171S4.4p Theorems about Zeros of Polynomial Functions

1

November 01, 2012

Oct 25­1:05 PM

CHAPTER 4: Polynomial and Rational Functions4.1 Polynomial Functions and Models4.2 Graphing Polynomial Functions4.3 Polynomial Division; The Remainder and Factor Theorems4.4 Theorems about Zeros of Polynomial Functions4.5 Rational Functions4.6 Polynomial and Rational Inequalities

MAT 171 Precalculus AlgebraDr. Claude Moore

Cape Fear Community College

The following lesson is a brief discussion of and suggestions relative to studying Chapter 4.171Session4171Session4 ( Package file )To view this lesson, click the globe to the lower left.

Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx

Click the globe to the left and visit SAS Curriculum Pathways for interactive programs on Synthetic Division. Enter the User Name given by your instructor and use Quick Launch # 2300.

Oct 25­1:05 PM

4.4 Theorems about Zeros of Polynomial Functions

• Find a polynomial with specified zeros.• For a polynomial function with integer coefficients, • find the rational zeros and the other zeros, if possible.• Use Descartes’ rule of signs to find information • about the number of real zeros of a polynomial • function with real coefficients.

The Fundamental Theorem of Algebra

Every polynomial function of degree

n, with n ≥ 1, has at least one zero in the system of complex numbers.

Oct 25­1:05 PM

The Fundamental Theorem of AlgebraExample: Find a polynomial function of degree 4 having

zeros 1, 2, 4i, and ­4i.

Solution: Such a polynomial has factors (x − 1),(x − 2), (x − 4i), and (x + 4i), so we have:

Let an = 1:

Oct 25­1:05 PM

Zeros of Polynomial Functions with Real Coefficients

Nonreal Zeros: If a complex number a + bi, b ≠ 0, is a zero of a polynomial function f(x) with real coefficients, then its conjugate, a − bi, is also a zero. (Nonreal zeros occur in conjugate pairs.)

Irrational Zeros: If where a, b, and c are rational and b is not a perfect square, is a zero of a polynomial function f(x) with rational coefficients, then its conjugate is also a zero.

Example

Suppose that a polynomial function of degree 6 with rational coefficients has ­3 + 2i, ­6i, and as three of its zeros. Find the other zeros.

Solution: The other zeros are the conjugates of the given zeros, ­3 ­ 2i, 6i, and There are no other zeros because the polynomial of degree 6 can have at most 6 zeros.

Oct 25­1:05 PM

Rational Zeros Theorem

Let the polynomial function be represented as

P(x) = anxn + an­1xn­1 + an­2xn­2 + ... + a1x + a0

where all the coefficients are integers.

Consider a rational number denoted by p/q, where p and q are relatively prime (having no common factor besides ­1 and 1). If p/q is a zero of P(x), then p is a factor of a0 and q is a factor of an.

Oct 25­1:05 PM

ExampleGiven f(x) = 2x3 − 3x2 − 11x + 6:a) Find the rational zeros and then the other zeros.b) Factor f(x) into linear factors.

Solution: a) Because the degree of f(x) is 3, there are at most 3

distinct zeros. The possibilities for p/q are:

Use synthetic division to help determine the zeros. It is easier to consider the integers before the fractions. We try 1:We try ­1:

Since f(1) = ­6, 1 is Since f(­1) = 12, ­1 is not a zero. not a zero.

171S4.4p Theorems about Zeros of Polynomial Functions

2

November 01, 2012

Oct 25­1:05 PM

Example continued

We try 3:

Since f(3) = 0, 3 is a zero. Thus x ­ 3 is a factor. Using the results of the division above, we can express f(x) as

f(x) = (x ­ 3)(2x2 + 3x ­ 2) .

We can further factor 2x2 + 3x ­ 2 as (2x ­ 1)(x + 2).

The rational zeros are −2, 3 and

The complete factorization of f(x) is:

f(x) = (2x ­ 1)(x ­ 3)(x + 2)

Oct 25­1:05 PM

Descartes’ Rule of Signs

Let P(x) be a polynomial function with real coefficients and a nonzero constant term.

The number of positive real zeros of P(x) is either:1. The same as the number of variations of sign in P(x), or2. Less than the number of variations of sign in P(x) by a positive even integer.

The number of negative real zeros of P(x) is either:3. The same as the number of variations of sign in P(­x), or4. Less than the number of variations of sign in P(­x) by a positive even integer.

A zero of multiplicity m must be counted m times.

Oct 25­1:05 PM

Example

What does Descartes’ rule of signs tell us about the number of positive real zeros and the number of negative real zeros?

There are two variations of sign, so there are either two or zero positive real zeros to the equation.

There are two variations of sign, so there are either two or zero negative real zeros to the equation.

Total Number of Zeros (or Roots) = 4:Possible number of zeros (or roots) by kind:Positive 2 2 0 0Negative 2 0 2 0Nonreal 0 2 2 4

Oct 25­1:26 PM

342/2. Find a polynomial function of degree 3 with the given numbers as zeros: ­1, 0, 4

342/4. Find a polynomial function of degree 3 with the given numbers as zeros: 2, i, ­i

Oct 25­1:26 PM

342/6. Find a polynomial function of degree 3 with

the given numbers as zeros: ­5, √3, ­ √3

342/8. Find a polynomial function of degree 3 with the given numbers as zeros: ­4, 1 ­ √5, 1 + √5

Oct 25­1:26 PM

342/14. Find a polynomial function of degree 4 with ­ 2 as a zero of multiplicity 1, 3 as a zero of multiplicity 2, and ­1 as a zero of multiplicity 1.

342/16. Find a polynomial function of degree 5 with ­1/2 as a zero of multiplicity 2, 0 as a zero of multiplicity 1, and 1 as a zero of multiplicity 2.

171S4.4p Theorems about Zeros of Polynomial Functions

3

November 01, 2012

Oct 25­1:26 PM

342/14. Find a polynomial function of degree 4 with ­ 2 as a zero of multiplicity 1, 3 as a zero of multiplicity 2, and ­1 as a zero of multiplicity 1.

342/16. Find a polynomial function of degree 5 with ­1/2 as a zero of multiplicity 2, 0 as a zero of multiplicity 1, and 1 as a zero of multiplicity 2.

Oct 25­1:26 PM

343/24. Suppose that a polynomial function of degree 4 with rational coefficients has the given numbers as zeros. Find the other zero(s): 6 ­ 5i, ­1 + √7

n = 4 means that the polynomial function has 4 roots (zeros) when we include complex solutions (roots or zeros).

x = 6 ­ 5i and ­1 + √7 are given as two roots (zeros).Since x = 6 ­ 5i is a root, we know that the conjugate x = 6 + 5i is a root.Since x = ­1 + √7 is a root, we know that the conjugate x = ­1 ­ √7 is a root.Therefore, the four (4) roots (zeros) are

x = 6 ­ 5i, 6 + 5i, x = ­1 + √7, and x = ­1 ­ √7.

Oct 25­1:26 PM

343/26. Suppose that a polynomial function of degree 5 with rational coefficients has the given numbers as zeros. Find the other zero(s): 3/4, ­ √3, 2i

n = 5 means that the polynomial function has 5 roots (zeros) when we include complex solutions (roots or zeros).

x = 3/4, x = ­√3, and x = 2i are given as three roots (zeros).

Since x = ­√3 is a root, we know that the conjugate x = √3 is a root.Since x = 2i is a root, we know that the conjugate x = ­2i is a root.Therefore, the five (5) roots (zeros) are

x = 3/4, x = ­√3, x = ­√3, x = 2i, and x = ­2i.

Oct 25­1:26 PM

343/29. Suppose that a polynomial function of degree 5 with rational coefficients has the given numbers as zeros. Find the other zero(s): 6, ­3 + 4i, 4 ­ √5

n = 5 means that the polynomial function has 5 roots (zeros) when we include complex solutions (roots or zeros).

x = 6, x = ­3 + 4i, and x = 4 ­ √5 are given as three roots (zeros).Since x = ­3 + 4i is a root, we know that the conjugate x = ­3 ­ 4i is a root.

Since x = 4 ­ √5 is a root, we know that the conjugate x = 4 + √5 is a root.Therefore, the five (5) roots (zeros) are

x = 6, x = ­3 + 4i, x = ­3 ­ 4i, x = 4 ­ √5, and x = 4 + √5.

Oct 29­3:33 PM

343/36. Find a polynomial function of lowest degree with rational coefficients that has the given numbers as some of its zeros: ­5i

Find the polynomial function with lowest degree (smallest n) that has x = ­5i as a root (zero). The conjugate x = 5i is also a root (zero) of the function.Thus, we can write f(x) = (x + 5i )(x ­ 5i ) = x2 + 25.

Oct 25­1:26 PM

343/44. Given that the polynomial function has the given zero, find the other zeros: f(x) = x3 ­ x2 + x ­ 1; 1

171S4.4p Theorems about Zeros of Polynomial Functions

4

November 01, 2012

Oct 25­1:26 PM

343/46. Given that the polynomial function has the given zero, find the other zeros: f(x) = x4 ­ 16; 2i

Oct 25­1:26 PM

343/47. Given that the polynomial function has the given zero, find the other zeros:

f(x) = x3 ­ 6x2 + 13x ­ 20; 4

Oct 25­1:26 PM

343/48. Given that the polynomial function has the given zero, find the other zeros: f(x) = x3 ­ 8; 2

Oct 25­1:26 PM

343/50. List all possible rational zeros of the function: f(x) = x7 + 37x5 ­ 6x2 + 12

Oct 25­1:26 PM

343/52. List all possible rational zeros of the function: f(x) = 3x3 ­ x2 + 6x ­ 9

Oct 25­1:26 PM

343/53. List all possible rational zeros of the function: f(x) = 15x6 + 47x2 + 2

171S4.4p Theorems about Zeros of Polynomial Functions

5

November 01, 2012

Oct 25­1:26 PM

343/54. List all possible rational zeros of the function: f(x) = 10x25 + 3x17 ­ 35x + 6

Oct 25­1:26 PM

343/62. For each polynomial function: a) Find the rational zeros and then the other zeros; that is, solve f(x) = 0. b) Factor f(x) into linear factors.

f(x) = 2x3 + 7x2 + 2x ­ 8

Other roots are (­3 ­ √41) / 4 and (­3 + √41) / 4.

Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx

Oct 25­1:26 PM

343/64. For each polynomial function: a) Find the rational zeros and then the other zeros; that is, solve f(x) = 0. b) Factor f(x) into linear factors.

f(x) = 3x4 ­ 4x3 + x2 + 6x ­ 2

Divide 3x2 ­ 6x + 6 = 0 by ­3 to yield x2 ­ 2x + 2 = 0. This gives x = 1+ i and x = 1­ i to go with x = ­1 and x = 1/3.

Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx

Oct 25­1:26 PM

343/66. For each polynomial function: a) Find the rational zeros and then the other zeros; that is, solve f(x) = 0. b) Factor f(x) into linear factors.

f(x) = x4 + 5x3 ­ 27x2 + 31x ­ 10

Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx

Divide 32 + 8x ­ 5 = 0 by ­3 to yield x2 ­ 2x + 2 = 0. This gives x = 1+ i and x = 1­ i to go with x = ­1 and x = 1/3.

Oct 25­1:26 PM

343/71. For each polynomial function: a) Find the rational zeros and then the other zeros; that is, solve f(x) = 0. b) Factor f(x) into linear factors.

f(x) = (1/3)x3 ­ (1/2)x2 ­ (1/6)x + 1/6

Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx

The quadratic (1/3)x2 ­ (1/3)x ­ 1/3 = 0 multiplied by 3 yields an equivalent equation x2 ­ x ­ 1 = 0. This gives

x = (1+ √5) / 2 and x = (1­ √5) / 2 to go with x = 1/2.

Oct 25­1:26 PM

344/74. Find only the rational zeros of the function.

f(x) = x4 ­ 3x3 ­ 9x2 ­ 3x ­ 10

Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx

171S4.4p Theorems about Zeros of Polynomial Functions

6

November 01, 2012

Oct 25­1:26 PM

344/76. Find only the rational zeros of the function.

f(x) = 2x3 + 3x2 + 2x + 3

Use this Excel program to calculate synthetic division problems. cfcc.edu/mathlab/synthetic­division.xlsx

Oct 25­1:26 PM

344/78. Find only the rational zeros of the function.

f(x) = x4 + 6x3 + 17x2 + 36x + 66

q ∈ ±1 p ∈ ±1, 2, 3, 6, 11, 22, 33, 66There are no positive real roots by Descrates Rule of Signs. I tried all 16 possible rational roots and found that none of them worked. So, f(x) has NO rational roots.

No Real Roots

Oct 25­1:26 PM

344/80. Find only the rational zeros of the function.

f(x) = x5 ­ 3x4 ­ 3x3 + 9x2 ­ 4x + 12

Oct 25­1:26 PM

344/84. What does Descartes rule of signs tell you about the number of positive real zeros and the number of negative real zeros of the function?

P(x) = ­3x5 ­ 7x3 ­ 4x ­ 5

Oct 25­1:26 PM

344/88. What does Descartes rule of signs tell you about the number of positive real zeros and the number of negative real zeros of the function?

g(z) = ­z10 + 8z7 + z3 + 6z ­ 1

Oct 25­1:26 PM

344/92. What does Descartes rule of signs tell you about the number of positive real zeros and the number of negative real zeros of the function?

Q(x) = x4 ­ 2x2 + 12x ­ 8

171S4.4p Theorems about Zeros of Polynomial Functions

7

November 01, 2012

Oct 25­1:26 PM

344/96. What does Descartes rule of signs tell you about the number of positive real zeros and the number of negative real zeros of the function?

f(x) = x4 ­ 9x2 ­ 6x + 4

Oct 25­1:26 PM

344/98. Sketch the graph of the polynomial function. Follow the procedure outlined on p. 317. Use the rational zeros theorem when finding the zeros. f(x) = 3x3 ­ 4x2 ­ 5x + 2

Oct 25­1:26 PM

344/100. Sketch the graph of the polynomial function. Follow the procedure outlined on p. 317. Use the rational zeros theorem when finding the zeros. f(x) = 4x4 ­ 37x2 + 9

Oct 25­1:26 PM

344/98. Sketch the graph of the polynomial function. Follow the procedure outlined on p. 317. Use the rational zeros theorem when finding the zeros. f(x) = 3x3 ­ 4x2 ­ 5x + 2

Oct 25­1:26 PM

344/100. Sketch the graph of the polynomial function. Follow the procedure outlined on p. 317. Use the rational zeros theorem when finding the zeros. f(x) = 4x4 ­ 37x2 + 9