4.4 Real Zeros of Polynomial Functions Understand the factor theorem Factor higher degree polynomials completely Analyze polynomials having multiple zeros Understand the rational zeros test and Descartes’ rule of signs Solve higher degree polynomial equations Understand the intermediate value theorem

Factor Theorem A polynomial f(x) has a factor x − k if and only if f(k) = 0.

Example: Applying the factor theorem (1 of 2) Use the graph and the factor theorem to list the factors of f(x). Assume that all zeros are integers. Solution The zeros or x-intercepts of f are −2, 1 and 3. Since f(−2) = 0, the factor theorem states that (x + 2) is a factor, and f(1) = 0 implies that (x − 1) is a factor and f(3) = 0 implies (x − 3) is a factor. Thus the factors are (x + 2), (x − 1), and (x − 3).

Example: Applying the factor theorem (2 of 2)

Zeros with Multiplicity If f(x) = (x + 2)², then the factor (x + 2) occurs twice and the zero −2 is called a zero of multiplicity 2. The polynomial g(x) = (x + 1)³(x – 2) has zeros −1 and 2 with multiplicities 3 and 1 respectively.

Complete Factored Form has n real zeros c1, c2, c3, …, cn, where a distinct zero is listed as many times as its multiplicity. Then f(x) can be written in complete factored form as f(x) = an(x − c1)(x − c2)(x − c3) ···(x − cn).

Example: Finding a complete factorization Write the complete factorization for the polynomial f(x) = 7x³ − 21x² − 7x + 21 with given zeros −1, 1 and 3. Solution Leading coefficient is 7. Zeros are −1, 1 and 3. The complete factorization: f(x) = 7(x + 1)(x − 1)(x − 3).

Example: Factoring a polynomial graphically Use the graph of f to factor f(x) = 2x³ − 4x² − 10x + 12. Solution Leading coefficient is 2. Zeros are −2, 1, and 3. The complete factorization: f(x) = 2(x + 2)(x − 1)(x − 3).

Example: Factoring a polynomial symbolically The polynomial f(x) = 2x³ − 2x² − 34x − 30 has a zero of −1. Express f(x) in complete factored form. Solution If −1 is a zero, by the factor theorem (x + 1) is a factor. Use synthetic division.

Rational Zeros Test (1 of 2)

Rational Zeros Test (2 of 2)

Example: Finding rational zeros of a polynomial (1 of 3) Find all rational zeros of f(x) = 6x³ − 5x² − 7x + 4 and factor f(x). Solution p is a factor of 4, q is a factor of 6 Any rational zero must occur in the list

Example: Finding rational zeros of a polynomial (2 of 3) Evaluate f(x) at each value in the list.

Example: Finding rational zeros of a polynomial (3 of 3)

Descartes’ Rule of Signs (1 of 2) Let P(x) define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of x. a. The number of positive real zeros either equals the number of variations in sign occurring in the coefficients of P(x) or is less than the number of variations by a positive even integer.

Descartes’ Rule of Signs (2 of 2) Let P(x) define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of x. b. The number of negative real zeros either equals the number of variations in sign occurring in the coefficients of P(−x) or is less than the number of variations by a positive even integer.

Example: Applying Descartes’ rule of signs (1 of 2) Solution P(x) has three variations in sign Thus, P(x) has 3, or 3 − 2 = 1 positive real zeros.

Example: Applying Descartes’ rule of signs (2 of 2) For negative zeros, consider the variations in sign for P (−x) P(−x) has one variation in sign, P(x) has only 1 negative real zero.

Polynomial Equations Factoring can be used to solve polynomial equations with degree greater than 2.

Example: Solving a polynomial equation (1 of 2) Solution

Example: Solving a polynomial equation (2 of 2)

Example: Solving a polynomial equation (3 of 3)

Example: Finding a solution graphically Since there is only one x-intercept the equation has one real solution: x  2.65

Intermediate Value Theorem Let (x1, y1) and (x2, y2) with y1 ≠ y2 and x1 < x2, be two points on the graph of a continuous function f. Then, on the interval x1 ≤ x ≤ x2, f assumes every value between y1 and y2 at least once.

Applications: Intermediate Value Theorem There are many examples of the intermediate value theorem. Physical motion is usually considered to be continuous. Suppose at one time a car is traveling at 20 miles per hour and at another time it is traveling at 40 miles per hour. It is logical to assume that the car traveled 30 miles per hour at least once between these times. In fact, by the intermediate value theorem, the car must have assumed all speeds between 20 and 40 miles per hour at least once.