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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.

Martin Sleziak
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Sarah
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  • 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. – Will Jagy Sep 22 '16 at 00:45
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    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. – vadim123 Sep 22 '16 at 00:46
  • @vadim123 What's not particularly nice about $1 \pm \sqrt{6}/2 \pm \sqrt{10}/2$? – Robert Israel Sep 22 '16 at 00:56
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    @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. – dxiv Sep 22 '16 at 01:02
  • 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. – Oscar Lanzi Sep 22 '16 at 01:19
  • Can't we do anything with the obvious factorization $(x^2−3x−4)(x^2−5x+6)(x^2+2x) = -30$ hence$(x+1)(x-4)(x-3)(x-2)(x)(x+2) = -30$? EDIT: oops I just saw stewbasic did that, and reduced it to a cubic in $(x-1)^2$ – smci Sep 22 '16 at 06:46
  • Tip: the resulted number-only value (with no x) is the key. You spit that into the numbers you need for the high factor elements. – Overmind Sep 22 '16 at 09:15

2 Answers2

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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.

stewbasic
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  • Nice. And the cubic (in $y^2$) does actually have an easy to find rational root, which leaves a plain quadratic. – dxiv Sep 22 '16 at 01:11
  • dammit I almost noticed the symmetry when was looking for a solution myself. – UmNyobe Sep 22 '16 at 08:35
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    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. – Claude Leibovici Sep 22 '16 at 09:40
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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".

Robert Israel
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