1. EXAMPLE 5 Use the result to write f (x) as a product of two factors. Then factor completely. f (x) = x 3 – 2x 2 – 23x + 60 The zeros are 3, – 5, and 4. Standardized Test Practice The correct answer is A. ANSWER = (x – 3)(x + 5)(x – 4) = (x – 3)(x 2 + x – 20)

2. EXAMPLE 1 List possible rational zeros List the possible rational zeros of f using the rational zero theorem. a. f (x) = x 3 + 2x 2 – 11x + 12 Factors of the constant term: + 1, + 2, + 3, + 4, + 6, + 12 Factors of the leading coefficient: + 1 Possible rational zeros: +, +, +, +, +, + 1 1 2 1 3 1 4 1 6 1 12 1 Simplified list of possible zeros: + 1, + 2, + 3, + 4, + 6, + 12

3. EXAMPLE 1 List possible rational zeros b. f (x) = 4x 4 – x 3 – 3x 2 + 9x – 10 Factors of the constant term: + 1, + 2, + 5, + 10 Factors of the leading coefficient: + 1, + 2, + 4 +, +, +, +, +, +, +, +, +, +, + Possible rational zeros: 1 1 2 1 5 1 10 1 1 2 2 2 5 2 2 1 4 2 4 5 4 + 10 4 Simplified list of possible zeros: + 1, + 2, + 5, + 10, +, +, + 5 2 1 4 1 2 5 4 +

4. EXAMPLE 2 Find all real zeros of f (x) = x 3 – 8x 2 +11x + 20. SOLUTION List the possible rational zeros. The leading coefficient is 1 and the constant term is 20. So, the possible rational zeros are: x = +, +, +, +, +, + 5 1 4 1 2 1 1 1 10 1 20 1 STEP 1 Find zeros when the leading coefficient is 1

5. EXAMPLE 2 STEP 2 Find zeros when the leading coefficient is 1 1 1 – 8 11 20 Test x =1 : 1 – 7 4 1 – 7 4 24 Test x = –1 : –1 1 –8 11 20 1 – 9 20 0 –1 9 20 1 is not a zero. ↑ –1 is a zero ↑ Test these zeros using synthetic division.

6. EXAMPLE 2 Because –1 is a zero of f, you can write f (x) = (x + 1)(x 2 – 9x + 20). STEP 3 f (x) = (x + 1) (x 2 – 9x + 20) Factor the trinomial in f (x) and use the factor theorem. The zeros of f are –1, 4, and 5. ANSWER = (x + 1)(x – 4)(x – 5) Find zeros when the leading coefficient is 1

7. EXAMPLE 1 Find the number of solutions or zeros a. How many solutions does the equation x 3 + 5x 2 + 4x + 20 = 0 have? SOLUTION Because x 3 + 5x 2 + 4x + 20 = 0 is a polynomial equation of degree 3,it has three solutions. (The solutions are – 5, – 2i, and 2i.)

8. EXAMPLE 1 Find the number of solutions or zeros b. How many zeros does the function f (x) = x 4 – 8x 3 + 18x 2 – 27 have? SOLUTION Because f (x) = x 4 – 8x 3 + 18x 2 – 27 is a polynomial function of degree 4, it has four zeros. (The zeros are – 1, 3, 3, and 3.)

9. EXAMPLE 2 Find all zeros of f (x) = x 5 – 4x 4 + 4x 3 + 10x 2 – 13x – 14. SOLUTION STEP 1 Find the rational zeros of f. Because f is a polynomial function of degree 5, it has 5 zeros. The possible rational zeros are + 1, + 2, + 7, and + 14. Using synthetic division, you can determine that – 1 is a zero repeated twice and 2 is also a zero. STEP 2Write f (x) in factored form. Dividing f (x) by its known factors x + 1, x + 1, and x – 2 gives a quotient of x 2 – 4x + 7. Therefore: f (x) = (x + 1) 2 (x – 2)(x 2 – 4x + 7) Find the zeros of a polynomial function

10. EXAMPLE 2 STEP 3 Find the complex zeros of f. Use the quadratic formula to factor the trinomial into linear factors. f(x) = (x + 1) 2 (x – 2) x – (2 + i 3 ) x – (2 – i 3 ) The zeros of f are – 1, – 1, 2, 2 + i 3, and 2 – i 3. ANSWER Find the zeros of a polynomial function

11. EXAMPLE 4 Use Descartes’ rule of signs Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for f (x) = x 6 – 2x 5 + 3x 4 – 10x 3 – 6x 2 – 8x – 8. The coefficients in f (x) have 3 sign changes, so f has 3 or 1 positive real zero(s). f ( – x) = ( – x) 6 – 2( – x) 5 + 3( – x) 4 – 10( – x) 3 – 6( – x) 2 – 8( – x) – 8 = x 6 + 2x 5 + 3x 4 + 10x 3 – 6x 2 + 8x – 8 SOLUTION

12. EXAMPLE 4 Use Descartes’ rule of signs The coefficients in f (– x) have 3 sign changes, so f has 3 or 1 negative real zero(s). The possible numbers of zeros for f are summarized in the table below.

13. EXAMPLE 5 Approximate real zeros Approximate the real zeros of f (x) = x 6 – 2x 5 + 3x 4 – 10x 3 – 6x 2 – 8x – 8. SOLUTION Use the zero (or root) feature of a graphing calculator, as shown below. From these screens, you can see that the zeros are x ≈ – 0.73 and x ≈ 2.73. ANSWER

14. EXAMPLE 6 Approximate real zeros of a polynomial model s (x) = 0.00547x 3 – 0.225x 2 + 3.62x – 11.0 What is the tachometer reading when the boat travels 15 miles per hour? A tachometer measures the speed (in revolutions per minute, or RPMs ) at which an engine shaft rotates. For a certain boat, the speed x of the engine shaft (in 100s of RPMs ) and the speed s of the boat (in miles per hour) are modeled by TACHOMETER

15. EXAMPLE 6 Approximate real zeros of a polynomial model Substitute 15 for s(x) in the given function. You can rewrite the resulting equation as: 0 = 0.00547x 3 – 0.225x 2 + 3.62x – 26.0 Then, use a graphing calculator to approximate the real zeros of f (x) = 0.00547x 3 – 0.225x 2 + 3.62x – 26.0. SOLUTION From the graph, there is one real zero: x ≈ 19.9. The tachometer reading is about 1990 RPMs.ANSWER