please answer a minimum of 3 OF 6 question pairs. you must answer one question out of each pair with a minimum of one paragraph.
dream86Copyright © Cengage Learning. All rights reserved.
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Copyright © Cengage Learning. All rights reserved.
Remainder Theorem and Factor Theorem
SECTION 3.1
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Remainder Theorem and Factor Theorem
If P is a polynomial function, then the values of x for which
P(x) is equal to 0 are called the zeros of P.
For instance, –1 is a zero of P(x) = 2x3 – x + 1 because
P(–1) = 2(–1)3 – (–1) + 1
= –2 + 1 + 1
= 0
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Division of Polynomials
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Division of Polynomials
To divide a polynomial by a monomial, divide each term of
the polynomial by the monomial. For instance,
To divide a polynomial by a binomial, we use a method
similar to that used to divide natural numbers.
Divide each term in the numerator by the denominator.
Simplify.
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Division of Polynomials
For instance, consider (6x3 – 16x2 + 23x – 5) (3x – 2).
Think .
Multiply: 2x2(3x – 2) = 6x3 – 4x2
Subtract and bring down the next term, 23x.
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Division of Polynomials
Think .
Multiply: –4x(3x – 2) = –12x2 + 8x
Subtract and bring down the next term, –5.
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Division of Polynomials
Quotient
Dividend
Divisor
Subtract to produce the remainder, 5.
Multiply: 5(3x – 2) = 15x – 10.
Think .
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Division of Polynomials
The division process ends when the expression in the
bottom row is of lesser degree than the divisor. The expression in the bottom row is the remainder, and
the polynomial in the top row is the quotient.
Thus (6x3 – 16x2 + 23x – 5) (3x – 2) = 2x2 – 4x + 5
with a remainder of 5.
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Division of Polynomials
Although there is nothing wrong with writing the answer as
we did, it is more common to write the answer as the
quotient plus the remainder divided by the divisor.
Using this method, we write
Dividend
Quotient
Remainder
Divisor
Divisor
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Division of Polynomials
In every division, the dividend is equal to the product of the
divisor and quotient, plus the remainder. That is,
6x3 – 16x2 + 23x – 5 = (3x – 2) (2x2 – 4x + 5) + 5
The preceding polynomial division concepts are summarized by the following theorem.
Division Algorithm for Polynomials
Let P(x) and D(x) be polynomials, with D(x) of lower degree than P(x) and D(x) of degree 1 or more.
Dividend
Divisor
Quotient
Remainder
+
=
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Division of Polynomials
Then there exist unique polynomials Q(x) and R(x) such
that
P(x) = D(x) Q(x) + R(x)
where R(x) is either 0 or of degree less than the degree of D(x). The polynomial P(x) is called the dividend, D(x) is the divisor, Q(x) is the quotient, and R(x) is the remainder.
Before dividing polynomials, make sure that each polynomial is written in descending order.
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Division of Polynomials
It is helpful to insert a 0 in the dividend for a
missing term (one whose coefficient is 0) so that like terms
align in the same column. This is demonstrated in
Example 1.
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Example 1 – Divide Polynomials
Divide:
Solution:
Write the numerator in descending order. Then divide.
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Example 1 – Solution
Thus
cont’d
Inserting 0x3 for the missing term helps align
like terms in the same column.
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Synthetic Division
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Synthetic Division
A procedure called synthetic division can expedite the
division process.
To apply the synthetic division procedure, the divisor must be a polynomial of the form x – c, where c is a constant.
In the synthetic division procedure, the variables that occur in the polynomials are not listed.
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Synthetic Division
To understand how synthetic division is performed, examine the following long division on the left and the related synthetic division on the right.
Long Division Synthetic Division
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Synthetic Division
In the long division, the dividend is 4x3 – 5x2 + 2x – 10 and
the divisor is x – 2.
Because the divisor is of the form x – c, with c = 2, the division can be performed by the synthetic division procedure. Observe that in the above synthetic division:
1. The constant c is listed as the first number in the first row, followed by the coefficients of the dividend.
2. The first number in the third row is the leading coefficient of the dividend.
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Synthetic Division
3. Each number in the second row is determined by
computing the product of c and the number in the third row of the preceding column.
4. Each of the numbers in the third row, other than the first number, is determined by adding the numbers directly above it.
The following explanation illustrates the steps used to find the quotient and remainder of (2x3 – 8x + 7) (x + 3) using synthetic division.
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Synthetic Division
The divisor x + 3 is written in x – c form as x – (–3), which indicates that c = –3. The dividend 2x3 – 8x + 7 is missing an x2 term.
If we insert 0x2 for the missing term, the dividend becomes 2x3 + 0x2 – 8x + 7.
Write the constant c, –3, followed by the coefficients of the dividend. Bring down the first coefficient in the first row, 2, as the first number of the third row.
Coefficients of the dividend
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Synthetic Division
Multiply c times the first number in the third row, 2, to produce the first number of the second row, –6. Add the 0 and the –6 to produce the next number of the third row, –6.
Multiply c times the second number in the third row, –6, to produce the next number of the second row, 18. Add the –8 and the 18 to produce the next number of the third row, 10.
Multiply c times the third number in the third row, 10, to produce the next number of the second row, –30. Add the 7 and the –30 to produce the last number of the third row, –23.
Coefficients of
the quotient
Remainder
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Synthetic Division
The last number in the bottom row, –23, is the remainder.
The other numbers in the bottom row are the coefficients of
the quotient.
The quotient of a synthetic division always has a degree that is one less than the degree of the dividend. Thus the quotient in this example is 2x2 – 6x + 10.
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Synthetic Division
The results of the synthetic division can be expressed in fractional form as
or as
In Example 2, we illustrate the compact form of synthetic division, obtained by condensing the process explained here.
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Example 2 – Use Synthetic Division to Divide Polynomials
Use synthetic division to divide x4 – 4x2 + 7x + 15 by x + 4.
Solution:
Because the divisor is x + 4, we perform synthetic division with c = –4.
The quotient is x3 – 4x2 + 12x – 41, and the remainder is 179.
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Remainder Theorem
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Remainder Theorem
The following theorem shows that synthetic division can be used to determine the value P(c) for a given polynomial P(x) and constant c.
Remainder Theorem
If a polynomial P(x) is divided by x – c, then the remainder equals P(c).
In Example 3, we use synthetic division and the Remainder Theorem to evaluate a polynomial function.
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Example 3 – Use the Remainder Theorem to Evaluate a Polynomial Function
Let P(x) = 2x3 + 3x2 + 2x – 2. Use the Remainder Theorem
to find P(c) for c = –2 and c = .
Solution:
Perform synthetic division with c = –2 and c = and examine the remainders.
The remainder is –10. Therefore, P(–2) = –10.
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Example 3 – Solution
The remainder is 0. Therefore,
The points (–2, –10) and
are on the graph of P.
cont’d
P(x) = 2x3 + 3x2 + 2x – 2
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Remainder Theorem
Using the Remainder Theorem to evaluate a polynomial
function is often faster than evaluating the polynomial
function by direct substitution.
For instance, evaluating
P(x) = x5 – 10x4 + 35x3 – 50x2 + 24x by substituting 7 for x requires the following work.
P(7) = (7)5 – 10(7)4 + 35(7)3 – 50(7)2 + 24(7)
= 16,807 – 10(2401) + 35(343) – 50(49) + 24(7)
= 16,807 – 24,010 + 12,005 – 2450 + 168
= 2520
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Factor Theorem
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Factor Theorem
Note from Example 3 that . Recall that is a zero
of P because P(x) = 0 when x = .
The following theorem shows the important relationship between a zero of a given polynomial function and a factor of the polynomial.
Factor Theorem
A polynomial P(x) has a factor (x – c) if and only if P(c) = 0. That is, (x – c) is a factor of P(x) if and only if c is a zero of P.
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Example 4 – Apply the Factor Theorem
Use synthetic division and the Factor Theorem to determine whether (x + 5) or (x – 2) is a factor of
P(x) = x4 + x3 – 21x2 – x + 20.
Solution:
The remainder of 0 indicates that –5 is a zero of P and (x + 5) is a factor of P(x).
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Example 4 – Solution
The remainder of –42 indicates that (x – 2) is not a factor of
P(x).
cont’d
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Factor Theorem
Here is a summary of the important role played by the remainder in the division of a polynomial by (x – c).
Remainder of a Polynomial Division
In the division of the polynomial P(x) by (x – c), the remainder is
equal to P(c).
0 if and only if (x – c) is a factor of P(x).
0 if and only if c is a zero of P.
If c is a real number, then the remainder of P(x) (x – c)
is 0 if and only if (c, 0) is an x-intercept of the graph of P.
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Copyright © Cengage Learning. All rights reserved.
Polynomial Functions of Higher Degree
SECTION 3.2
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Polynomial Functions of Higher Degree
Table 3.1 summarizes information about graphs of polynomial functions of degree 0, 1, or 2.
In this section, we will focus on polynomial functions of degree 3 or higher.
Table 3.1
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Polynomial Functions of Higher Degree
These functions can be graphed by the technique of plotting points; however, some additional knowledge about polynomial functions will make graphing easier.
All polynomial functions have graphs that are smooth continuous curves.
The terms smooth and continuous are defined rigorously in calculus, but for the present, a smooth curve is a curve that does not have sharp corners, like the graph shown in Figure 3.5a.
Continuous, but not smooth
Figure 3.5(a)
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Polynomial Functions of Higher Degree
A continuous curve does not have a break or hole, like the
graph shown in Figure 3.5b.
Note
The general form of a
polynomial is given by
anxn + an – 1xn – 1 + . . . + a0
The coefficients an, an – 1, . . . , a0 are
all real numbers unless specifically
stated otherwise.
Not continuous
Figure 3.5(b)
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Far-Left and Far-Right Behavior
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Far-Left and Far-Right Behavior
The graph of a polynomial function may have several up and down fluctuations; however, the graph of every polynomial function eventually will increase or decrease without bound as | x | becomes larger.
The leading term anxn is said to be the dominant term of the polynomial function
P(x) = anxn + an – 1xn – 1 + . . . + a1x + a0
because, as | x | becomes larger, the absolute value of anxn will be much larger than the absolute value of any of the other terms.
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Far-Left and Far-Right Behavior
Because of this condition, you can determine the far-left and far-right behavior of the polynomial by examining the leading coefficient an and the degree n of the polynomial.
Note
The leading term of a polynomial function in x is the nonzero term that contains the largest power of x. The leading coefficient of a polynomial function is the coefficient of the leading term.
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Far-Left and Far-Right Behavior
Table 3.2 shows the far-left and far-right behavior of a polynomial function P with leading term anxn.
Table 3.2
The Leading Term Test
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Example 1 – Determine the Far-Left and Far-Right Behavior of a Polynomial Function
Examine the leading term to determine the far-left and far-right behavior of the graph of each polynomial function.
a. P(x) = x3 – x b. S(x) = x4 – x2 + 2
c. T(x) = –2x3 + x2 + 7x – 6 d. U(x) = 9 + 8x2 – x4
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Example 1(a) – Solution
Because an = 1 is positive and n = 3 is odd, the graph of P goes down to the far left and up to the far right.
See Figure 3.6.
Figure 3.6
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Example 1(b) – Solution
Because an = is positive and n = 4 is even, the graph of S goes up to the far left and up to the far right.
See Figure 3.7.
Figure 3.7
cont’d
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Example 1(c) – Solution
Because an = –2 is negative and n = 3 is odd, the graph of T goes up to the far left and down to the far right.
See Figure 3.8.
Figure 3.8
cont’d
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Example 1(d) – Solution
The leading term of U is –x4 and the leading coefficient is –1.
Because an = –1 is negative
and n = 4 is even, the graph
of U goes down to the far left
and down to the far right.
See Figure 3.9.
Figure 3.9
cont’d
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Real Zeros of a Polynomial Function
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Real Zeros of a Polynomial Function
Sometimes the real zeros of a polynomial function can be determined by using the factoring procedures.
We illustrate this concept in the next example.
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Example 3 – Factor to Find the Real Zeros of a Polynomial Function
Factor to find the three real zeros of P(x) = x3 + 3x2 – 4x.
Solution:
P can be factored as shown below.
P(x) = x3 + 3x2 – 4x
= x(x2 + 3x – 4)
= x(x – 1)(x + 4)
Thus by the Factor Theorem, the real zeros of P(x) are x = 0, x = 1, and x = –4.
Factor out the common factor x.
Factor the trinomial x2 + 3x – 4.
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Example 3 – Solution
The graph of P has x-intercepts at (0, 0), (1, 0), and (–4, 0).
cont’d
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Intermediate Value Theorem
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Intermediate Value Theorem
The following theorem states an important property of polynomial functions.
Intermediate Value Theorem
If P is a polynomial function and P(a) ≠ P(b) for a < b, then P takes on every value between P(a) and P(b) in the interval [a, b].
The Intermediate Value Theorem is often used to verify the existence of a zero of a polynomial function in an interval.
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Intermediate Value Theorem
The essential idea is to find two values a and b such that the polynomial function is positive at one of the values and negative at the other.
Then you can conclude by the Intermediate Value Theorem that the function has a zero between a and b.
Stated in geometric terms, if the points (a, P(a)) and (b, P(b)) are on opposite sides of the x-axis, then the graph of the polynomial function P must cross the x-axis at least once between a and b.
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Intermediate Value Theorem
See Figure 3.14.
Figure 3.14
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Example 4 – Apply the Intermediate Value Theorem
Use the Intermediate Value Theorem to verify that P(x) = x3 – x – 2 has a real zero between 1 and 2.
Solution:
Use substitution or synthetic division to evaluate P(1) and P(2).
P(x) = x3 – x – 2
P(1) = (1)3 – (1) – 2
= 1 – 1 – 2
= –2
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Example 4 – Solution
And
P(x) = x3 – x – 2
P(2) = (2)3 – (2) – 2
= 8 – 2 – 2
= 4
Because P(1) and P(2) have opposite signs, we know by the Intermediate Value Theorem that the polynomial function P has at least one real zero between 1 and 2.
cont’d
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Example 4 – Solution
The graph of P crosses the x-axis between x = 1 and x = 2.
Thus P has a real zero between 1 and 2.
cont’d
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Real Zeros, x-Intercepts, and Factors of a Polynomial Function
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Real Zeros, x-Intercepts, and Factors of a Polynomial Function
The next theorem summarizes important relationships among the real zeros of a polynomial function, the x-intercepts of its graph, and its factors; this theorem can be written in the form (x – c), where c is a real number.
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Real Zeros, x-Intercepts, and Factors of a Polynomial Function
Polynomial Functions, Real Zeros, Graphs, and Factors (x – c)
If P is a polynomial function and c is a real number, then all of the following statements are equivalent in the following sense: If any one statement is true, then they are all true, and if any one statement is false, then they are all false.
(x – c) is a factor of P.
x = c is a real solution of P(x) = 0.
x = c is a real zero of P.
(c, 0) is an x-intercept of the graph of y = P(x).
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Real Zeros, x-Intercepts, and Factors of a Polynomial Function
Sometimes it is possible to make use of the preceding theorem and a graph of a polynomial function to find factors of the function. For example, the graph of
S(x) = x3 – 2x2 – 5x + 6
is shown in Figure 3.15.
The x-intercepts are (–2, 0), (1, 0),
and (3, 0). Hence –2, 1, and 3 are
zeros of S, and [x – (–2)], (x – 1),
and (x – 3) are all factors of S.
Figure 3.15
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Copyright © Cengage Learning. All rights reserved.
Zeros of Polynomial Functions
SECTION 3.3
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Multiple Zeros of a Polynomial Function
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Multiple Zeros of a Polynomial Function
We know that if P is a polynomial function then the values of x for which P(x) is equal to 0 are called the zeros of P or the roots of the equation P(x) = 0.
A zero of a polynomial function may be a multiple zero.
For example, P(x) = x2 + 6x + 9 can be expressed in factored form as (x + 3)(x + 3).
Setting each factor equal to zero yields x = –3 in both cases.
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Multiple Zeros of a Polynomial Function
Thus P(x) = x2 + 6x + 9 has a zero of –3 that occurs twice.
The following definition will be most useful when we are discussing multiple zeros.
Definition of Multiple Zeros of a Polynomial Function
If a polynomial function P has (x – r ) as a factor exactly k times, then r is a zero of multiplicity k of the polynomial function P.
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Multiple Zeros of a Polynomial Function
Example
The graph of the polynomial function
P(x) = (x – 5)2(x + 2)3(x + 4)
is shown in Figure 3.18.
P(x) = (x – 5)2(x + 2)3(x + 4)
Figure 3.18
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Multiple Zeros of a Polynomial Function
This polynomial function has
5 as a zero of multiplicity 2.
–2 as a zero of multiplicity 3.
– 4 as a zero of multiplicity 1.
A zero of multiplicity 1 is generally referred to as a simple zero.
When searching for the zeros of a polynomial function, it is important that we know how many zeros to expect. For the work in this section, the following result is helpful.
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Multiple Zeros of a Polynomial Function
Number of Zeros of a Polynomial Function
A polynomial function P of degree n has at most n zeros, where each zero of multiplicity k is counted k times.
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Rational Zero Theorem
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Rational Zero Theorem
The rational zeros of polynomial functions with integer coefficients can be found with the aid of the following theorem.
Rational Zero Theorem
If P(x) = anxn + an – 1xn – 1 + . . . + a1x + a0 has integer
coefficients (an 0) and is a rational zero (in simplest
form) of P, then
p is a factor of the constant term a0.
q is a factor of the leading coefficient an.
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Rational Zero Theorem
The Rational Zero Theorem often is used to make a list of all possible rational zeros of a polynomial function.
The list consists of all rational numbers of the form ,
where p is an integer factor of the constant term a0 and q is an integer factor of the leading coefficient an.
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Example 1 – Apply the Rational Zero Theorem
Use the Rational Zero Theorem to list all possible rational zeros of
P(x) = 4x4 + x3 – 40x2 + 38x + 12
Solution:
List all integers p that are factors of 12 and all integers q that are factors of 4.
p: 1, 2, 3, 4, 6, 12
q: 1, 2, 4
Form all possible rational numbers using 1, 2, 3, 4, 6, and 12 for the numerator and 1, 2, and 4 for the denominator.
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Example 1 – Solution
By the Rational Zero Theorem, the possible rational zeros are
It is not necessary to list a factor that is already listed in reduced form.
For example, is not listed because it is equal to
cont’d
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Upper and Lower Bounds for Real Zeros
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Upper and Lower Bounds for Real Zeros
A real number b is called an upper bound of the zeros of the polynomial function P if no zero is greater than b.
A real number b is called a lower bound of the zeros of P if no zero is less than b.
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Upper and Lower Bounds for Real Zeros
The following theorem is often used to find positive upper bounds and negative lower bounds for the real zeros of a polynomial function.
Upper- and Lower-Bound Theorem
Let P be a polynomial function with real coefficients. Use synthetic division to divide P by x – b, where b is a nonzero real number.
Upper bound a. If b > 0 and the leading coefficient of P is positive, then b is an upper bound for the real zeros of P provided none of the numbers in the bottom row of the synthetic division are negative.
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Upper and Lower Bounds for Real Zeros
b. If b > 0 and the leading coefficient of P is negative, then b is an upper bound for the real zeros of P provided none of the numbers in the bottom row of the synthetic division are positive.
Lower bound If b < 0 and the numbers in the bottom row of the synthetic division alternate in sign (the number zero can be considered positive or negative as needed to produce an alternating sign pattern), then b is a lower bound for the real zeros of P.
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Upper and Lower Bounds for Real Zeros
Upper and lower bounds are not unique.
For example, if b is an upper bound for the real zeros of P, then any number greater than b is also an upper bound.
Likewise, if a is a lower bound for the real zeros of P, then any number less than a is also a lower bound.
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Example 2 – Find Upper and Lower Bounds
According to the Upper- and Lower-Bound Theorem, what is the smallest positive integer that is an upper bound and the largest negative integer that is a lower bound of the real zeros of
P(x) = 2x3 + 7x2 – 4x – 14?
Solution:
To find the smallest positive-integer upper bound, use synthetic division with 1, 2, . . . , as test values.
No negative numbers
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Example 2 – Solution
According to the Upper- and Lower-Bound Theorem, 2 is the smallest positive-integer upper bound.
Now find the largest negative-integer lower bound.
According to the Upper- and Lower-Bound Theorem, –4 is the largest negative-integer lower bound.
cont’d
Alternating signs
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Zeros of a Polynomial Function
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Zeros of a Polynomial Function
Guidelines for Finding the Zeros of a Polynomial Function with Integer Coefficients
1. Gather general information Determine the degree n of the polynomial function. The number of distinct zeros of the polynomial function is at most n. Apply Descartes’ Rule of Signs to find the possible number of positive zeros and possible number of negative zeros.
2. Check suspects Apply the Rational Zero Theorem to list rational numbers that are possible zeros. Use synthetic division to test numbers in your list. If you find an upper or a lower bound, then eliminate from your list any number that is greater than the upper bound or less than the lower bound.
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Zeros of a Polynomial Function
3. Work with the reduced polynomials Each time a zero is found, you obtain a reduced polynomial.
If a reduced polynomial is of degree 2, find its zeros either by factoring or by applying the quadratic formula.
If the degree of a reduced polynomial is 3 or greater, repeat the preceding steps for this polynomial.
Example 4 illustrates the procedure discussed in the preceding guidelines.
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Example 4 – Find the Zeros of a Polynomial Function
Find the zeros of P(x) = 3x4 + 23x3 + 56x2 + 52x + 16.
Solution:
1. Gather general information The degree of P is 4. Thus the number of zeros of P is at most 4. By Descartes’ Rule of Signs, there are no positive zeros and there are four, two, or no negative zeros.
2. Check suspects By the Rational Zero Theorem, the
possible negative rational zeros of P are
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Example 4 – Solution
Use synthetic division to test the possible rational zeros. The following work shows that – 4 is a zero of P.
3. Work with the reduced polynomials Because – 4 is a zero, (x + 4) and the first reduced polynomial (3x3 + 11x2 + 12x + 4) are both factors of P.
Thus
P(x) = (x + 4)(3x3 + 11x2 + 12x + 4)
cont’d
Coefficients of the first
reduced polynomial
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Example 4 – Solution
All remaining zeros of P must be zeros of 3x3 + 11x2 + 12x + 4.
The Rational Zero Theorem indicates that the only possible negative rational zeros of 3x3 + 11x2 + 12x + 4 are
Synthetic division is again used to test possible zeros.
cont’d
Coefficients of the second
reduced polynomial
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Example 4 – Solution
Because –2 is a zero, (x + 2) is also a factor of P. Thus
P(x) = (x + 4)(x + 2)(3x2 + 5x + 2)
The remaining zeros of P must be zeros of 3x2 + 5x + 2.
3x2 + 5x + 2 = 0
(3x + 2)(x + 1) = 0
x = and x = –1
The zeros of P(x) = 3x4 + 23x3 + 56x2 + 52x + 16 are –4, –2, and –1.
cont’d
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89
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Fundamental Theorem of Algebra
SECTION 3.4
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Fundamental Theorem of Algebra
The German mathematician Carl Friedrich Gauss (1777–1855) was the first to prove that every polynomial function has at least one complex zero.
This concept is so basic to the study of algebra that it is called the Fundamental Theorem of Algebra.
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Fundamental Theorem of Algebra
As you consider each of the following theorems, keep in mind that the terms complex coefficients and complex zeros include real coefficients and real zeros because the set of real numbers is a subset of the set of complex numbers.
Fundamental Theorem of Algebra
If P is a polynomial function of degree n 1 with complex coefficients, then P has at least one complex zero.
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Number of Zeros of a Polynomial Function
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Number of Zeros of a Polynomial Function
Let P be a polynomial function of degree n 1 with complex coefficients. The Fundamental Theorem implies that P has a complex zero—say, c1.
The Factor Theorem implies that
P(x) = (x – c1)Q(x)
where Q(x) is a polynomial of degree 1 less than the degree of P.
Recall that the polynomial Q(x) is called a reduced polynomial. Assuming that the degree of Q(x) is 1 or more, the Fundamental Theorem implies that it also must have a zero.
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Number of Zeros of a Polynomial Function
A continuation of this reasoning process leads to the following theorem.
Linear Factor Theorem
If P is a polynomial function of degree n 1 with leading coefficient an 0,
P(x) = anxn + an – 1xn – 1 + … + a1x1 + a0
then P has exactly n linear factors
P(x) = an(x – c1)(x – c2) … (x – cn)
where c1, c2,…, cn are complex numbers.
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Number of Zeros of a Polynomial Function
The following theorem follows directly from the Linear Factor Theorem.
Number of Zeros of a Polynomial Function Theorem
If P is a polynomial function of degree n 1, then P has exactly n complex zeros, provided each zero is counted according to its multiplicity.
The Linear Factor Theorem and the Number of Zeros of a Polynomial Function Theorem are referred to as existence theorems.
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Number of Zeros of a Polynomial Function
They state that an nth-degree polynomial will have n linear factors and n complex zeros, but they do not provide any information on how to determine the linear factors or the zeros.
In Example 1, we use previously developed methods to actually find the linear factors and zeros of some polynomial functions.
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Example 1 – Find the Zeros and Linear Factors of a Polynomial Function
Find all the zeros of each of the following polynomial functions, and write each function as a product of its leading coefficient and its linear factors.
a. P(x) = x4 – 4x3 + 8x2 – 16x + 16
b. S(x) = 2x4 + x3 + 39x2 + 136x – 78
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Example 1(a) – Solution
By the Linear Factor Theorem, P will have four linear factors and thus four zeros.
The possible rational zeros are 1, 2, 4, 8, and 16.
Use synthetic division to show that 2 is a zero of multiplicity 2.
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Example 1(a) – Solution
The final reduced polynomial is x2 + 4.
Solve x2 + 4 = 0 to find the remaining zeros.
x2 + 4 = 0
x2 = –4
The four zeros of P are 2, 2, –2i, and 2i.
cont’d
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Example 1(a) – Solution
The leading coefficient of P is 1.
Thus the linear factored form of P is
P(x) = 1(x – 2)(x – 2)(x – (–2i ))(x – 2i )
or, after simplifying,
P(x) = (x – 2)2(x + 2i )(x – 2i )
cont’d
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Example 1(b) – Solution
By the Linear Factor Theorem, S will have four linear factors and thus four zeros.
The possible rational zeros are
Use synthetic division to show that –3 and are zeros of S.
cont’d
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Example 1(b) – Solution
The final reduced polynomial is 2x2 – 4x + 52.
The remaining zeros can be found by using the quadratic formula to solve
2x2 – 4x + 52 = 0
x2 – 2x + 26 = 0
Divide each side by 2.
cont’d
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The four zeros are –3, , 1 + 5i, and 1 – 5i.
The leading coefficient of S is 2.
cont’d
Example 1(b) – Solution
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Thus the linear factored form of S is
or, after simplifying,
cont’d
Example 1(b) – Solution
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Conjugate Pair Theorem
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Conjugate Pair Theorem
You may have noticed that the complex zeros of the polynomial function in Example 1 were complex conjugates.
The following theorem shows that this is not a coincidence.
Conjugate Pair Theorem
If a + bi (b 0) is a complex zero of a polynomial function with real coefficients, then the conjugate a – bi is also a complex zero of the polynomial function.
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Example 2 – Use the Conjugate Pair Theorem to Find Zeros
Find all the zeros of P(x) = x4 – 4x3 + 14x2 – 36x + 45, given that 2 + i is a zero.
Solution:
Because the coefficients are real numbers and 2 + i is a zero, the Conjugate Pair Theorem implies that 2 – i also must be a zero.
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Example 2 – Solution
Using synthetic division with 2 + i and 2 – i, we have
The coefficients of the
reduced polynomial
The coefficients of the next
reduced polynomial
cont’d
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Example 2 – Solution
The resulting reduced polynomial is x2 + 9, which has 3i and –3i as zeros.
Therefore, the four zeros of x4 – 4x3 + 14x2 – 36x + 45 are 2 + i, 2 – i, 3i, and –3i.
cont’d
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Conjugate Pair Theorem
A graph of P(x) = x4 – 4x3 + 14x2 – 36x + 45 is shown in Figure 3.20.
Because the polynomial in Example 2 is a fourth-degree polynomial and because we have verified that P has four nonreal solutions, it comes as no surprise that the graph does not intersect the x-axis.
Figure 3.20
P(x) = x4 – 4x3 + 14x2 – 36x + 45
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Conjugate Pair Theorem
When performing synthetic division with complex numbers, it is helpful to write the coefficients of the given polynomial as complex coefficients.
For instance, –10 can be written as –10 + 0i.
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Conjugate Pair Theorem
Recall that the real zeros of a polynomial function P are the x-coordinates of the x-intercepts of the graph of P.
This important connection between the real zeros of a polynomial function and the x-intercepts of the graph of the polynomial function is the basis for using a graphing utility to solve equations.
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Finding a Polynomial Function with Given Zeros
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Finding a Polynomial Function with Given Zeros
Many of the problems in this section dealt with the process of finding the zeros of a given polynomial function.
Example 5 considers the reverse process: finding a polynomial function when the zeros are given.
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Example 5 – Determine a Polynomial Function Given Its Zeros
Find each polynomial function.
a. A polynomial function of degree 3 that has 1, 2, and –3 as zeros
b. A polynomial function of degree 4 that has real coefficients and zeros 2i and 3 – 7i
Solution:
a. Because 1, 2, and –3 are zeros, (x – 1), (x – 2),
and (x + 3) are factors.
The product of these factors produces a polynomial function with the indicated zeros.
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Example 5 – Solution
P(x) = (x – 1)(x – 2)(x + 3)
= (x2 – 3x + 2)(x + 3)
= x3 – 7x + 6
b. By the Conjugate Pair Theorem, the polynomial function also must have –2i and 3 + 7i as zeros.
The product of the factors x – 2i, x – (–2i), x – (3 – 7i), and x – (3 + 7i) produces the desired polynomial function.
cont’d
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Example 5 – Solution
P(x) = (x – 2i)(x + 2i)[x – (3 – 7i)][x – (3 + 7i)]
= (x2 + 4)(x2 – 6x + 58)
= x4 – 6x3 + 62x2 – 24x + 232
cont’d
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Finding a Polynomial Function with Given Zeros
A polynomial function that has a given set of zeros is not unique.
For example, P(x) = x3 – 7x + 6 has zeros 1, 2, and –3, but so does any nonzero multiple of P, such as S(x) = 2x3 – 14x + 12.
This concept is illustrated in Figure 3.22.
The graphs of the two polynomial functions are different, but they have the same zeros.
Figure 3.22
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Copyright © Cengage Learning. All rights reserved.
Graphs of Rational Functions and Their Applications
SECTION 3.5
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Vertical and Horizontal Asymptotes
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Vertical and Horizontal Asymptotes
If P(x) and Q(x) are polynomials, then the function F given by
is called a rational function. The domain of F is the set of all real numbers except those for which Q(x) = 0.
For example, let
Setting the denominator equal to 0, we have
2x3 + x2 – 15x = 0
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Vertical and Horizontal Asymptotes
x(2x – 5)(x + 3) = 0
The denominator is 0 for x = 0, x = , and x = –3.
Thus the domain of F is the set of all real numbers except 0, , and –3.
The graph of is
given in Figure 3.23.
Figure 3.23
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Vertical and Horizontal Asymptotes
The graph shows that G has the following properties:
The graph has an x-intercept at (–1, 0) and a y-intercept at .
The graph does not exist when x = 2.
Note the behavior of the graph as x takes on values that are close to 2 but less than 2.
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Vertical and Horizontal Asymptotes
Mathematically, we say that “x approaches 2 from the left.” From this table and the graph, it appears that as x approaches 2 from the left, the functional values G(x) decrease without bound.
In this case, we say that “G(x) approaches negative infinity.”
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Vertical and Horizontal Asymptotes
Now observe the behavior of the graph as x takes on values that are close to 2 but greater than 2. Mathematically, we say that “x approaches 2 from the right.”
From this table and the graph, it appears that as x approaches 2 from the right, the functional values G(x) increase without bound.
In this case, we say that “G(x) approaches positive infinity.”
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Vertical and Horizontal Asymptotes
Now consider the values of G(x) as x increases without bound.
The following table gives values of G(x) for selected values of x.
As x increases without bound, the values of G(x) become closer to 1.
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Vertical and Horizontal Asymptotes
Now let the values of x decrease without bound.
The following table gives the values of G(x) for selected values of x.
As x decreases without bound, the values of G(x) become closer to 1.
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Vertical and Horizontal Asymptotes
When we are discussing functional values that increase or decrease without bound, it is convenient to use mathematical notation. The notation
f (x) as x a+
means that the functional values f (x) increase without bound as x approaches a from the right.
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Vertical and Horizontal Asymptotes
Recall that the symbol does not represent a real number but is used merely to describe the concept of a variable taking on larger and larger values without bound. See Figure 3.24(a).
Figure 3.24(a)
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Vertical and Horizontal Asymptotes
The notation
f (x) as x a –
means that the function values f (x) increase without bound as x approaches a from the left. See Figure 3.24(b).
Figure 3.24(b)