Hawkes Learning Systems: College Algebra

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Hawkes Learning Systems:College Algebra

Section 5.3: Locating Real Zeros of Polynomials




Objectives

o The Rational Zero Theorem.

o Descartes’ Rule of Signs.

o Bounds of real zeros.

o The Intermediate Value Theorem.




The Rational Zero Theorem

o It can be proven (using ideas from a branch of mathematics called abstract algebra) that there is no formula based on elementary mathematical operations that identifies all the zeros of an arbitrary polynomial of degree five or higher.

o However, there are some tools that give us hints about where to look for zeros of a given polynomial. One such tool is the Rational Zero Theorem.





If is a

polynomial with integer coefficients, then any

rational zero of must be of the form , where

. is a factor of the constant term and is a

factor of the leading coefficient .

11 1 0...n n

n nf x a x a x a x a

fpq

p 0a q

na





CautionThe Rational Zero Theorem does not necessarily find even a single zero of a polynomial; instead it identifies a list of rational numbers that could potentially be zeros. In order to determine if any of the potential zeros actually are zeros, we have to test them. Additionally, the theorem says nothing about irrational zeros or complex zeros. If a given polynomial has some zeros that are either irrational or else have a non-zero imaginary part, we must resort to other means to find them.




Example 1: Factoring Polynomials

For the following polynomial, list all of the potential rational zeros. Then write the polynomial in factored form and identify the actual zeros. 3 22 9 17 6f x x x x Step 1: List the factors of the constant term, , and the leading coefficient, .

62

0Factors of : 1,2,3,6a 3Factors of : 1,2a

Step 2: By the Rational Zero Theorem, any rational zero must be one of the numbers generated by dividing factors of a0 by a3.

1 3Rational Zeros: 1,2,3,6, ,

2 2

Possible




Example 1: Factoring Polynomials (Cont.)

Step 3: Use synthetic division to identify which of the potential zeros actually are zeros. You may have to try several potential zeros before finding an actual zero.

2 9 17 1 6

2211

116

60

Step 4: Once we determine that is a zero, then we can finish by factoring

.

1

22 11 6x x

22 11 6 1f x x x x

2 1 6 1x x x

1Actual Zeros: ,6, 12





For the following polynomial, list all of the potential rational zeros. Then write the polynomial in factored form and identify the actual zeros.

4 3 24 37 2 93 54g x x x x x

0Factors of : 1,2,3,6,9,18,27,54a

4Factors of : 1,2,4a

1 3 9 27 1 3 9 27Possible Rational Zeros: 1, 2,3,6,9,18,27,54, , , , , , , ,

2 2 2 2 4 4 4 4





3 4 37 2 93 544

43

40

3028

2172

54

02

48

326436

720

2 34 32 36 24

g x x x x x

34 9 1 24

g x x x x x

3Actual Zeros: 9,1, 2,

4

4 3 24 37 2 93 54g x x x x x

Use synthetic division to find the actual zeros.




Descartes’ Rule of Signs

Let be a polynomial with real coefficients, and assume . A variation in sign of f is a change in the sign of one coefficient of f to the next, either from positive to negative or vice versa. 1. The number of positive real zeros of f is either the

number of variations in sign of or is less than this by a positive even integer.

2. The number of negative real zeros of f is either the number of variations in sign of or is less than this by a positive even integer.

11 1 0...n n

n nf x a x a x a x a

0 0a

f x

f x




Descartes’ Rule of Signs

o Note that in order to apply Descartes’ Rule of Signs, it is critical to first write the terms of the polynomial in descending order.

o Note also that unless the number of variations in sign is 0 or 1, the rule does not give us a definitive answer for the number of zeros to expect.

o For example, if the number of variations in sign of . is , we know only that there will be positive real zeros. f x 4 4, 2, 0or




Example 3: Using Descartes’ Rule of Signs

Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros of each of the following polynomials. Then use the Rational Zeros Theorem and other means to find the zeros, if possible.

3 26 3 10f x x x x

Sign change

Sign change

There are two variations in sign in the function , so we know there are 2 or 0 positive real zeros.

f x

3 26 3 10f x x x x 3 26 3 10x x x

Sign change

There is only one variation in sign in the function . so we know there is exactly 1 negative real zero.

f x




Example 3: Using Descartes’ Rule of Signs (Cont.)

3 26 3 10f x x x x

0Factors of : 1, 2,5,10a

Factors of : 1na

Possible Rational Zeros: 1, 2,5,10Since we know there is exactly 1 negative real zero, use synthetic division to find this first.

1 1 6 3 10

117

710

100

21 7 10f x x x x

1 5 2x x x

Zeros : 1,5,2




Example 4: Using Descartes’ Rule of Signs

Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros of the following polynomial. Then use the Rational Zeros Theorem and other means to find the zeros, if possible.

3 22 6 10 30f x x x x Sign change

There is exactly 1 positive real zero.

3 22 6 10 30f x x x x

3 22 6 10 30f x x x x

Sign change

Sign change

There are either 2 or 0 negative real zeros.




Example 4: Using Descartes’ Rule of Signs (Cont.)

3 22 6 10 30f x x x x

0Factors of : 1,2,3,5,6,10,15,30a

Factors of : 1,2na

1 3 5 15Possible Rational Zeros: 1, 2,3,5,6,10,15,30, , , ,

2 2 2 2

3 2 6 10 30

260

010

300

23 2 10f x x x

2 3 5 5x x x

Zeros: 3, 5, 5




Upper and Lower Bounds of Zeros

Assume is a polynomial with real coefficients, a positive leading coefficient, and degree . Let and be fixed numbers, with . Then: 1. No real zero of is larger than (we say is an upper bound

of the zeros of ) if the last row in the synthetic division of . by contains no negative numbers. That is, is an upper bound of the zeros if the quotient and remainder have no negative coefficients when is divided by

2. No real zero of is smaller than (we say is a lower bound of the zeros of ) if the last row in the synthetic division of . by has entries that alternate in sign ( can be counted as either positive or negative, as necessary).

f x1 a b

0a b f b bf

f x b

f x .x b

x b

f a af

f x x a 0




Example 5: Locating Bounds With Synthetic Division

This example revisits the polynomial from Example 3. Use synthetic division to identify upper and lower bounds of the real zeros of the following polynomial.

f x

3 26 3 10f x x x x

5 1 6 3 10

151

52

100

We can begin by trying out any positive number whatsoever as an upper bound. Since there are negative numbers in the last row, 5 is not an upper bound.

6 1 6 3 10

160

03

1828

The number 6 is an upper bound according to the theorem, as all of the coefficients in the last row are non-negative.




Example 5: Locating Bounds With Synthetic Division

1 1 6 3 10

117

710

100

We find that is a lower bound, as the signs in the last row alternate. Note: 0 can be positive or negative, as necessary.

1

To summarize, we know that all real zeros of f lie in the interval , meaning there is no need to test the numbers out of the list of potential rational zeros . We have already determined in Example 3 that the zeros of are , all of which do indeed lie in the interval .

1,6 2, 5, 10,10 1,2,5,10

f 1,5,2 1,6




Upper and Lower Bounds of Zeros

CautionDon’t read more into the Upper and Lower Bounds Theorem than is actually there, as the theorem is not always powerful enough to indicate the numbers that are the best upper or lower bounds. The tradeoff for this weakness in the theorem is that it is quickly and easily applied.




The Intermediate Value Theorem

o The last technique for locating zeros that we study makes use of a property of polynomials called continuity.

o One consequence of continuity is that the graph of a continuous function has no “breaks” in it.

o This means that, assuming that the function can be graphed at all, it can be drawn without lifting the drawing tool.




The Intermediate Value Theorem

Assume that is a polynomial with real coefficients, and that and are real numbers with . If and differ in sign, then there is at least one point such that and . That is, at least one zero of lies between and .

f xa b

a b f a f bc

a c b 0f c f a b




Example 6: Approximating Zeros

Show that has a zero between and . 35 9f x x x 1 2

1 5 0 1 9

555

56

63

We could determine and by direct computation, but for many polynomials, it is quicker to use synthetic division as illustrated. Remember that the remainder upon dividing by is thus and . The critical point is that these two values are opposite in sign, so a zero of must lie between and .

2 5 0 1 9

51010

2021

4233

1f 2f

f x x k ;f k 1 3f 2 33f

f 1 2




Example 7: Approximating Zeros

Find an approximation (to the nearest tenth) of the zero between 1 and 2 for the function from Example 6, . 35 9f x x x

1.1 5 0 1 9

55.55.5

6.057.05

7.7551.245

1.2 5 0 1 9

566

7.28.2

9.840.84

Since the result of is negative and the result of

is positive, we know that the actual zero must lie between and .

1.15 5 0 1 9

55.755.75

6.61257.6125

8.7543750.245625

The value of midway between and is negative, so the zero must lie between and . We have shown that the zero, to the nearest tenth, is 1.2.

1.1f

1.2f

1.1 1.2

f1.1 1.2

1.15 1.2

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Hawkes Learning Systems: College Algebra