Which of the following polynomials has the greatest real root?

(A) $x^{19}+2018x^{11}+1$
(B) $x^{17}+2018x^{11}+1$
(C) $x^{19}+2018x^{13}+1$
(D) $x^{17}+2018x^{13}+1$
(E) $2019x+2018$

Enter the letter of the polynomial with the greatest real root.
By Descartes' Rule of Signs, none of the polynomials has a positive root, and each one has exactly one negative root.  Furthermore, each polynomial is positive at $x = 0$ and negative at $x = -1,$ so each real root lies between $-1$ and 0.  Also, each polynomial is increasing on the interval $(-1,0).$

Let $r_A$ and $r_B$ be the roots of the polynomials in options A and B, respectively, so
\[r_A^{19} + 2018r_A^{11} + 1 = r_B^{17} + 2018r_B^{11} + 1 = 0,\]so $r_A^{19} = r_B^{17}.$  Since $r_A \in (-1,0),$ $r_B^{17} = r_A^{19} > r_A^{17},$ so $r_B > r_A.$

Similarly, let $r_C$ and $r_D$ be the roots of the polynomials in options C and D, respectively, so
\[r_C^{19} + 2018r_C^{13} + 1 = r_D^{17} + 2018r_D^{13} + 1 = 0,\]so $r_C^{19} = r_D^{17}.$  Since $r_C \in (-1,0),$ $r_D^{17} = r_C^{19} > r_C^{17},$ so $r_D > r_C.$

Since
\[r_B^{17} + 2018r_B^{11} + 1 = r_D^{17} + 2018r_D^{13} + 1 = 0,\]we have that $r_B^{11} = r_D^{13}.$  Since $r_D \in (-1,0),$ $r_B^{11} = r_D^{13} > r_D^{11},$ so $r_B > r_D.$

Therefore, the largest root must be either $r_B$ or the root of $2019x + 2018 = 0,$ which is $-\frac{2018}{2019}.$

Let $f(x) = x^{17} + 2018x^{11} + 1,$ so $f(r_B) = 0.$  Note that
\[f \left( -\frac{2}{3} \right) = -\frac{2^{17}}{3^{17}} - 2018 \cdot \frac{2^{11}}{3^{11}} + 1.\]We claim that $2018 \cdot 2^{11} > 3^{11}.$  Since $2^2 > 3,$ $2^{22} > 3^{11}.$  Then
\[2018 \cdot 2^{11} = 1009 \cdot 2^{22} > 3^{11}.\]From $2018 \cdot 2^{11} > 3^{11},$ $2018 \cdot \frac{2^{11}}{3^{11}} > 1,$ so
\[f \left( -\frac{2}{3} \right) = -\frac{2^{17}}{3^{17}} - 2018 \cdot \frac{2^{11}}{3^{11}} + 1 < 0.\]Since $f(x)$ is an increasing function, we can conclude that $r_B > -\frac{2}{3} > -\frac{2018}{2019}.$  Therefore, the answer is $\boxed{\text{(B)}}.$