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Find the roots of the polynomial $A(x)=x^6-21x^5+175x^4-735x^3+1624x^2-1764x+720$.

Using the Rational Root Theorem, one can find that the given polynomial factors as $A(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$. Is there something "fancy" that we can note (maybe about the coefficients) in order to escape the trial and error required by the Rational Root Theorem?

amWhy
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yinivem462
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    Not really, I don't think. One can at least use (a simple case of) Descartes's rule of signs to rule out negative roots. – Greg Martin Jan 24 '23 at 20:25
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    One can observe that $21=\sum_{i=1}^6i$ and $720=\prod_{i=1}^6i$, then make an educated guess. – Théophile Jan 24 '23 at 20:28
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    One can make "educated guesses" and apply other ad hoc observations. But certainly the Rational Root Theorem is a general statement, and why not apply it. You can also reduce modulo a prime, to find the possible roots, see this similar question from yesterday, if you like it. – Dietrich Burde Jan 24 '23 at 20:47
  • @DietrichBurde, can you clarify what does "reduce modulo a prime" mean and it's application when trying to find roots of a polynomial? I am not familiar with this method and would love to read about it (maybe the OP as well). – Math Student Jan 24 '23 at 21:03
  • @KaloyanK. Please click on the question of yesterday. In my answer I took $p=2$ (then a factorisation is almost trivial to see, and one of the roots must be also one for the original problem). – Dietrich Burde Jan 24 '23 at 21:08
  • How do you define "fancy"? – amWhy Jan 24 '23 at 21:51
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    You need six numbers that multiply together to make $720$. They can't be large unless most of them are $1$, so try $\pm 1$ first. You find $1$ works but only once. Similarly once you find the single root at $2$, you know the rest are at least $3$ in absolute value, so the product is already at least $162$. You don't have to worry about anything larger than $10$ now because the product will be too large. – Ross Millikan Jan 24 '23 at 22:28
  • I think $6!=720$ is common enough to recognize and start to suspect it's $\prod_{n=1}^6(x-n)$. One sanity check that is consistent before working it out is to reduce the coefficients mod $7$, since this will make $x^6-1 \mod 7$ if our guess is correct, which isn't too hard to do in your head by looking at most of the coefficients. – Merosity Jan 24 '23 at 23:36

1 Answers1

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Two methods come to mind here

  1. The signs of the coefficients are strictly alternating, so all real roots (let alone rational ones) must be positive. Thereby the trials are cut in half.

  2. Suppose you have identified the roots $1$ and $2$ and removed them through synthetic division. You are left with the fourth-degree polynomial

$x^4-18x^3+119x-342x+360=0.$

Now try to equate this with $(x^2-ax+b)^2$ where the cubic terms render $2a=18$ and the quadratic terms render $a^2+2b=119$. Thus $a=9$ and $b=19$. We then get actually

$(x^2-9x+19)^2=x^4-18x^3+119x-342x+\color{blue}{361},$

just missing our target polynomial but still we can render

$(x^2-9x+19)^2=1.$

Thus $x^2-9x+18=0$ or $x^2-9x+20=0$, and solving these quadratic equations gives all four remaining roots.

A more sophisticated version of the method, involving the solution of an eighth-degree equation with roots containing square-root radicals, is given here.

Oscar Lanzi
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