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Polynomial And Rational Functions
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2. 3.4 Real Zeros Of Polynomials
Copyright © Cengage Learning. All rights reserved.

3. Objectives Rational Zeros of Polynomials
Descartes’ Rule of Signs and Upper and Lower Bounds for Roots
Using Algebra and Graphing Devices to Solve Polynomial Equations

4. Real Zeros Of Polynomials
The Factor Theorem tells us that finding the zeros of a polynomial is really the same thing as factoring it into linear factors. In this section we study some algebraic methods that help us to find the real zeros of a polynomial and thereby factor the polynomial. We begin with the rational zeros of a polynomial.

5. Rational Zeros of Polynomials

6. Rational Zeros of Polynomials
To help us understand the next theorem, let’s consider the polynomial P (x) = (x – 2)(x – 3)(x – 4) = x3 – x2 – 14x From the factored form we see that the zeros of P are 2, 3, and –4. When the polynomial is expanded, the constant 24 is obtained by multiplying (–2)  (–3)  4. This means that the zeros of the polynomial are all factors of the constant term.
Factored form
Expanded form

7. Rational Zeros of Polynomials
The following generalizes this observation We see from the Rational Zeros Theorem that if the leading coefficient is 1 or –1, then the rational zeros must be factors of the constant term.

8. Example 1 – Using the Rational Zeros Theorem
Find the rational zeros of P (x) = x3 – 3x + 2.
Solution:
Since the leading coefficient is 1, any rational zero must be a divisor of the constant term 2.
So the possible rational zeros are 1 and 2. We test each of these possibilities.
P (1) = (1)3 – 3(1) + 2
= 0

9. Example 1 – Solution P (–1) = (–1)3 – 3(–1) + 2
cont’d
P (–1) = (–1)3 – 3(–1) + 2
= 4
P (2) = (2)3 – 3(2) + 2
P (–2) = (–2)3 – 3(–2) + 2
= 0
The rational zeros of P are 1 and –2.

10. Rational Zeros of Polynomials
The following box explains how we use the Rational Zeros Theorem with synthetic division to factor a polynomial.

11. Example 2 – Finding Rational Zeros
Factor the polynomial P(x) = 2x3 + x2 – 13x + 6, and find all its zeros.
Solution:
By the Rational Zeros Theorem the rational zeros of P are of the form
The constant term is 6 and the leading coefficient is 2, so

12. Example 2 – Solution
cont’d
The factors of 6 are 1, 2, 3, 6, and the factors of 2 are 1, 2. Thus, the possible rational zeros of P are
Simplifying the fractions and eliminating duplicates, we get the following list of possible rational zeros:

13. Example 2 – Solution
cont’d
To check which of these possible zeros actually are zeros, we need to evaluate P at each of these numbers. An efficient way to do this is to use synthetic division.

14. Example 2 – Solution
cont’d
From the last synthetic division we see that 2 is a zero of P and that P factors as P(x) = 2x3 + x2 – 13x + 6
= (x – 2)(2x2 + 5x – 3)
= (x – 2)(2x – 1)(x + 3)
From the factored form we see that the zeros of P are
2, , and –3.
Given polynomial
From synthetic division
Factor 2x2 + 5x – 3

15. Descartes’ Rule of Signs and Upper and Lower Bounds for Roots

16. Descartes’ Rule of Signs and Upper and Lower Bounds for Roots
To describe this rule, we need the concept of variation in sign. If P (x) is a polynomial with real coefficients, written with descending powers of x (and omitting powers with coefficient 0), then a variation in sign occurs whenever adjacent coefficients have opposite signs. For example, has three variations in sign.

17. Descartes’ Rule of Signs and Upper and Lower Bounds for Roots

18. Example 4 – Using Descartes’ Rule
Use Descartes’ Rule of Signs to determine the possible number of positive and negative real zeros of the polynomial P (x) = 3x6 + 4x5 + 3x3 – x – 3
Solution:
The polynomial has one variation in sign, so it has one positive zero. Now P (–x) = 3(–x)6 + 4(–x)5 + 3(–x)3 – (–x) – 3

19. Example 4 – Solution
cont’d
So P (–x) has three variations in sign. Thus, P(x) has either three or one negative zero(s), making a total of either two or four real zeros.

20. Descartes’ Rule of Signs and Upper and Lower Bounds for Roots
We say that a is a lower bound and b is an upper bound for the zeros of a polynomial if every real zero c of the polynomial satisfies a  c  b. The next theorem helps us to find such bounds for the zeros of a polynomial.

21. Example 5 – Upper and Lower Bounds for Zeros of a Polynomial
Show that all the real zeros of the polynomial P (x) = x4 – 3x2 + 2x – 5 lie between –3 and 2.
Solution:
We divide P (x) by x – 2 and x + 3 using synthetic division.

22. Example 5 – Solution
cont’d
By the Upper and Lower Bounds Theorem, –3 is a lower bound and 2 is an upper bound for the zeros.
Since neither –3 nor 2 is a zero (the remainders are not 0 in the division table), all the real zeros lie between these numbers.

23. Using Algebra and Graphing Devices to Solve Polynomial Equations

24. Example 7 – Solving a Fourth-Degree Equation Graphically
Find all real solutions of the following equation, rounded to the nearest tenth x4 + 4x3 – 7x2 – 2x – 3 = 0
Solution:
To solve the equation graphically, we graph P (x) = 3x4 + 4x3 – 7x2 – 2x – 3
First we use the Upper and Lower Bounds Theorem to find two numbers between which all the solutions must lie.

25. Example 7 – Solution
cont’d
This allows us to choose a viewing rectangle that is certain to contain all the x-intercepts of P. We use synthetic division and proceed by trial and error.
To find an upper bound, we try the whole numbers, 1, 2, 3, , as potential candidates. We see that 2 is an upper bound for the solutions.

26. Example 7 – Solution
cont’d
Now we look for a lower bound, trying the numbers –1, –2, and –3 as potential candidates. We see that –3 is a lower bound for the solutions.
Thus, all the solutions lie between –3 and 2.
So the viewing rectangle [–3, 2] by [–20, 20] contains all the x-intercepts of P.

27. Example 7 – Solution
cont’d
The graph in Figure 3 has two x-intercepts, one between –3 and –2 and the other between 1 and 2.
Zooming in, we find that the solutions of the equation, to the nearest tenth, are –2.3 and 1.3.
y = 3x4 + 4x3 – 7x2 – 2x – 3
Figure 3