Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Zeros of a Polynomial Function Learn to find the possible number of positive and negative zeros of polynomials. Learn to find the bounds on the real zeros of polynomials. Learn basic facts about the complex zeros of polynomials. Learn to use the Conjugate Pairs Theorem to find zeros of polynomials. SECTION

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley DESCARTE’S RULE OF SIGNS Let F(x) be a polynomial function with real coefficients and with terms written in descending order. 1.The number of positive zeros of F is either equal to the number of variations of sign of F(x) or less than that number by an even integer.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley DESCARTE’S RULE OF SIGNS When using Descarte’s Rule, a zero of multiplicity m should be counted as m zeros. 2.The number of negative zeros of F is either equal to the number of variations of sign of F(–x) or less than that number by an even integer.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Using Descarte’s Rule of Signs Find the possible number of positive and negative zeros of Solution There are three variations in sign in f (x). + to – – to ++ to – The number of positive zeros is either 3 or (3 – 2 =) 1.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Using Descarte’s Rule of Signs Solution continued There are two variations in sign in f (–x). + to –– to + The number of negative zeros is either 2 or (2 – 2 =) 0.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley RULES FOR BOUNDS Let F(x) be a polynomial function with real coefficients and a positive leading coefficient. Suppose F(x) is synthetically divided by x – k. Then 1.If k > 0, and each number in the last row is either zero or positive, then k is an upper bound on the zeros of F(x).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley RULES FOR BOUNDS If 0 appears in the last row of the synthetic division, then it may be assigned either a positive or a negative sign in determining whether the signs alternate. 2.If k < 0, and numbers in the last row alternate in sign, then k is a lower bound on the zeros of F(x).

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Finding the Bounds on the Zeros Find upper and lower bounds on the zeros of Solution The possible zeros are: ±1, ±2, ±3, and ±6. There is only one variation in sign so there is one positive zero. Use synthetic division until last row is all positive or 0. Not 1, not 2, but 3.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Finding the Bounds on the Zeros Solution continued So 3 is upper bound. Use synthetic division until last row alternates in sign. Not –1, but –2. So –2 is lower bound.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FUNDAMENTAL THEOREM OF ALGEBRA Every polynomial with complex coefficients a n, a n – 1, …, a 1, a 0 has at least one complex zero.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FACTORIZATION THEOREM FOR POLYNOMIALS If P(x) is a complex polynomial of degree n ≥ 1, it can be factored into n (not necessarily distinct) linear factors of the form where a, r 1, r 2, …, r n are complex numbers.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley NUMBER OF ZEROS THEOREM A polynomial of degree n has exactly n zeros, provided that a zero of multiplicity k is counted k times.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Constructing a Polynomial Whose Zeros are Given Find a polynomial P(x) of degree 4 with a leading coefficient of 2 and zeros –1, 3, i, and – i. Write P(x) Solution a. Since P(x) has degree 4, we write a.in completely factored form; b.by expanding the product found in part a.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Constructing a Polynomial Whose Zeros are Given Solution continued b. by expanding the product found in part a.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley CONJUGATE PAIRS THEOREM If P(x) is a polynomial function whose coefficients are real numbers and if z = a + bi is a zero of P, then its conjugate, is also a zero of P.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley ODD–DEGREE POLYNOMIALS WITH REAL ZEROS Any polynomial P(x) of odd degree with real coefficients must have at least one real zero.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Using the Conjugate Pairs Theorem A polynomial P(x) of degree 9 with real coefficients has the following zeros: 2, of multiplicity 3; 4 + 5i, of multiplicity 2; and 3 – 7i. Write all nine zeros of P(x). 2, 2, 2, 4 + 5i, 4 – 5i, 4 + 5i, 4 – 5i, 3 + 7i, 3 – 7i Solution Since complex zeros occur in conjugate pairs, the conjugate 4 – 5i of 4 + 5i is a zero of multiplicity 2, and the conjugate 3 + 7i of 3 – 7i is a zero of P(x). The nine zeros of P(x) are:

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FACTORIZATION THEOREM FOR A POLYNOMIAL WITH REAL COEFFICIENTS Every polynomial with real coefficients can be uniquely factored over the real numbers as a product of linear factors and/or prime quadratic factors.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Finding the Complex Real Zeros of a Polynomial Given that 2 – i is a zero of Solution The conjugate of 2 – i, 2 + i is also a zero. So P(x) has linear factors: find the remaining zeros.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Finding the Complex Real Zeros of a Polynomial Solution continued Divide P(x) by x 2 – 4x + 5

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Finding the Complex Real Zeros of a Polynomial Solution continued We can multiply these factors to find P(x). The zeros of P(x) are 1 (of multiplicity 2), 2 – i, and 2 + i.

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Finding the Zeros of a Polynomial Find all zeros of the polynomial P(x) = x 4 – x 3 + 7x 2 – 9x – 18. Solution Possible zeros are: ±1, ±2, ±3, ±6, ±9, ±18 Use synthetic division to find that 2 is a zero. (x – 2) is a factor of P(x). Solve

Slide Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Finding the Zeros of a Polynomial Solution continued The four zeros of P(x) are –1, 2, –3i, and 3i.