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

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)