TEKS

5.7
2A.2.A, 2A.8.B;
Apply the Fundamental
Theorem of Algebra
P.1.D, P.3.B

Before You found zeros using the rational zero theorem.

Now You will classify the zeros of polynomial functions.
Why? So you can determine boat speed, as in Example 6.

Key Vocabulary The equation x3 2 5x2 2 8x 1 48 5 0, which becomes (x 1 3)(x 2 4)2 5 0 when
• repeated solution factored, has only two distinct solutions: 23 and 4. Because the factor x 2 4
• irrational appears twice, however, you can count the solution 4 twice. So, with 4 counted as
conjugates, p. 267 a repeated solution, this third-degree equation has three solutions: 23, 4, and 4.
• complex conjugates, The previous result is generalized by the fundamental theorem of algebra, first
p. 278 proved by the German mathematician Karl Friedrich Gauss (1777–1855).

KEY CONCEPT For Your Notebook

The Fundamental Theorem of Algebra
Theorem: If f(x) is a polynomial of degree n where n > 0, then the equation
f (x) 5 0 has at least one solution in the set of complex numbers.
Corollary: If f (x) is a polynomial of degree n where n > 0, then the equation
f(x) 5 0 has exactly n solutions provided each solution repeated twice is
counted as 2 solutions, each solution repeated three times is counted as
3 solutions, and so on.

The corollary to the fundamental theorem of algebra also implies that an

nth-degree polynomial function f has exactly n zeros.

EXAMPLE 1 Find the number of solutions or zeros

a. How many solutions does the equation x 3 1 5x 2 1 4x 1 20 5 0 have?
b. How many zeros does the function f (x) 5 x4 2 8x 3 1 18x2 2 27 have?

Solution
a. Because x 3 1 5x 2 1 4x 1 20 5 0 is a polynomial equation of degree 3,
it has three solutions. (The solutions are 25, 22i, and 2i.)
b. Because f (x) 5 x4 2 8x 3 1 18x2 2 27 is a polynomial function of
degree 4, it has four zeros. (The zeros are 21, 3, 3, and 3.)

✓ GUIDED PRACTICE for Example 1

1. How many solutions does the equation x4 1 5x 2 2 36 5 0 have?

2. How many zeros does the function f(x) 5 x 3 1 7x2 1 8x 2 16 have?

5.7 Apply the Fundamental Theorem of Algebra 379

 EXAMPLE 2 Find the zeros of a polynomial function
Find all zeros of f (x) 5 x 5 2 4x 4 1 4x 3 1 10x 2 2 13x 2 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 61, 62, 67,
and 614. Using synthetic division, you can determine that 21 is a zero
repeated twice and 2 is also a zero.
STEP 2 Write f(x) in factored form. Dividing f (x) by its known factors x 1 1,
x 1 1, and x 2 2 gives a quotient of x2 2 4x 1 7. Therefore:
f(x) 5 (x 1 1)2 (x 2 2)(x2 2 4x 1 7)
STEP 3 Find the complex zeros of f. Use the quadratic formula to factor the
trinomial into linear factors.
} }
f(x) 5 (x 1 1)2(x 2 2) F x 2 (2 1 i Ï3 ) GF x 2 (2 2 i Ï 3 ) G
} }
c The zeros of f are 21, 21, 2, 2 1 i Ï 3 , and 2 2 i Ï3 .

BEHAVIOR NEAR ZEROS The graph of f in Example 2 is y

10
shown at the right. Note that only the real zeros appear
as x-intercepts. Also note that the graph is tangent to the (21, 0)
x-axis at the repeated zero x 5 21, but crosses the x-axis at 1 (2, 0) x
the zero x 5 2. This concept can be generalized as follows:
• When a factor x 2 k of a function f is raised to an odd
power, the graph of f crosses the x-axis at x 5 k.
• When a factor x 2 k of a function f is raised to an even
power, the graph of f is tangent to the x-axis at x 5 k.

✓ GUIDED PRACTICE for Example 2

Find all zeros of the polynomial function.

3. f (x) 5 x 3 1 7x2 1 15x 1 9 4. f (x) 5 x5 2 2x4 1 8x2 2 13x 1 6

}
REVIEW COMPLEX COMPLEX CONJUGATES Also in Example 2, notice that the zeros 2 1 i Ï 3 and
}
NUMBERS 2 2 i Ï3 are complex conjugates. This illustrates the first theorem given below.
For help with complex A similar result applies to irrational zeros of polynomial functions, as shown in
conjugates, see p. 278. the second theorem below.

KEY CONCEPT For Your Notebook

Complex Conjugates Theorem
If f is a polynomial function with real coefficients, and a 1 bi is an imaginary
zero of f, then a 2 bi is also a zero of f.

Irrational Conjugates Theorem

Suppose f is a polynomial function with rational coefficients, and a and b are
} } }
rational numbers such that Ï b is irrational. If a 1 Ïb is a zero of f, then a 2 Ïb
is also a zero of f.

380 Chapter 5 Polynomials and Polynomial Functions

 EXAMPLE 3 Use zeros to write a polynomial function
Write a polynomial function f of least degree that has rational coefficients, a
}
leading coefficient of 1, and 3 and 2 1 Ï5 as zeros.

Solution
} }
Because the coefficients are rational and 2 1 Ï5 is a zero, 2 2 Ï5 must also be
a zero by the irrational conjugates theorem. Use the three zeros and the factor
theorem to write f(x) as a product of three factors.
} }
f (x) 5 (x 2 3) F x 2 (2 1 Ï 5 ) G F x 2 (2 2 Ï 5 ) G Write f (x) in factored form.
} }
5 (x 2 3) F(x 2 2) 2 Ï 5 GF(x 2 2) 1 Ï5 G Regroup terms.

5 (x 2 3)[(x 2 2)2 2 5] Multiply.

2
5 (x 2 3)[(x 2 4x 1 4) 2 5] Expand binomial.
2
5 (x 2 3)(x 2 4x 2 1) Simplify.

5 x 3 2 4x2 2 x 2 3x2 1 12x 1 3 Multiply.

3 2
5 x 2 7x 1 11x 1 3 Combine like terms.

CHECK You can check this result by evaluating f (x) at each of its three zeros.
f (3) 5 33 2 7(3)2 1 11(3) 1 3 5 27 2 63 1 33 1 3 5 0 ✓
} } } }
f (2 1 Ï5 ) 5 (2 1 Ï5 ) 2 7(2 1 Ï5 ) 1 11(2 1 Ï 5 ) 1 3
3 2

} } }
5 38 1 17Ï 5 2 63 2 28Ï5 1 22 1 11Ï 5 1 3
50✓
} }
Since f (2 1 Ï5 ) 5 0, by the irrational conjugates theorem f (2 2 Ï5 ) 5 0. ✓

✓ GUIDED PRACTICE for Example 3

Write a polynomial function f of least degree that has rational coefficients, a

leading coefficient of 1, and the given zeros.
} }
5. 21, 2, 4 6. 4, 1 1 Ï 5 7. 2, 2i, 4 2 Ï 6 8. 3, 3 2 i

DESCARTES’ RULE OF SIGNS French mathematician René Descartes (1596–1650)

found the following relationship between the coefficients of a polynomial
function and the number of positive and negative zeros of the function.

KEY CONCEPT For Your Notebook

Descartes’ Rule of Signs
Let f (x) 5 anxn 1 an 2 1xn 2 1 1 . . . 1 a2x2 1 a1x 1 a0 be a polynomial function
with real coefficients.
• The number of positive real zeros of f is equal to the number of changes in
sign of the coefficients of f (x) or is less than this by an even number.
• The number of negative real zeros of f is equal to the number of changes in
sign of the coefficients of f (2x) or is less than this by an even number.

5.7 Apply the Fundamental Theorem of Algebra 381

 EXAMPLE 4 Use Descartes’ rule of signs
Determine the possible numbers of positive real zeros, negative real zeros, and
imaginary zeros for f(x) 5 x 6 2 2x 5 1 3x 4 2 10x 3 2 6x 2 2 8x 2 8.

Solution
f(x) 5 x6 2 2x5 1 3x4 2 10x3 2 6x2 2 8x 2 8

The coefficients in f (x) have 3 sign changes, so f has 3 or 1 positive real zero(s).
f(2x) 5 (2x) 6 2 2(2x) 5 1 3(2x)4 2 10(2x) 3 2 6(2x)2 2 8(2x) 2 8
5 x6 1 2x5 1 3x4 1 10x3 2 6x2 1 8x 2 8

The coefficients in f (2x) 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.

Positive Negative Imaginary Total

real zeros real zeros zeros zeros
3 3 0 6

3 1 2 6

1 3 2 6

1 1 4 6

✓ GUIDED PRACTICE for Example 4

Determine the possible numbers of positive real zeros, negative real zeros, and
imaginary zeros for the function.
9. f (x) 5 x 3 1 2x 2 11 10. g(x) 5 2x4 2 8x 3 1 6x 2 2 3x 1 1

APPROXIMATING ZEROS All of the zeros of the function in Example 4 are

irrational or imaginary. Irrational zeros can be approximated using technology.

EXAMPLE 5 Approximate real zeros

Approximate the real zeros of f (x) 5 x 6 2 2x 5 1 3x 4 2 10x 3 2 6x 2 2 8x 2 8.

Solution
ANOTHER WAY
Use the zero (or root) feature of a graphing calculator, as shown below.
In Example 5, you can
also approximate the
zeros of f using the
calculator’s trace feature.
However, this generally
gives less precise results
than the zero (or root)
Zero Zero
feature. X=-.7320508 Y=0 X=2.7320508 Y=0

c From these screens, you can see that the zeros are x ø 20.73 and x ø 2.73.

382 Chapter 5 Polynomials and Polynomial Functions

 EXAMPLE 6 Approximate real zeros of a polynomial model
TACHOMETER 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
s(x) 5 0.00547x 3 2 0.225x2 1 3.62x 2 11.0
What is the tachometer reading when the boat travels
15 miles per hour?

Solution
Substitute 15 for s(x) in the given function. You can rewrite
the resulting equation as:
0 5 0.00547x 3 2 0.225x2 1 3.62x 2 26.0
Then, use a graphing calculator to approximate the real
zeros of f(x) 5 0.00547x 3 2 0.225x2 1 3.62x 2 26.0.
Zero
From the graph, there is one real zero: x ø 19.9. X=19.863247 Y=0

c The tachometer reading is about 1990 RPMs.

✓ GUIDED PRACTICE for Examples 5 and 6

11. Approximate the real zeros of f(x) 5 3x5 1 2x4 2 8x 3 1 4x 2 2 x 2 1.

12. WHAT IF? In Example 6, what is the tachometer reading when the boat
travels 20 miles per hour?

5.7 EXERCISES HOMEWORK

KEY
5 WORKED-OUT SOLUTIONS
on p. WS1 for Exs. 15, 37, and 61
5 TAKS PRACTICE AND REASONING
Exs. 9, 33, 51, 52, 63, 64, 66, and 67

SKILL PRACTICE
1. VOCABULARY Copy and complete: For the equation (x 2 1)2 (x 1 2) 5 0,
a(n) ? solution is 1 because the factor x 2 1 appears twice.

2. ★ WRITING Explain the difference between complex conjugates and

WRITING
irrational conjugates.

EXAMPLE 1 NUMBER OF SOLUTIONS OR ZEROS Identify the number of solutions or zeros.

on p. 379
3. x4 1 2x 3 2 4x 2 1 x 2 10 5 0 4. 5y 3 2 3y 2 1 8y 5 0
for Exs. 3–9
5. 9t 6 2 14t 3 1 4t 2 1 5 0 6. f (z) 5 27z 4 1 z2 2 25
7. g(s) 5 12s 7 2 9s6 1 4s5 2 s 3 2 20s 1 50 8. h(x) 5 2x12 1 7x8 1 5x4 2 8x 1 6

9. ★ MULTIPLE CHOICE How many zeros does the function

TAKS REASONING
f (x) 5 16x 2 22x 3 1 6x6 1 19x5 2 3 have?
A 1 B 3 C 5 D 6

5.7 Apply the Fundamental Theorem of Algebra 383

EXAMPLE 2 FINDING ZEROS Find all zeros of the polynomial function.
on p. 380
10. f (x) 5 x4 2 6x 3 1 7x2 1 6x 2 8 11. f (x) 5 x4 1 5x 3 2 7x2 2 29x 1 30
for Exs. 10–19
12. g(x) 5 x4 2 9x2 2 4x 1 12 13. h(x) 5 x 3 1 5x2 2 4x 2 20
14. f (x) 5 x4 1 15x2 2 16 15. f (x) 5 x4 1 x 3 1 2x2 1 4x 2 8
16. h(x) 5 x4 1 4x 3 1 7x2 1 16x 1 12 17. g(x) 5 x4 2 2x 3 2 x2 2 2x 2 2
18. g(x) 5 4x4 1 4x 3 2 11x2 2 12x 2 3 19. h(x) 5 2x4 1 13x 3 1 19x2 2 10x 2 24

EXAMPLE 3 WRITING POLYNOMIAL FUNCTIONS Write a polynomial function f of least degree

on p. 381 that has rational coefficients, a leading coefficient of 1, and the given zeros.
for Exs. 20–32
20. 1, 2, 3 21. 22, 1, 3 22. 25, 21, 2 23. 23, 1, 6
24. 2, 2i, i 25. 3i, 2 2 i 26. 21, 2, 23i 27. 5, 5, 4 1 i
} } } } }
28. 4, 2Ï 5 , Ï 5 29. 24, 1, 2 2 Ï 6 30. 22, 21, 2, 3, Ï 11 31. 3, 4 1 2i, 1 1 Ï 7

32. ERROR ANALYSIS Describe and correct the

error in writing a polynomial function with f(x) 5 (x 2 2)[x 2 (1 1 i)]
rational coefficients and zeros 2 and 1 1 i.
5 x(x 2 1 2 i) 2 2(x 2 1 2 i)
33. ★ OPEN-ENDED MATH Write a polynomial
TAKS REASONING 5 x2 2 x 2 ix 2 2x 1 2 1 2i
function of degree 5 with zeros 1, 2, and 2i.
5 x2 2 (3 1 i)x 1 (2 1 2i)

EXAMPLE 4 CLASSIFYING ZEROS Determine the possible numbers of positive real zeros,
on p. 382 negative real zeros, and imaginary zeros for the function.
for Exs. 34–41
34. f (x) 5 x4 2 x2 2 6 35. g(x) 5 2x 3 1 5x2 1 12
36. g(x) 5 x 3 2 4x2 1 8x 1 7 37. h(x) 5 x5 2 2x 3 2 x2 1 6x 1 5
38. h(x) 5 x5 2 3x 3 1 8x 2 10 39. f (x) 5 x5 1 7x4 2 4x 3 2 3x2 1 9x 2 15
40. g(x) 5 x6 1 x5 2 3x4 1 x 3 1 5x2 1 9x 2 18 41. f (x) 5 x 7 1 4x4 2 10x 1 25

EXAMPLE 5 APPROXIMATING ZEROS Use a graphing calculator to graph the function. Then
on p. 382 use the zero (or root) feature to approximate the real zeros of the function.
for Exs. 42–49
42. f (x) 5 x 3 2 x2 2 8x 1 5 43. f (x) 5 2x4 2 4x2 1 x 1 8
44. g(x) 5 x 3 2 3x2 1 x 1 6 45. h(x) 5 x4 2 5x 2 3
46. h(x) 5 3x 3 2 x2 2 5x 1 3 47. g(x) 5 x4 2 x 3 1 2x2 2 6x 2 3
48. f (x) 5 2x6 1 x4 1 31x2 2 35 49. g(x) 5 x5 2 16x 3 2 3x2 1 42x 1 30

50. REASONING Two zeros of f(x) 5 x 3 2 6x2 2 16x 1 96 are 4 and 24. Explain
why the third zero must also be a real number.

51. ★ SHORT RESPONSE Describe the possible numbers of positive real, negative
TAKS REASONING
real, and imaginary zeros for a cubic function with rational coefficients.

52. ★ MULTIPLE CHOICE Which is not a possible classification of the zeros of

TAKS REASONING
f (x) 5 x5 2 4x3 1 6x2 1 12x 2 6 according to Descartes’ rule of signs?
A 3 positive real zeros, 2 negative real zeros, and 0 imaginary zeros
B 3 positive real zeros, 0 negative real zeros, and 2 imaginary zeros
C 1 positive real zero, 4 negative real zeros, and 0 imaginary zeros
D 1 positive real zero, 2 negative real zeros, and 2 imaginary zeros

5 WORKED-OUT SOLUTIONS 5 TAKS PRACTICE

384 Chapter 5 Polynomials
on p. WS1 and Polynomial Functions AND REASONING
 CLASSIFYING ZEROS Determine the numbers of positive real zeros, negative real
zeros, and imaginary zeros for the function with the given degree and graph.
Explain your reasoning.
53. Degree: 3 54. Degree: 4 55. Degree: 5

y y y

10 10 10
3 x 1 x 1 x

CHALLENGE Show that the given number is a zero of the given function but that
the conjugate of the number is not a zero.
56. f (x) 5 x 3 2 2x2 1 2x 1 5i; 2 2 i 57. g(x) 5 x 3 1 2x2 1 2i 2 2; 21 1 i

58. Explain why the results of Exercises 56 and 57 do not contradict the complex
conjugate theorem on page 380.

PROBLEM SOLVING
EXAMPLE 6 59. BUSINESS For the 12 years that a grocery store has been open, its annual
on p. 383 revenue R (in millions of dollars) can be modeled by the function
for Exs. 59–62
R 5 0.0001(2t 4 1 12t 3 2 77t 2 1 600t 1 13,650)
where t is the number of years since the store opened. In which year(s) was
the revenue $1.5 million?
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60. ENVIRONMENT From 1990 to 2003, the number N of inland

lakes in Michigan infested with zebra mussels can be
modeled by the function
N 5 20.028t 4 1 0.59t 3 2 2.5t 2 1 8.3t 2 2.5
where t is the number of years since 1990. In which year
did the number of infested inland lakes first reach 120?
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61. PHYSIOLOGY A study group found that a person’s score S on a step-climbing

exercise test was related to his or her amount of hemoglobin x (in grams per
100 milliliters of blood) by this function:
S 5 20.015x3 1 0.6x2 2 2.4x 1 19
Given that the normal range of hemoglobin is 12–18 grams per 100 milliliters of
blood, what is the most likely amount of hemoglobin for a person who scores 75?
62. POPULATION From 1890 to 2000, the American Indian, Eskimo, and Aleut
population P (in thousands) can be modeled by the function
P 5 0.0035t 3 2 0.235t 2 1 4.87t 1 243
where t is the number of years since 1890. In which year did the population
first reach 722,000?

5.7 Apply the Fundamental Theorem of Algebra 385

 63. ★ SHORT RESPONSE A 60-inch-long bookshelf is warped under 180 pounds of
TAKS REASONING
books. The deflection d of the bookshelf (in inches) is given by
d 5 (2.724 3 1027)x4 2 (3.269 3 1025)x 3 1 (9.806 3 1024)x2
where x is the distance (in inches) from the bookshelf’s left end. Approximate
the real zeros of the function on the domain 0 ≤ x ≤ 60. Explain why all your
answers make sense in this situation.

64. ★ EXTENDED RESPONSE You plan to save $1000 each year towards buying a
TAKS REASONING
used car in four years. At the end of each summer, you deposit $1000 earned
from summer jobs into your bank account. The table shows the value of your
deposits over the four year period. In the table, g is the growth factor 1 1 r
where r is the annual interest rate expressed as a decimal.

Year 1 Year 2 Year 3 Year 4

2
Value of 1st deposit 1000 1000g 1000g 1000g3

Value of 2nd deposit — 1000 ? ?

Value of 3rd deposit — — 1000 ?

Value of 4th deposit — — — 1000

a. Apply Copy and complete the table.

b. Model Write a polynomial function that gives the value v of your
account at the end of the fourth summer in terms of g.
c. Reasoning You want to buy a car that costs about $4300. What growth
factor do you need to obtain this amount? What annual interest rate do
you need? Explain how you found your answers.

65. CHALLENGE A monument with the dimensions shown X

is to be built using 1000 cubic feet of marble. What X
is the value of x?
FT
X
FT
FT X FT

TAKS PRACTICE at classzone.com

M IXED R EVIEW FOR TAKS
REVIEW 66. TAKS PRACTICE Which of the following is the solution of this system of
Lesson 3.2; linear equations? TAKS Obj. 4
TAKS Workbook
22x 1 3y 5 20
4x 1 4y 5 215

25 , 5
A 1 2}4 22
} B 1 }52 , 2}
25
4 2
C 25 , 35
1}2 4 2
} D No solution

REVIEW 67. TAKS PRACTICE What is the approximate 3 in.

TAKS Preparation volume of the bird feeder shown? TAKS Obj. 8
p. 608;
F 156 in.3 G 184 in.3 6.5 in.
TAKS Workbook
H 212 in.3 J 269 in.3
3 in.

386 Chapter 5EXTRA PRACTICE

Polynomials and Polynomial Functions
for Lesson 5.7, p. 1014 ONLINE QUIZ at classzone.com