Slide 23
a’s are real numbers, an 0, and a0 0.
1. f(x) has two change-of-signs; thus, f(x) has two or zero positive real roots.
2. f(-x) = -4x3 - 5x2 + 6 has one change-of-signs; thus, f(x) has one negative real root.
Example 1: Descartes’s Rule of Signs
Slide 24
Factor out x; f(x) = x(4x2 - 5x + 6) = xg(x)
1. g(x) has two change-of-signs; thus, g(x) has two or zero positive real roots.
2. g(-x) = 4x2 + 5x + 6 has zero change-of-signs; thus, g(x) has no negative real root.
Example 2: Descartes’s Rule of Signs
Slide 25
rational zero = p/q,
in reduced form, and p and q are factors of a0 and an, respectively.
Slide 26
f(x) = 4x3 - 5x2 + 6 p {1, 2, 3, 6}
q {1, 2, 4}
p/q {1, 2, 3, 6, 1/2, 1/4, 3/2, 3/4} represents all possible rational roots of f(x) = 4x3 - 5x2 + 6 .
Example 3: Rational Zero Test
Slide 27
f(x) is a polynomial with real coefficients and an > 0 with f(x) (x - c), using synthetic division:
1. If c > 0 and each # in last row is either positive or zero, c is an upper bound.
2. If c < 0 and the #’s in the last row alternate positive and negative, c is an lower bound.
Slide 28
2x3 - 3x2 - 12x + 8 divided by x + 3
-3 2 -3 -12 8 -6 27 -45
2 -9 15 -37
c = -3 < 0 and #’s in last row alternate positive/negative. Thus, x = -3 is a lower bound to real roots.
Example 4: Upper and Lower Bound