Download Section 4.6

Section 4.6
The Fundamental Theorem of Algebra
Pairs of complex numbers of the forms
𝑎 + 𝑏𝑖 and 𝑎 − 𝑏𝑖, where 𝑏 ≠ 0
Notes:
4. Determine the possible number of positive or negative real zeros, and imaginary
zeros for f(x).
6
2
f ( x)  2 x  3 x  x  1
Positive Zeros:
2 sign changes in f(x) , so (2 or 0)
f ( x)  2( x)6  3( x)2  ( x)  1
 2 x 6  3x 2  x  1
Negative Zeros: 2 sign changes in f(-x), so (2 or 0)
Imaginary Zeros:
(2, 4, 6)
𝑦 = (𝑥 + 4)(𝑥 − 1)(𝑥 − 7)
𝑦 = (𝑥 + 4)(𝑥 2 − 8𝑥 + 7)
𝑦 = (𝑥 − 10)(𝑥 − (− 5))(𝑥 − ( 5))
𝑦 = (𝑥 − 10)(𝑥 + 5)(𝑥 − 5)
𝑦 = 𝑥 3 − 8𝑥 2 + 7𝑥 + 4𝑥 2 − 32𝑥 + 28
𝑦 = (𝑥 − 10)(𝑥 2 − 25)
𝑦 = 𝑥 3 − 4𝑥 2 − 25𝑥 + 28
y = (x − 10)(𝑥 2 − 5)
𝑦 = 𝑥 3 − 10𝑥 2 − 5𝑥 + 50
𝐴𝑙𝑙 𝑟𝑎𝑑𝑖𝑐𝑎𝑙𝑠 𝑎𝑛𝑑 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦 #′ 𝑠 ℎ𝑎𝑣𝑒 𝑡𝑜 ℎ𝑎𝑣𝑒 𝑎 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑡ℎ𝑒𝑦
come in pairs
,3 + 𝑖
𝑦 = (𝑥 − 8)(𝑥 − 3 − 𝑖 )(𝑥 − 3 + 𝑖 )
𝑦 = (𝑥 − 8)(𝑥 − 3 + 𝑖)(𝑥 − 3 − 𝑖)
+𝟏
𝑦 = (𝑥 − 8)(𝑥 2 − 3𝑥 − ix − 3x + 9 + 3i + ix − 3i − 𝑖 2 )
𝑦 = (𝑥 − 8)(𝑥 2 − 6x + 10)
𝑦 = 𝑥 3 − 6𝑥 2 + 10𝑥 − 8𝑥 2 + 48𝑥 − 80
𝑦 = 𝑥 3 − 14𝑥 2 + 58𝑥 − 80
See next slide for problem
𝑦 = 𝑥(𝑥 − 2 − 2 )(𝑥 − 2 + 2 )(𝑥 − 2 + 3𝑖 )(𝑥 − 2 − 3𝑖 )
𝑦 = 𝑥(𝑥 − 2 + 2)(𝑥 − 2 − 2)(𝑥 − 2 − 3𝑖) (𝑥 − 2 + 3i)
𝑦 = 𝑥(𝑥 2 − 2𝑥 − 𝑥 2 − 2𝑥 + 4 + 2 2 + 𝑥 2 − 2 2 − 4)(𝑥 2 − 2𝑥 + 3𝑖𝑥 − 2𝑥 + 4 − 6𝑖 − 3𝑖𝑥 + 6𝑖 − 9𝑖 2 )
𝑦 = 𝑥(𝑥 2 − 4𝑥 + 2)(𝑥 2 − 4𝑥 + 13)
𝑦 = 𝑥(𝑥 4 − 8𝑥 3 + 31𝑥 2 − 60𝑥 + 26)
𝑦 = 𝑥 5 − 8𝑥 4 + 31𝑥 3 − 60𝑥 2 + 26𝑥
A French mathematician Rene’ Descartes found the following relationship between the coefficients of a polynomial function
and the number of positive and negative zeros of the function.
Descartes’ Rule of Signs
Let f(x) =
be a polynomial function with real coefficients.
*The number of positive real zeros of f is equal to the number of changes in sign of
the coefficients of f(x) or is less than this by an even number.
*The number of negative real zeros of f is equal to the number of changes in sign of
the coefficients of f(-x) or is less than this by an even number.
Example:
4. Determine the possible number of positive or negative real zeros, and imaginary
zeros for f(x).
6
2
f ( x)  2 x  3 x  x  1
Positive Zeros:
2 sign changes in f(x) , so (2 or 0)
f ( x)  2( x)6  3( x)2  ( x)  1
 2 x 6  3x 2  x  1
Negative Zeros: 2 sign changes in f(-x), so (2 or 0)
Imaginary Zeros:
(2, 4, 6)