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Saudi Arabian Mathematical Competitions

Algebra Difficulty 5.0 AIME, harder Prove it Saudi Arabia

Let $a_{1}$ , $a_{2}$ , $a_{3}$ , $a_{4}$ , $a_{5}$ be nonzero real numbers. Prove that the polynomial
$P(X) = \prod_{k=0}^{4} \left( a_{k+1} X^{4} + a_{k+2} X^{3} + a_{k+3} X^{2} + a_{k+4} X + a_{k+5} \right),$
where $a_{5+i} = a_{i}$ for $i = 1, 2, 3, 4$ , has a root with negative real part.

Solution

Assume, to the contrary, that all roots of the polynomial $P(X)$ have nonegative real parts. We deduce that the real parts of the sums of the roots of its factors
$a_{k} X^{4} + a_{k+1} X^{3} + a_{k+2} X^{2} + a_{k+3} X + a_{k+4}$
for $k = 1, 2, 3, 4$ , are nonegative. Therefore, by Vieta's relations, we have
$\operatorname{Re} \left( -\frac{a_{k+1}}{a_{k}} \right) \geq 0$
for $k = 1, 2, 3, 4$ . Hence
$-1 = \operatorname{Re} \prod_{k=1}^{4} \left( -\frac{a_{k+1}}{a_{k}} \right) = \prod_{k=1}^{4} \operatorname{Re} \left( -\frac{a_{k+1}}{a_{k}} \right) \geq 0,$
which is a contradiction.

This proves that the polynomial $P(X)$ has a root with negative real part.

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