2021 Fall OMC 10 #14

Problem:

A cubic polynomial $P(x)=a{x}^3+bx+c$ with $a\ne 0$ has three roots $r$, $s$, and $t$. Which of the following polynomials has roots $r+s,s+t,$ and $t+r$?

$\mathbf{\text{(A)}}a{x}^3-bx-c\mathbf{\text{(B)}}a{x}^3-bx+c\mathbf{\text{(C)}}a{x}^3+bx-c\mathbf{\text{(D)}}a{x}^3+b{x}^2-c\mathbf{\text{(E)}}a{x}^3+b{x}^2+c$

Solution:

Write the polynomial as $a{x}^3+0x^2+bx+c$. 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 $\mathbf{\text{(C)}}$.