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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)
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Topics in the Theory of Polynomial
Functions (I)
The Intermediate Value Theorem
If P(x) defines a polynomial function with only real
coefficients, and if, for real numbers a and b, the values P(a)
and P(b) are opposite in sign, then there exists at least one
real zero between a and b.
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Applying the Intermediate Value
Theorem
Example Show that the polynomial function defined by
P(x) = x3 − 2x2 − x + 1 has a real zero between 2 and 3.
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Applying the Intermediate Value
Theorem
Example Show that the polynomial function defined by
P(x) = x3 − 2x2 − x + 1 has a real zero between 2 and 3.
Analytic Solution
Evaluate P(2) and P(3).
P(2) = 23 − 2(2)2 − 2 + 1 = −1
P(3) = 33 − 2(3)2 − 3 + 1 = 7
Since P(2) = −1 and P(3) = 7 differ in sign, the intermediate
value theorem assures us that there is a real zero between
2 and 3.
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Applying the Intermediate Value
Theorem
We see that the zero lies between 2.246
and 2.247 since there is a sign change in
the function values.
Caution If P(a) and P(b) do not differ in
sign, it does NOT imply that there is no
zero between a and b.
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Division of Polynomials
Example Divide the polynomial 3x3 − 2x + 5 by x − 3.
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Division of Polynomials
Example Divide the polynomial 3x3 − 2x + 5 by x − 3.
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Division of Polynomials
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Division of Polynomials
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Synthetic Division
Example Divide the polynomial 3x3 − 2x + 5 by x − 3.
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Synthetic Division
This abbreviated form of long division is called synthetic
division.
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Using Synthetic Division
Example Use synthetic division to divide 5x3 − 6x2 − 28x + 8
by x + 2.
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Using Synthetic Division
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The Remainder Theorem
Remainder Theorem
If a polynomial P(x) is divided by x − k, the remainder is equal to
P(k).
Example Use the remainder theorem and synthetic division to
find P(−2) if P(k) = − x4 + 3x2 − 4x − 5.
Note: Can be solved by substitution and or by synthetic division
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k is Zero of a Polynomial Function if
P(k) = 0
Example Decide whether the given number is a zero of P.
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k is Zero of a Polynomial Function if
P(k) = 0
Example Decide whether the given number is a zero of P.
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k is Zero of a Polynomial Function if
P(k) = 0
Graphical Solution
Y1 = P(x) in part (a)
Y1(2) = P(2) in part (a)
Y2 = P(x) in part (b)
Y2 (−2) = P(−2) in part(b)
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The Factor Theorem
The Factor Theorem
The polynomial P(x) has a factor x − k if and only if P(k) = 0.
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Example using the Factor Theorem
Determine whether the second polynomial is a factor of the
first.
P(x) = 4x3 + 24x2 + 48x + 32; x + 2
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Example using the Factor Theorem
Determine whether the second polynomial is a factor of the
first.
P(x) = 4x3 + 24x2 + 48x + 32; x + 2
Solution Use synthetic division with k = −2.
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Relationships Among x-Intercepts,
Zeros, and Solutions
b) Graph P in a suitable viewing window and locate the xintercepts.
c) Solve the polynomial equation 2x3 + 5x2 − x − 6 = 0.
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Relationships Among x-Intercepts,
Zeros, and Solutions
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Relationships Among x-Intercepts,
Zeros, and Solutions
b) The calculator will determine the x-intercepts: −2, −1.5,
and 1.
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Topics in the Theory of Polynomial
Functions
Example Find a polynomial having zeros 3 and 2 + i that
satisfies the requirement P(−2) = 4.
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Topics in the Theory of Polynomial
Functions
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
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Zeros of a Polynomial Function
Number of Zeros Theorem
A function defined by a polynomial of degree n has at most n
distinct (unique) complex zeros.
Example
Find all complex zeros of P(x) = x4 − 7x3 + 18x2 − 22x + 12
given that 1 − i is a zero.
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Zeros of a Polynomial Function
Number of Zeros Theorem
A function defined by a polynomial of degree n has at most n
distinct (unique) complex zeros.
Example
Find all complex zeros of P(x) = x4 − 7x3 + 18x2 − 22x + 12
given that 1 − i is a zero.
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Zeros of a Polynomial Function
Using the Conjugate Zeros Theorem, 1 + i is also a zero.
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Multiplicity of a Zero
• The multiplicity of the zero refers to the number of times
a zero appears.
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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.
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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.
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Rough Sketch using Multiplicities
and End Behavior
Draw a rough sketch of the polynomial
The following figure illustrates some conclusions.
a.
P(x) = (x + 3)2(x − 2)2
c.
P(x) = -x2(x − 1)(x + 2)2
b.
P(x) = (x + 3)2(x − 2)3
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Sketching a Graph of a Polynomial
Function by Hand
Example Sketch P(x) = −2(x + 4)2(x + 3)(x − 1)2 by hand.
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Sketching a Graph of a Polynomial
Function by Hand
Example Sketch P(x) = −2(x + 4)2(x + 3)(x − 1)2 by hand.
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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The Rational Zeros Theorem
Example P(x) = 6x4 + 7x3 − 12x2 − 3x + 2
a) List all possible rational zeros.
b) Use a graph to eliminate some of the possible zeros
listed in part (a).
c) Find all rational zeros and factor P(x).
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The Rational Zeros Theorem
b) 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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The Rational Zeros Theorem
c) Show that 1 and −2 are zeros.
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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.
a) 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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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.
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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
a) 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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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.
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