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$$ X^5-55X+21$$ Prove that the given polynomial has 2 roots which satisfy the condition: $$X_1X_2=1$$ and find them. I have tried to make use of Viette's relations ,but couldnt get to a satisfying result.

Razvan Paraschiv
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3 Answers3

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Hint: Check the polynomial GCD of the polynomial and its reverse, $21X^5-55X^4+1$.

ccorn
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The solution needn't be pulled out of hat like magic. Rather, we can discover it by exploiting innate reciprocation symmetry, i.e. if both $\,x_1$ and $\,x_2 = x_1^{-1}$ are roots of $f(x)$ then

$ 0 = f(x_2) = f(x_1^{-1})\,\Rightarrow\,0 = x_1^5f(x_1^{-1}) = 1\!-55x_1^4\!+21x_1^5 = \tilde f(x_1) = \,$ reverse of $f(x_1)$

Therefore $\,x_1\,$ is a root of $\,g(x) := \gcd(f(x),\tilde f(x))\,$ since $\,g(x) = a(x) f(x) + b(x) \tilde f(x)$

by the Bezout gcd identity. By Euclid the gcd $= x^2\!-3x +\color{#c00}1$ whose roots have product $= \color{#c00}1$.

Remark $\ $ Such inversion symmetry often proves useful, e.g. see here.

Generally it is wise to attempt to exploit any innate symmetry before diving head-first into difficult brute force calculations. For example, for problems of the above sort it is much more efficient to compute polynomial gcds than to compute polynomial factorizations. Follow the prior link for many more examples of such symmetry exploits.

Community
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Bill Dubuque
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$$ x^5-55x+21=(x^2-3x+1)(x^3+3x^2+8x+21)$$

It's not hard to prove that $x^3+3x^2+8x+21$ has only 1 real root, negative (just for curiosity, we won't need him), while $x^2-3x+1$ has 2 positive roots: $\frac{3\pm\sqrt{5}}{2}$.

It turns out that these 2 roots are good for us: $$\frac{3+\sqrt{5}}{2}\cdot\frac{3-\sqrt{5}}{2}=1$$

user340508
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    By what means did you manage to write the given polynomial into that form? – Razvan Paraschiv May 27 '16 at 20:26
  • I used a computer program, but without it I would try as follows: there is no general method of finding the roots of polynomials of degree $>4$ (Galois theorem), so the natural way is to look for a decomposition; since there are no easily seen "single roots", it remains to look for a decomposition of the form (deg 2 polynomial)*(deg 3 polynomial). – user340508 May 27 '16 at 20:39
  • so: $x^5-55x+21=(x^2+ax+b)(x^3+cx^3+dx+e)$ and we're dealing with a system of equation; $be=21$ so we may start with the most simply solutions - integers: $3\cdot 7$, $7\cdot 3$, $1\cdot 21$ or $21\cdot 1$, here the third one works fine (for these $b,e$ we can find other coefficients) – user340508 May 27 '16 at 20:39
  • it was too long for a comment, so I split it in two comments – user340508 May 27 '16 at 20:40
  • There is a hint inside the problem: words "find them". It means that the roots can be found and written -- this is a reason to look for a decomposition :-) – user340508 May 27 '16 at 20:44