1. Warm up Write the quadratic f(x) in vertex form.

2. Polynomial Functions Descartes Rule of Signs

3. Compare the products of: = =

4. Questions: Do you know that the variation in the signs is f (x). It can determine the number of positive and negative zeros?

5. Descartes’ Rule of Signs

6. In some cases, the following rule is helpful in eliminating candidates from lengthy lists of possible rational roots. – It was discovered by the French philosopher and mathematician René Descartes around 1637.

7. Variation in Sign To describe this rule, we need the concept of variation in sign. – Suppose P(x) is a polynomial with real coefficients, written with descending powers of x (and omitting powers with coefficient 0). – A variation in sign occurs whenever adjacent coefficients have opposite signs.

8. Variation in Sign For example, P(x) = 5x 7 – 3x 5 – x 4 + 2x 2 + x – 3 has three variations in sign.

9. Descartes’ Rule of Signs Let P be a polynomial with real coefficients. 1.The number of positive real zeros of P(x) is either equal to the number of variations in sign in P(x) or is less than that by an even whole number. 2.The number of negative real zeros of P(x) is either equal to the number of variations in sign in P(-x) or is less than that by an even whole number.

10. EXAMPLE : Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f (x)  x 3  2x 2  5x + 4. Solution 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f (x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f (  x). We obtain this equation by replacing x with  x in the given function. f (  x)  (  x) 3  2(  x) 2  x  4 f (x)  x 3  2x 2  5x + 4 This is the given polynomial function. Replace x with  x.   x 3  2x 2  5x + 4

11. EXAMPLE : Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f (x)  x 3  2x 2  5x + 4. Solution Now count the sign changes. There are three variations in sign. The number of negative real zeros of f is either equal to the number of sign changes, 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or 3  2  1 negative real zero. f (  x)   x 3  2x 2  5x + 4 1 2 3

12. Using Descartes’ Rule Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros of the polynomial P(x) = 3x 6 + 4x 5 + 3x 3 – x – 3 – The polynomial has one variation in sign. – So, it has one positive zero.

13. Using Descartes’ Rule Now, P(–x) = 3(–x) 6 + 4(–x) 5 +3(–x) 3 – (–x) – 3 = 3x 6 – 4x 5 – 3x 3 + x – 3 – Thus, P(–x) has three variations in sign. – So, P(x) has either three or one negative zero(s), making a total of either two or four real zeros.

14. Example 1: State the number of positive and negative real zeros for

15. Example 1: State the number of positive and negative real zeros for

16. Your Turn: Find the possible number of positive negative and complex solutions for each of the following: