AMC12 2018 A

Answer (B): By Descartes’ Rule of Signs, none of these polynomials has a positive root, and each one has exactly one negative root. Because each polynomial is positive at $x=0$ and negative at $x=-1$, it follows that each has exactly one root between $-1$ and $0$. Note also that each polynomial is increasing throughout the interval $(-1,0)$. Because $x^{19} > x^{17}$ for all $x$ in the interval $(-1,0)$, it follows that the polynomial in choice A is greater than the polynomial in choice B on that interval, which implies that the root of the polynomial in choice A is less than the root of the polynomial in choice B. Because $x^{13} > x^{11}$ for all $x$ in the interval $(-1,0)$, it follows that the polynomial in choice C is greater than the polynomial in choice A on that interval, which implies that the root of the polynomial in choice C is less than the root of the polynomial in choice A and therefore less than the root of the polynomial in choice B. The same reasoning shows that the root of the polynomial in choice D is less than the root of the polynomial in choice B. Furthermore, $2018 > 2018x^6$ on the interval $(-1,0)$, so $x^6+2018 > 2019x^6$, from which it follows that $x^{11}(x^6+2018) < 2019x^{17}$. Therefore the polynomial in choice B is less than $2019x^{17}+1$ on the interval $(-1,0)$. The polynomial in choice E has root $-\left(1-\frac{1}{2019}\right)$. Bernoulli’s Inequality shows that $(1+x)^{17} > 1+17x$ for all $x>-1$, which implies that \[ -2019\left(1-\frac{1}{2019}\right)^{17}+1 < -2019\left(1-\frac{17}{2019}\right)+1 = -2001 < 0, \] so the polynomial in choice B is negative at the root of the polynomial in choice E. This shows that the root of the polynomial in choice B is greater than the root in choice E. Because the unique real root of the polynomial in choice B is greater than the unique root of the polynomial in each of the other choices, that polynomial has the greatest real root.