Section 2.5 Zeros of Polynomial Functions

1. Section 2.5Zeros of Polynomial Functions

2. Objectives: • Use the Rational Zero Theorem to find possible rational zeros • Find zeros of a polynomial function • Solve polynomial equations • Use the Linear Factorization Theorem to find polynomials with given zeros • Use Descartes’s Rule of Signs

3. The Rational Zero Theorem If f (x)= anxn+ an-1xn-1+…+ a1x + a0has integer coefficients, and if (where is reduced to lowest terms) is a rational zero off, then p is a factor of the constant term a0, and q is a factor of the leading coefficient an. The Rational Zero Theorem - theorem that gives a list of possiblerational zeros of a polynomial function. Not every number in the list will be a zero of the function, but every rational zero of the polynomial function will be a part of the list. The Rational Zero Theorem

4. Solution:The constant term is – 2and the leading coefficient is 15. Divide 1 and 2 by 1. Divide 1 and 2 by 3. Divide 1 and 2 by 5. Divide 1 and 2 by 15. There are 16 possible rational zeros. The actual solution set to f(x)=15x3 + 14x2 – 3x – 2 = 0 is {-1, -1/3, 2/5}, which contains 3 of the 16 possible solutions. Example 1 List all possible rational zeros of f(x)=15x3 + 14x2 – 3x – 2.

5. 2 1 - 3 - 4 12 2 - 2 - 12 1 - 1 - 6 0 Example 2 • 1) List all possible rational zeros of • Find one of the zeros of the function using synthetic division, then factor the remaining polynomial. • Find all remaining zeros of the function. How can the graph below help you find the zeros?

6. 3 6 - 19 2 3 18- 3 - 3 6 - 1 - 1 0 Example 3 • 1) List all possible rational zeros of • Starting with the integers, find one zero of the function using synthetic division, then factor the remaining polynomial. • Find remaining zeros of the function.

7. 2 -1 1 - 1 7 - 9 - 18 2 2 18 18 1 1 9 9 0 - 10- 9 1 0 9 0 Example 4 • List all possible rational roots of • Starting with the integers, find two roots of the equation using synthetic division. The graph below will help you easily find those roots. • Factor the remaining polynomial. Find all the roots of the equation. • The graph below will NOT help you find the imaginary roots. Why?

8. 2 • 0 -6 - 8 24 2 4 - 4 -24 1 2 -2 -12 0 x-intercept: 2 Example 5 Solve:x4–6x2–8x + 24 = 0 1. Begin by listing all possible rational roots (zeros): 2. The graph of f (x) = x4–6x2–8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. The zero remainder indicates that 2 is a root of x4-6x2- 8x + 24 = 0.

9. (x – 2)(x3 + 2x2– 2x – 12) = 0 After first synthetic division. • 1 2 -2 -12 2 8 12 1 4 6 0 The zero remainder indicates that 2 is a root ofx3 + 2x2-2x - 12 = 0. Example 5 Solve:x4–6x2– 8x + 24 = 0 3. Now we can rewrite the given equation in factored form. x4–6x2 + 8x + 24 = 0 This is the given equation. x – 2 = 0 or x3 + 2x2–2x – 12 = 0 Set each factor equal to zero. 4. However, take a second look at the figure of the graph of x4–6x2–8x + 24 = 0. Because the graph touches x-axis at 2, this means that 2 is a root of even multiplicity. So, 2 must also be a root of x3 + 2x2–2x–12 = 0, confirmed by the following synthetic division. These are the coefficients of x3 + 2x2-2x - 12 = 0.

10. (x – 2)(x3 + 2x2– 2x – 12) = 0 After first synthetic division. (x – 2)(x – 2)(x2+4x + 6) = 0 After second synthetic division. Example 5 Solve:x4–6x2–8x + 24 = 0 5. Putting everything together: x4–6x2– 8x + 24 = 0 This is the given equation. x – 2 = 0 x – 2 = 0 x2+4x + 6 = 0 Set each factor equal to zero. x = 2 x = 2 x2+4x + 6 = 0 Solve. The solution set of the original equation is

11. Properties of Polynomial Equations As we noticed from Examples 4 and 5: • Ifa polynomial equation is of degree n, then, counting multiple roots separately, the equation has n roots (zeros). This result is called the Fundamental Theorem of Algebra: • If a + biis a root (zero) of a polynomial equation with real coefficients (b 0), then the complex number a–biis also a root (zero). • Complex imaginary roots, if they exist, occur in conjugate pairs.

13. (# 28) Example 6 Find a 3rd degree polynomial function with real coefficients, where and are zeros, and .

14. Let f (x)= anxn+ an-1xn-1+ … + a2x2+ a1x + a0be a polynomial with real coefficients. 1. The number of positive real zeros of f iseither equal to the number of sign changes of f (x)or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero. 2. The number of negative real zeros of fiseither equal to the number of sign changes of f (-x) or is less than that number by an even integer. If f (-x) has only one variation in sign, then f has exactly one negative real zero. Descartes’s Rule of Signs

16. Descartes’s Rule of Signs

17. 1 4 - 1 5 - 2 - 6 4 3 8 6 4 3 8 6 0 - 3 0 - 6 4 0 8 0 Example 7 • How many positive and negative zeros are there? • Find all Possible Rational Zeros. • Use a graphing utility to find the first real zero of the function. Confirm using synthetic division. • Find all remaining zeros of the function.

18. Practice 1 List all possible rational zeros of the function (a) (b) (c) (d)

19. Practice 2 Find a third-degree polynomial function with real coefficients that have as zeros and such that . (a) (b) (c) (d)

20. Practice 3 What are the zeros of the function ? Find the first zero using a graphing utility. (a) (b) (c) (d)