Difference between revisions of "2024 AMC 8 Problems/Problem 22"

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Factoring <math>x^5+2x^4+3x^3+3x^2+2x+1</math> yields <math>(x+1)(x^2+1)(x^2+x+1)</math>. We can easily find one of the solutions is <math>x=-1</math>. Using the quadratic formula on the rest of the factors yields <math>-i, i, \frac{-1-i sqrt(3)}{2}</math>

Revision as of 13:18, 21 January 2024

Problem 22

What is the sum of the cubes of the solutions cubed of $x^5+2x^4+3x^3+3x^2+2x+1=0$?

Solution

Factoring $x^5+2x^4+3x^3+3x^2+2x+1$ yields $(x+1)(x^2+1)(x^2+x+1)$. We can easily find one of the solutions is $x=-1$. Using the quadratic formula on the rest of the factors yields $-i, i, \frac{-1-i sqrt(3)}{2}$