Use Descartes' rule of signs to find the number of possible positive, negative, and nonreal roots for the following equation. $$3 x ^ { 3 } + 8 x ^ { 2 } - 4 x + 10 = 0$$
A) $\quad 1$ positive; 1 negative; 1 nonreal
B) $\quad 0$ positive; 0 negative; 3 nonreal
C) $\quad$ positive; 0 or 2 negative; 0 or 2 nonreal
D) $\quad 0$ or 2 positive; 1 negative; 0 or 2 nonreal
E) none of these

Question

Quizplus · Accepted Answer

Descartes' rule of signs for positive roots: Changing the sign of the coefficients from positive to negative or vice versa occurs twice (from $3x^3$ to $8x^2$ and from $-4x$ to $+10$), indicating 0 or 2 positive roots. For negative roots, substitute $x$ with $-x$ to get $3(-x)^3 + 8(-x)^2 - 4(-x) + 10 = -3x^3 + 8x^2 + 4x + 10$, which has one sign change (from $-3x^3$ to $8x^2$), indicating 1 negative root. Since the equation is cubic, it must have 3 roots in total. If there are 0 or 2 positive roots and 1 negative root, the remaining roots (if any) must be nonreal, leading to 0 or 2 nonreal roots.

Use Descartes' Rule of Signs to Find the Number of Possible