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Chapter 3

Polynomial Functions

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Chapter 3: Polynomial Functions

3.1 Complex Numbers

3.2 Quadratic Functions and Graphs

3.3 Quadratic Equations and Inequalities

3.4 Further Applications of Quadratic Functions and Models

3.5 Higher-Degree Polynomial Functions and Graphs

3.6 Topics in the Theory of Polynomial Functions (I)

3.7 Topics in the Theory of Polynomial Functions (II)

3.8 Polynomial Equations and Inequalities; Further Applications and Models

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3.7 Topics in the Theory of Polynomial Functions (II)

Complex Zeros and the Fundamental Theorem of Algebra

It can be shown that if a + bi is a zero of a polynomial function with real coefficients, then so is its complex conjugate, a – bi.

Conjugate Zeros Theorem

If P(x) defines a polynomial function having only real coefficients, and if a + bi is a zero of P(x), then the conjugate a – bi is also a zero of P(x).

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3.7 Topics in the Theory of Polynomial Functions (II)

Example Find a polynomial having zeros 3 and 2 + i that

satisfies the requirement P(–2) = 4.

Solution Since 2 + i is a zero, so is 2 – i. A general solution is

Since P(–2) = 4, we have

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3.7 Fundamental Theorem of Algebra

If P(x) is a polynomial of degree 1 or more, then there is some number k such that P(k) = 0. In other words,

where Q(x) can also be factored resulting in

Fundamental Theorem of Algebra

Every function defined by a polynomial of degree 1 or more has at least one complex (real or imaginary) zero.

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3.7 Zeros of a Polynomial Function

Example

Find all complex zeros of

given that 1 – i is a zero.

Solution

Number of Zeros Theorem

A function defined by a polynomial of degree n has at most n distinct (unique) complex zeros.

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3.7 Zeros of a Polynomial Function

Using the Conjugate Zeros Theorem, 1 + i is also a zero.

The zeros of x² – 5x + 6 are 2 and 3. Thus,

and has four zeros: 1 – i, 1 + i, 2, and 3.

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3.7 Multiplicity of a Zero

The multiplicity of the zero refers to the number of times a zero appears.

e.g.

– x = 0 leads to a single zero

– (x + 2)² leads to a zero of –2 with multiplicity two

– (x – 1)³ leads to a zero of 1 with multiplicity three

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3.7 Polynomial Function Satisfying Given Conditions

Example Find a polynomial function with real coefficients

of lowest possible degree having a zero 2 of multiplicity 3, a

zero 0 of multiplicity 2, and a zero i of single multiplicity.

Solution By the conjugate zeros theorem, –i is also a zero.

This is one of many such functions. Multiplying P(x) by any

nonzero number will yield another function satisfying these

conditions.

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3.7 Multiplicity of Zeros

Observe the behavior around the zeros of the polynomials

The following figure illustrates some conclusions.

By observing the dominating term and noting the parity of

multiplicities of zeros of a polynomial in factored form, we

can draw a rough graph of a polynomial by hand.

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3.7 Sketching a Graph of a Polynomial � Function by Hand

Example

Solution The dominating term is –2x⁵, so the end behavior

will rise on the left and fall on the right. Because –4 and 1 are

x-intercepts determined by zeros of even multiplicity, the

graph will be tangent to the x-axis at these x-intercepts. The

y-intercept is –96.

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3.7 The Rational Zeros Theorem

Example

List all possible rational zeros.
Use a graph to eliminate some of the possible zeros listed in part (a).
Find all rational zeros and factor P(x).

The Rational Zeros Theorem

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3.7 The Rational Zeros Theorem

Solution

(a)

From the graph, the zeros

are no less than –2 and no

greater than 1. Also, –1 is

clearly not a zero since the

graph does not intersect the

x-axis at the point (-1,0).

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3.7 The Rational Zeros Theorem

(c) Show that 1 and –2 are zeros.

Solving the equation 6x²+ x – 1 = 0, we get x = –1/2, 1/3.

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3.7 Descartes’ Rule of Sign

Descartes’ Rule of Signs

Let P(x) define a polynomial function with real coefficients

and a nonzero constant term, with terms in descending

powers of x.

The number of positive real zeros either equals the number of variations in sign occurring in the coefficients of P(x) or is less than the number of variations by a positive even integer.

(b) The number of negative real zeros either equals the number of variations in sign occurring in the coefficients of P(−x) or is less than the number of variations by a positive even integer.

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3.7 Applying Descartes’ Rule of Signs

Example Determine the possible number of positive real zeros and negative real zeros of P(x) = x⁴ – 6x³ + 8x² + 2x – 1.

We first consider the possible number of positive zeros by observing that P(x) has three variations in signs.

+ x⁴ – 6x³ + 8x² + 2x – 1

Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive real zeros.

For negative zeros, consider the variations in signs for P(−x).

P(−x) = (−x)⁴ – 6(−x)³ + 8(−x)² + 2(−x) − 1

= x⁴ + 6x³ + 8x² – 2x – 1

Since there is only one variation in sign, P(x) has only one negative real root.

1

2

3

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3.7 Boundedness Theorem

Boundedness Theorem

Let P(x) define a polynomial function of degree n ≥ 1 with

real coefficients and with a positive leading coefficient. If

P(x) is divided synthetically by x – c, and

if c > 0 and all numbers in the bottom row of the synthetic division are nonnegative, then P(x) has no zero greater than c;

(b) if c < 0 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then P(x) has no zero less than c.

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3.7 Using the Boundedness Theorem

Example Show that the real zeros of P(x) = 2x⁴ – 5x³ + 3x + 1 satisfy the following conditions.

(a) No real zero is greater than 3.

(b) No real zero is less than –1.

Solution

a) c > 0

b) c < 0

All are nonegative.

The numbers alternate in sign.

No real zero greater than 3.

No zero less than −1.

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