[b]p17.[/b] Let the roots of the polynomial $f(x) = 3x^3 + 2x^2 + x + 8 = 0$ be $p, q$, and $r$. What is the sum $\frac{1}{p} +\frac{1}{q} +\frac{1}{r}$ ?


[b]p18.[/b] Two students are playing a game. They take a deck of five cards numbered $1$ through $5$, shuffle them, and then place them in a stack facedown, turning over the top card next to the stack. They then take turns either drawing the card at the top of the stack into their hand, showing the drawn card to the other player, or drawing the card that is faceup, replacing it with the card on the top of the pile. This is repeated until all cards are drawn, and the player with the largest sum for their cards wins. What is the probability that the player who goes second wins, assuming optimal play?


[b]p19.[/b] Compute the sum of all primes $p$ such that $2^p + p^2$ is also prime.


[b]p20.[/b] In how many ways can one color the $8$ vertices of an octagon each red, black, and white, such that no two adjacent sides are the same color?


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. We start with the polynomial \( f(x) = 3x^3 + 2x^2 + x + 8 \). The roots of this polynomial are \( p, q, \) and \( r \).
2. We need to find the sum \( \frac{1}{p} + \frac{1}{q} + \frac{1}{r} \).
3. Using the identity for the sum of the reciprocals of the roots of a polynomial, we have:
   \[
   \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{pq + qr + rp}{pqr}
   \]
4. By Vieta's formulas, for a polynomial \( ax^3 + bx^2 + cx + d = 0 \), the sums and products of the roots are given by:
   \[
   p + q + r = -\frac{b}{a}, \quad pq + qr + rp = \frac{c}{a}, \quad pqr = -\frac{d}{a}
   \]
5. For our polynomial \( 3x^3 + 2x^2 + x + 8 \), we identify \( a = 3 \), \( b = 2 \), \( c = 1 \), and \( d = 8 \).
6. Using Vieta's formulas, we find:
   \[
   pq + qr + rp = \frac{c}{a} = \frac{1}{3}
   \]
   \[
   pqr = -\frac{d}{a} = -\frac{8}{3}
   \]
7. Substituting these values into the formula for the sum of the reciprocals, we get:
   \[
   \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = \frac{pq + qr + rp}{pqr} = \frac{\frac{1}{3}}{-\frac{8}{3}} = \frac{1}{3} \cdot -\frac{3}{8} = -\frac{1}{8}
   \]

The final answer is \(\boxed{-\frac{1}{8}}\).