1. Rational Root Theorem

4. Finding Zeros of a Polynomial Function Use the Rational Zero Theorem to find all possible rational zeros. Use Synthetic Division to try to find one rational zero (the remainder will be zero). If “n” is a rational zero, factor the original polynomial as (x – n)q(x). Test remaining possible rational zeros in q(x). If one is found, then factor again as in the previous step. Continue in this way until all rational zeros have been found. See if additional irrational or non-real complex zeros can be found by solving a quadratic equation.

5. Finding Rational Zeros So which one do you pick? Pick any. Find one that is a zero using synthetic division... Possible zeros are + 1, + 2, + 4, + 8

6. Let’s try 1. Use synthetic division 1 1 1 –10 8 1 2 –8 1 2 –8 0 1 is a zero of the function The depressed polynomial is x 2 + 2x – 8 Find the zeros of x 2 + 2x – 8 by factoring or (by using the quadratic formula)… (x + 4)(x – 2) = 0 x = –4, x = 2 The zeros of f(x) are 1, –4, and 2

7. Use synthetic division Find all possible rational zeros of:

8. Example Continued This new factor has the same possible rational zeros: Check to see if -1 is also a zero of this: Conclusion:

9. Example Continued This new factor has as possible rational zeros: Check to see if -1 is also a zero of this: Conclusion:

10. Example Continued Check to see if 1 is a zero: Conclusion:

11. Example Continued Check to see if 2 is a zero: Conclusion:

12. Example Continued Summary of work done:

13. Using The Linear Factorization Theorem We can now use the Linear Factorization Theorem for a fourth- degree polynomial.

14. Using The Linear Factorization Theorem

15. Descartes Rule of Signs is a method for determining the number of sign changes in a polynomial function.

16. Polynomial FunctionSign Changes Conclusion 3 2 1 There is one positive real zero. Descarte’s Rule of Signs and Positive Real Zeros

18. Descarte’s Rule of Signs Example Determine the possible number of positive real zeros and negative real zeros of P(x) = x 4 – 6x 3 + 8x 2 + 2x – 1. We first consider the possible number of positive zeros by observing that P(x) has three variations in signs. + x 4 – 6x 3 + 8x 2 + 2x – 1 Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive real zeros. For negative zeros, consider the variations in signs for P(  x). P(  x) = (  x) 4 – 6(  x) 3 + 8(  x) 2 + 2(  x)  1 = x 4 + 6x 3 + 8x 2 – 2x – 1 Since there is only one variation in sign, P(x) has only one negative real root. 1 2 3 Total number of zeros 4 Positive:3 1 Negative:1 1 Nonreal:0 2