Theory of equations (Descartes' rule of signs)
Equations and inequalities · Mathematics
Study notes
Q: How many positive and negative real roots can x³ - 2x² - 5x + 6 = 0 have? Step 1: P(x) = x³ - 2x² - 5x + 6. Signs: + to - (1), - to - (none), - to + (2). Two changes. Step 2: Positive real roots: 2 or 0. Step 3: P(-x) = -x³ - 2x² + 5x + 6. Signs: - to - (none), - to + (1), + to + (none). One change. Step 4: Negative real roots: exactly 1. (Actual roots: 3, 1, -2: two positive, one negative. ✓)