Use Descartes' rule of signs to find the number of possible positive, negative, and nonreal roots for the following equation. $$7 x ^ { 3 } + 5 x ^ { 2 } - 8 x + 8 = 0$$
A) $\quad 0$ or 2 positive; 1 negative; 0 or 2 nonreal
B) $\quad 0$ positive; 0 negative; 3 nonreal
C) $\quad 1$ positive; 0 or 2 negative; 0 or 2 nonreal
D) $\quad$ l positive; 1 negative; 1 nonreal
E) none of these

Question

Quizplus · Accepted Answer

Descartes' rule of signs for the given polynomial $7x^3 + 5x^2 - 8x + 8 = 0$ indicates 2 sign changes, suggesting 2 or 0 positive roots. For negative roots, substituting $x$ with $-x$ gives $7(-x)^3 + 5(-x)^2 - 8(-x) + 8 = -7x^3 + 5x^2 + 8x + 8$, which has 1 sign change, indicating exactly 1 negative root. The total number of roots for a cubic equation is 3, so if there are 3 real roots (2 positive, 1 negative), there are 0 nonreal roots. If there are fewer real roots, the remaining must be nonreal, maintaining the total at 3. Therefore, the correct answer is 0 or 2 positive; 1 negative; 0 or 2 nonreal roots.

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