2024 DMC Mock 10 Problems/Problem 21

Problem

Given that $r,s,t$ are the three solutions of $x^3-2x^2-3x+5=0$ , find $r^4+s^4+t^4$ .

$\textbf{(A)}\ 38\qquad\textbf{(B)}\ 39\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 41\qquad\textbf{(E)}\ 42$

Solution

Let $P_n=r^n+s^n+t^n,S_1=r+s+t,S_2=rs+st+ts,$ and $S_3=rst$ . Then by Vieta's Formulas, $S_1=2,S_2=-3,$ and $S_3=-5$ . From Newton's Relations, we find: $P_2=P_1S_1-2S_2$ $P_3=P_2S_1-P_1S_2+3S_3$ $P_4=P_3S_1-P_2S_2+P_1S_3$ Since $P_1=S_1$ , we have $P_1=2$ . From the first equation, $P_2=10$ ; from the second equation, $P_3=11$ ; and from the third equation, $P_4=42$ . Thus the answer is $\boxed{\textbf{(E)}\ 42}$ .

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2024 DMC Mock 10 Problems/Problem 21

Problem

Solution