1 / 6

Advanced Algebra Notes Section 5.7: Apply the Fundamental Theorem of Algebra

Advanced Algebra Notes Section 5.7: Apply the Fundamental Theorem of Algebra. The equation x 3 – 5x 2 – 8x + 48 = 0, when solved has the solutions -3, and 4 (Double Root). There are 3 solutions to this third degree equation.

suki
Télécharger la présentation

Advanced Algebra Notes Section 5.7: Apply the Fundamental Theorem of Algebra

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Advanced Algebra Notes Section 5.7: Apply the Fundamental Theorem of Algebra The equation x3 – 5x2 – 8x + 48 = 0, when solved has the solutions -3, and 4 (Double Root). There are 3 solutions to this third degree equation. The previous result is generalized by _________________________________, first proved by the German mathematician Karl Freidrich Gauss. Fundamental Theorem of Algebra If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has at least one solution in the set of complex numbers. Corollary: If f(x) is a polynomial of degree n where n > 0, then the equation f(x) = 0 has exactly n solutions provided each solution that is repeated is counted that many times. (Double Root = 2, etc.) The corollary implies that an nth-degree polynomial function f has exactly n zeros. Fundamental Theorem of Algebra

  2. Examples: Find the zeros of a polynomial function. • 1. f(x) = x3 + x2 – 3x - 3 , How many zeros does this function have: ______. What • are they? 3

  3. 5 • 2. f(x) = x5 - 4x3 + x2 – 4 , How many zeros does this function have: ______. What • are they? Do the same list because the is the same.

  4. Behavior Near Zeros When a factor x – k of a function f is raised to an odd power, the graph of f crosses the x-axis at x = k. When a factor x – k of a function f is raised to an even power, the graph of f is tangent to the x-axis at x = k. Note that the solutions from example 2 above are complex conjugates. This is an illustration of our next theorem. Complex Conjugates Theorem If f is a polynomial function with real coefficients, and a + bi is an imaginary zero of f, then a – bi is also a zero of f. Irrational Conjugates Theorem Suppose f is a polynomial function with real coefficients, and a & b are rational numbers such that is irrational. If a + is a zero of f, then a - is also a zero of f.

  5. Example: • Write a polynomial function f of least degree that has rational coefficients, a leading • coefficient of 1, and 2 and -2 – 5i as zeros.

  6. 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: • Determine the possible number of positive or negative real zeros, and imaginary • zeros for f(x). Positive Zeros: 2 sign changes in f(x) , so (2 or 0) Negative Zeros: 2 sign changes in f(-x), so (2 or 0) Imaginary Zeros: (2, 4, 6)

More Related