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on August 07, 2023

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hi

welcome to my blog

i'm sets

i'm a rising sophomore

not sure why you would be interested but i'll be posting daily updates about my life (especially math and physics) on this blog

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Mathematical Beauty

2021 Fall OMC 10 #14

Problem: A cubic polynomial $P(x)=ax^3+bx+c$ with $a \neq 0$ has three roots $r$, $s$, and $t$. Which of the following polynomials has roots $r+s, s+t,$ and $t+r$? $\textbf{(A)} ax^3-bx-c \hspace{0.5cm}\textbf{(B)} ax^3-bx+c \hspace{0.5cm}\textbf{(C)} ax^3+bx-c \hspace{0.5cm} \textbf{(D)} ax^3+bx^2-c \hspace{0.5cm}\textbf{(E)} ax^3+bx^2+c$ Solution: Write the polynomial as $ax^3 + 0x^2 + bx + c$. By Vieta's, we have $r+s+t = 0$ and $rst = \frac{-c}{a}$ and $rs+st+tr=\frac{b}{a}$. If a polynomial has roots $r+s$, $s+t$, and $t+r$, we can write its roots as $-t, -r, -s$. The sum of the roots is still $0$, but the product is now the negative of before, namely $\frac{c}{a}$. Lastly, the sum of the product of any two of these roots is still unchanged. Using another round of Vieta's, all of this corresponds to answer option $\textbf{(C)}$.

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