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Difference between revisions of "2021 JMC 10 Problems/Problem 22"

(Created page with "==Problem== Let <math>r_1,r_2,r_3,r_4</math> be the roots of <math>P(x)= x^4+4x^3-3x^2+2x-1.</math> Suppose <math>Q(x)</math> is the monic polynomial with all six roots in th...")
 
(No difference)

Latest revision as of 15:51, 1 April 2021

Problem

Let $r_1,r_2,r_3,r_4$ be the roots of $P(x)= x^4+4x^3-3x^2+2x-1.$ Suppose $Q(x)$ is the monic polynomial with all six roots in the form $r_{i}+r_{j}$ for integers $1\le i < j \le 4.$ What is the coefficient of the $x^4$ term in the polynomial $Q(x)?$

$\textbf{(A) } 32 \qquad \textbf{(B) } 36 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 56$

Solution

Note that $\sum (r_{a}+r_{b})(r_{c}+r_{d})$ has $15\cdot 4=60$ terms. We can also see that all terms are in the form $r_{i}r_{j}$ or $(r_{i})^2$. The only way $(r_{i})^2$ are produced is $(r_{i}+r_{j})(r_{i}+r_{k})$ and there are $\tbinom{3}{2}=3$ pairs of these for a given value of $i$.


So three copies of each $(r_{i})^2$ is produced, and there are $\tfrac{60-3\cdot 4}{6}=8$ copies of $r_{i}r_{j}$ by symmetry. By Vieta's, our desired answer is equal to \[3\sum (r_{i})^2 + 8\sum r_{i}r_{j} = 42.\]

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