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hester-barber
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THE FUNDAMENTAL THEOREM OF ALGEBRA
Descartes’ Rule of Signs
If f(x) is a polynomial function with real coefficients, then
*The number of positive real zeros is equal to the number of variations in sign of f(x), or less by some even number.
*The number of negative real zeros is equal to the number of variations in sign of f(-x), or less by some even number.
Describe the zeros
f(x) = 6x5 + 8x2 – 10x – 15 1 positive, 2 or 0 negativeDoes this function have imaginary zeros?f(x) = -11x4 + 20x3 + 3x2 – x + 18
3 or 1 positive, 1 negative
Does this function have imaginary zeros?
The Fundamental Theorem of Algebra (Gauss) says that every polynomial has at least one zero in the complex numbers. The linear factorization theorem (which is derived from it) states that every polynomial f(x) = axn + … can be factored as a(x – c1)(x – c2) …(x – cn). Therefore every polynomial has exactly n zeros among the complex numbers (counting multiple roots).
Which kind(s) of zeros appear on graphs?
How would you factor x2 – 5 over the real numbers?
Can you see why x2 + 4 = (x + 2i)(x – 2i) when factored over the complex numbers?
Now factor x4 – x2 – 20.
Write x5 + x3 + 2x2 – 12x + 8 as the product of linear factors and list all zeros.
Answer: f(x) = (x – 1)(x – 1)(x + 2)(x + 2i)(x – 2i)
Zeros are 1, 1, -2, 2i, and -2i
Complex zeros occur in conjugate pairs!
Write a 4th degree polynomial with roots -1, -1, and 3i.
Answer: f(x) = x4 + 2x3 + 10x2 + 18x + 9
How would you write the factor if 2 + 3i was a root???
Find all of the zeros of f(x) = x4 – 3x3 + 6x2 + 2x – 60, given that 1 + 3i is a zero of f.
Answer : 1 + 3i, 1 – 3i, 3, -2.