1. Fundamental Theorem of Algebra If f(x) is a polynomial of degree n, where n≥1, then the equation f(x) = 0 has at least one complex root. Date: 2.6 Topic: Complex Zeros and the Fundamental Theorem of Algebra

2. Properties of Polynomial Equations If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots: x 4 +3x 2 +2x-1 has possibly 4 roots If a+bi is a root of the equation, then a-bi is also a root.

3. Text Example State how many complex and real zeros the function has f(x) = x 4  5x 3  x 2  3x + 6 Graph the function on a graphing calculator to find out how many are real zeros. Degree is 4, therefore there are 4 complex zeros. Imaginary roots never cross the x-axis. How many x-intercepts are there?2 Because there are 2 x-intercepts there are 2 real zeros.

4. Linear Factorization Theorem If f (x)  a n x n  a n-1 x n-1  …  a 1 x  a 0 and n≥1, and a n ≠ 0, then f(x) =a n (x - c 1 )(x - c 2 )…(x - c n ) where c 1, c 2, …, c n are complex numbers (possibly real roots with multiplicity greater than 1)

5. f(x)= (x + 2) 2 (x - 4) 2 f(x)= (x - (-2)) 2 (x - 4) 2 Use the given zeros and multiplicity in the equation. f(x)= (x 2 + 4x + 4) (x 2 - 8x + 16) Multiply. f(x)= x 4 - 4x 3 - 12x 2 + 32x + 64 Multiply. Example Write a polynomial function with real coefficents whose zeros and multiplicity are -2(multiplicity 2) and 4(mulitiplicity 2)

6. f(x)= a n (x + 2) (x - 2) (x - i) (x + i) f(x)= a n (x - (-2)) (x - 2) (x - i) (x - (-i)) Use the given zeros in the equation. If i is a zero, so is -i f(x)= a n (x 2 - 4) (x 2 + 1) Multiply. f(x)= a n (x 4 - 3x 2 - 4) Multiply. Example Find a fourth-degree polynomial function with real coefficents that has -2, 2, and i as zeros and such that f(3) = -150 50a n = -150 a n = -3 f(x)= -3x 4 + 9x 2 + 12 a n (3 4 - 33 2 - 4) = -150 f(x)= -3(x 4 - 3x 2 - 4) substitute a n in the equation. Find a n, use that f(3) = -150

7. Complete Student Checkpoint Find a third-degree polynomial function f(x) with real coefficients that has -3 and i as zeros and such that f(1)=8

8. Text Example Solve: x 4  6x 2  8x + 24  0. Solution The graph of f (x)  x 4  6x 2  8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. x-intercept: 2 The zero remainder indicates that 2 is a root of x 4  6x 2  8x + 24  0. 2 1 0  6  8 24 2 4  4  24 1 2  2  120 Day 2

9. Solution Now we can rewrite the given equation in factored form. (x – 2)(x 3  2x 2  2x  12)  0 This is the result obtained from the synthetic division. x – 2  0 or x 3  2x 2  2x  12  Set each factor equal to zero. x 4  6x 2  8x + 24  0 This is the given equation. Text Example cont. Solve: x 4  6x 2  8x + 24  0. 2 1 0  6  8 24 2 4  4  24 1 2  2  120

10. Solution We can use the same approach to look for rational roots of the polynomial equation x 3  2x 2  2x  12 = 0, listing all possible rational roots. However, take a second look at the figure of the graph of x 4  6x 2  8x + 24  0. Because the graph turns around at 2, this means that 2 is a root of even multiplicity (x - 2) 2. Thus, 2 must also be a root of x 3  2x 2  2x  12 = 0, confirmed by the following synthetic division. x-intercept: 2 2 1 2  2  12 2 8 12 1 4 6 0 These are the coefficients of x 3  2x 2  2x  12 = 0. The zero remainder indicates that 2 is a root of x 3  2x 2  2x  12 = 0. Text Example cont. Solve: x 4  6x 2  8x + 24  0. So, x 3  x 2 - 2x  12 = (x – 2)(x 2  4x  6)

11. Solution Now we can solve the original equation as follows. (x – 2)(x 3  2x 2  2x  12)  0 This was obtained from the first synthetic division. x 4  6x 2  8x + 24  0 This is the given equation. (x – 2)(x – 2)(x 2  4x  6)  0 This was obtained from the second synthetic division. x – 2  0 or x – 2  0 or x 2  4x  6  Set each factor equal to zero. x  2 x  2 x = -2+ i  2 Solve, use the quadratic formula to solve x 2  4x  6  Text Example cont. Solve: x 4  6x 2  8x + 24  0. The solution set of the original equation is: {2, -2+ i  2}

12. Complete Student Checkpoint Factor as the product of factors that are irreducible over a. the rational numbers b. the real numbers c. the complex imaginary numbers hint: factor hint: factor the binomial that can be factored into real numbers hint: factor the binomial that can be factored into imaginary numbers

13. Complete Student Checkpoint Solve: p/q =+1, +13 are possible zeros Plug them in to find the zeros (use synthetic division): f(1) = 0 1 is a zero 1 -6 22 -30 13 1 1 -5 17 -13 1 -5 17 -13 0 1 -5 17 -13 1 1 -4 13 1 -4 13 0 Use possible zeros and synthetic division to factor again Use quadratic formula:

14. Complex Zeros and the Fundamental Theorem of Algebra