1. OBJECTIVES: 1. USE THE FUNDAMENTAL THEOREM OF ALGEBRA 2. FIND COMPLEX CONJUGATE ZEROS. 3. FIND THE NUMBER OF ZEROS OF A POLYNOMIAL. 4. GIVE THE COMPLETE FACTORIZATION OF POLYNOMIAL EXPRESSIONS. 4.6The Fundamental Theorem of Algebra

2. Example #1 Finding a Polynomial Given Its Zeros Find a polynomial f (x) of degree 4 such that −3, −1, and 2 are zeros, 2 is a multiplicity of 2, and f (1) = 32. First we must write the zeros as factors, repeating one of them twice. This is the simplest polynomial that has the given zeros. Any polynomial that is multiplied by a constant will have the same zeros, so we represent this constant with a variable, a, and we get the general polynomial: To find the specific polynomial we must use the fact that f (1) = 32.

3. Example #1 Finding a Polynomial Given Its Zeros Find a polynomial f (x) of degree 4 such that −3, −1, and 2 are zeros, 2 is a multiplicity of 2, and f (1) = 32. Plugging in 1 for x and setting it equal to 32 we get: And finally for the specific polynomial:

4. Example #2 Conjugate Zeros Find the zeros. From the graph it is clear this polynomial does not have any real zeros, but it does have two complex zeros. Complex zeros always come in pairs called conjugates.

5. Example #3 A Polynomial with Specific Zeros Find a polynomial with real coefficients whose zeros include the numbers −3 and 2 − i. If 2 – i is a zero, then so must be 2 + i as they always come in pairs. Rewrite all zeros into factor form. Multiply the complex factors first.

6. Example #4 Completely Factor Completely factor the polynomial function over the set of real numbers given 2 − i is a zero of the function. f(x) = x 4 − 4x 3 − 2x 2 + 28x − 35 Once again, given 2 – i is a zero, so must be 2 + i. Writing them as factors and multiplying them together we get: This implies that the original polynomial is divisible by x 2 – 4x + 5.

7. Example #4 Completely Factor Completely factor over the real numbers. f(x) = x 4 − 4x 3 − 2x 2 + 28x − 35 (x 4 – 4x 3 + 5x 2 ) –7x 2 + 28x – 35 (–7x 2 + 28x – 35) 0 x2 x2 0x 3 – 7x 2 + 28x + 0x (0x 3 – 0x 2 + 0x) − 7

8. Example #4 Completely Factor Completely factor over the real numbers. f(x) = x 4 − 4x 3 − 2x 2 + 28x − 35 The quotient is still a quadratic that can still be broken down: So the fully factored form of the polynomial function is:

9. Example #5 Completely Factor Completely factor the polynomial function. f(x) = x 4 − 5x 3 + 2x 2 + 22x − 20 Given no information, the best way to factor this polynomial would be using a graph. It appears that − 2 and 1 are zeros of the function. Using synthetic division we get:

10. Example #5 Completely Factor Completely factor the polynomial function. f(x) = x 4 − 5x 3 + 2x 2 + 22x − 20 The remaining quadratic is not factorable and requires the quadratic formula. And now the completely factored form: