The polynomial x³ - 3x² - 6x + 8 has x = 1 as one of its roots. What is the sum of the other two roots?

Since $x = 1$ is a root, $x - 1$ is a factor. Dividing yields $x^2 - 2x - 8$, which you can confirm: $(x - 1)(x^2 - 2x - 8) = x^3 - 3x^2 - 6x + 8$. The other roots solve $x^2 - 2x - 8 = 0$, which factors as $(x - 4)(x + 2) = 0$, so they are 4 and $-2$. Their sum is 2, choice C, matching Vieta's formula $-(-2) = 2$. Choice A ($-2$) is one of the roots, or the sum with a wrong sign. Choice B (8) is the constant term of the cubic. Choice D ($-3$) is the cubic's $x^2$-coefficient.