Consider polynomial $f(x)={{x}^{2}}-\alpha x+1$ with $\alpha \in \mathbb{R}.$ a) For $\alpha =\frac{\sqrt{15}}{2}$, let write $f(x)$ as the quotient of two polynomials with nonnegative coefficients. b) Find all value of $\alpha $ such that $f(x)$ can be written as the quotient of two polynomials with nonnegative coefficients.
Problem
Source: VMO 2019
Tags: factoring polynomials, polynomial, algebra
Pathological
16.04.2019 00:32
https://artofproblemsolving.com/community/c6h1470023p8526130
It easily solves part (a) since $\frac{\sqrt{15}}{2} < 2$.
As for part (b), it shows that $0 \le \alpha < 2$ all work. Furthermore, it's trivial that $\alpha < 0$ work since $f(x) = \frac{x^2 - \alpha x + 1}{1}$ is such a way.
khanhnx
25.07.2019 11:38
For part a, we consider 2 polynomial $P(x) = x^{16} + \dfrac{223}{256} x^8 + 1$ and $Q(x) = \left(x^2 + \dfrac{\sqrt{15}}{2} x + 1\right)\left(x^4 + \dfrac{7}{4} x^2 + 1\right)\left(x^8 + \dfrac{17}{16} x^4 + 1\right)$ Just note that: $x^{16} + \dfrac{223}{256} x^8 + 1 = \left(x^2 - \dfrac{\sqrt{15}}{2} + 1\right)\left(x^2 + \dfrac{\sqrt{15}}{2} x + 1\right)\left(x^4 + \dfrac{7}{4} x^2 + 1\right)\left(x^8 + \dfrac{17}{16} x^4 + 1\right)$ then we are done
yayups
10.08.2019 08:32
I think this is a blatant rip off of TSTST 2017/3...
Justin_Thomas_2005
17.05.2021 19:24
part b let consider a sequence: u1=a, u_n+1= 2-u_n^2. It's easy to find out that a>-2