)^2 + 1$ but that's not quite obvious to guess. $\endgroup$ – dxiv Sep 22 '16 at 1:02
  • $\begingroup$ Here's a method that does not require any "tricks". When you solve a quartic equation by the general method, look for a rational root of the resolvent cubic. If this exists, the roots will take a simplified form. $\endgroup$ – Oscar Lanzi Sep 22 '16 at 1:19

Let $P(x)$ denote the LHS. Note that the first summand factors nicely: $$ P(x)=(x-2)(x-3)(x-4)x(x+1)(x+2)+30. $$ From this it is clear that $P(x)$ is symmetric around $x=1$. Explicitly, setting $y=x-1$ we have $$\begin{eqnarray*} P(x)&=&(y-1)(y-2)(y-3)(y+1)(y+2)(y+3)+30\\ &=&(y^2-1)(y^2-4)(y^2-9)+30. \end{eqnarray*}$$ Note that this is a cubic in $y^2$, so it is possible to solve for $y^2$ using the formula for roots of a cubic, and hence find the roots of $P$ in terms of radicals.

  • $\begingroup$ Nice. And the cubic (in $y^2$) does actually have an easy to find rational root, which leaves a plain quadratic. $\endgroup$ – dxiv Sep 22 '16 at 1:11
  • $\begingroup$ dammit I almost noticed the symmetry when was looking for a solution myself. $\endgroup$ – UmNyobe Sep 22 '16 at 8:35
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    $\begingroup$ This is a beautiful use of symmetry ! And, since there is a very simple root for the cubic in $y^2$, the solutions are quite nice (if one enjoys radicals). Thanks for this answer. $\endgroup$ – Claude Leibovici Sep 22 '16 at 9:40

Your $6$'th degree polynomial does not have any rational roots (which you could show using the Rational Roots Theorem). It turns out to have a quadratic factor $x^2-2x-5$. It factors completely into factors of the form $x - (a + b \sqrt{6} + c \sqrt{10})$ with $a$, $b$, $c$ rational, but I don't see how you could guess that "by hand".


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