Download Section 3-2 Finding Rational Zeros of Polynomials

3-2 Finding Rational Zeros of Polynomials
195
Section 3-2 Finding Rational Zeros of Polynomials
Factor Theorem
Fundamental Theorem of Algebra
Imaginary Zeros
Rational Zeros
In this section we develop some important properties of polynomials with arbitrary coefficients. Then we consider the problem of finding all the rational zeros
of a polynomial with rational coefficients. In some cases, this process will also
enable us to find irrational or imaginary zeros.
1
Let P(x) ⫽ x3 ⫺ 5x2 ⫹ 2x ⫹ 8
(A) (i) Find P(2).
(ii) What is the remainder if P(x) is divided by x ⫺ 2?
(iii) Is x ⫺ 2 a factor of P(x)?
(B) Repeat part A for x ⫽ ⫺1, x ⫽ 1, x ⫽ 3, and x ⫽ 4.
What relationship between factors and zeros is suggested by these results?
Factor Theorem
The division algorithm (Theorem 1 in Section 3-1)
P(x) ⫽ (x ⫺ r)Q(x) ⫹ R
may, because of the remainder theorem (Theorem 2 in Section 3-1), be written in
a form where R is replaced by P(r):
P(x) ⫽ (x ⫺ r)Q(x) ⫹ P(r)
It is easy to see that x ⫺ r is a factor of P(x) if and only if P(r) ⫽ 0; that is, if
and only if r is a zero of the polynomial P(x). This result is known as the factor
theorem:
1
FACTOR THEOREM
If r is a zero of the polynomial P(x), then x ⫺ r is a factor of P(x). Conversely, if x ⫺ r is a factor of P(x), then r is a zero of P(x).
The relationship between zeros, roots, factors, and x intercepts is fundamental to the study of polynomials. With the addition of the factor theorem, we now
know that the following statements are equivalent for any polynomial P(x):
1. r is a root of the equation P(x) ⫽ 0.
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3 POLYNOMIAL AND RATIONAL FUNCTIONS
2. r is a zero of P(x).
3. x ⫺ r is a factor of P(x).
If, in addition, the coefficients of P(x) are real numbers and r is a real number,
then we can add a fourth statement to this list:
4. r is an x intercept of the graph of P(x).
EXAMPLE
1
Solutions
Factors, Zeros, Roots, and Intercepts
(A)
(B)
(C)
(D)
Use the factor theorem to show that x ⫹ 1 is a factor of P(x) ⫽ x25 ⫹ 1.
What are the zeros of P(x) ⫽ 3(x ⫺ 5)(x ⫹ 2)(x ⫺ 3)?
What are the roots of x4 ⫺ 1 ⫽ 0?
What are the x intercepts of the graph of P(x) ⫽ x4 ⫺ 1?
(A) Since x ⫹ 1 ⫽ x ⫺ (⫺1), we have r ⫽ ⫺1 and
P(r) ⫽ P(⫺1) ⫽ (⫺1)25 ⫹ 1 ⫽ ⫺1 ⫹ 1 ⫽ 0
FIGURE 1
x intercepts of P(x) ⫽ x4 ⫺ 1.
Hence, ⫺1 is a zero of P(x) ⫽ x25 ⫹ 1. By the factor theorem, x ⫺ (⫺1) ⫽
x ⫹ 1 is a factor of x25 ⫹ 1.
(B) Since (x ⫺ 5), (x ⫹ 2), and (x ⫺ 3) are all factors of P(x), 5, ⫺2, and 3 are
zeros of P(x).
(C) Factoring the left side, we have
x4 ⫺ 1 ⫽ 0
(x2 ⫺ 1)(x2 ⫹ 1) ⫽ 0
(x ⫺ 1)(x ⫹ 1)(x ⫺ i)(x ⫹ i) ⫽ 0
Thus, the roots of x4 ⫺ 1 ⫽ 0 are 1, ⫺1, i, and ⫺i.
(D) From part C, the zeros of P(x) are 1, ⫺1, i, and ⫺i. However, x intercepts
must be real numbers. Thus the x intercepts of the graph of P(x) ⫽ x4 ⫺ 1
are 1 and ⫺1 (see Fig. 1).
MATCHED PROBLEM
1
(A) Use the factor theorem to show that x ⫺ 1 is a factor of P(x) ⫽ x54 ⫺ 1.
(B) What are the zeros of
P(x) ⫽ 2(x ⫹ 3)(x ⫹ 7)(x ⫺ 8)(x ⫹ 1)?
(C) What are the roots of x2 ⫹ 4 ⫽ 0?
(D) What are the x intercepts of P(x) ⫽ x2 ⫹ 4?
Fundamental Theorem of Algebra
Theorem 2, often referred to as the fundamental theorem of algebra, requires
verification that is beyond the scope of this book, so we state it without proof.
3-2 Finding Rational Zeros of Polynomials
197
FUNDAMENTAL THEOREM OF ALGEBRA
Every polynomial P(x) of degree n ⬎ 0 has at least one zero.
2
If P(x) ⫽ an x n ⫹ an⫺1x n⫺1 ⫹ ⴢ ⴢ ⴢ ⫹ a1x ⫹ a0 is a polynomial of degree
n ⬎ 0 with complex coefficients, then, according to Theorem 2, it has at least one
zero, say r1. According to the factor theorem, x ⫺ r1 is a factor of P(x). Thus,
P(x) ⫽ (x ⫺ r1)Q1(x)
where Q1(x) is a polynomial of degree n ⫺ 1. If n ⫺ 1 ⫽ 0, then Q1(x) ⫽ an. If
n ⫺ 1 ⬎ 0, then, by Theorem 2, Q1(x) has at least one zero, say r2. And
Q1(x) ⫽ (x ⫺ r2)Q2(x)
where Q2(x) is a polynomial of degree n ⫺ 2. Thus,
P(x) ⫽ (x ⫺ r1)(x ⫺ r2)Q2(x)
If n ⫺ 2 ⫽ 0, then Q2(x) ⫽ an. If n ⫺ 2 ⬎ 0, then Q2(x) has at least one zero,
say r3. And
Q2(x) ⫽ (x ⫺ r3)Q3(x)
where Q3(x) is a polynomial of degree n ⫺ 3.
We continue in this way until Qk(x) is of degree zero—that is, until k ⫽ n.
At this point, Qn(x) ⫽ an, and we have
P(x) ⫽ (x ⫺ r1)(x ⫺ r2) ⴢ ⭈ ⭈ ⭈ ⴢ (x ⫺ rn)an
Thus, r1, r2, . . . , rn are n zeros, not necessarily distinct, of P(x). Is it possible for
P(x) to have more than these n zeros? Let’s assume that r is a number different
from the zeros above. Then
P(r) ⫽ an(r ⫺ r1)(r ⫺ r2) ⴢ ⭈ ⭈ ⭈ ⴢ (r ⫺ rn) ⫽ 0
since r is not equal to any of the zeros. Hence, r is not a zero, and we conclude
that r1, r2, . . . , rn are the only zeros of P(x). We have just sketched a proof of
Theorem 3.
3
n ZEROS THEOREM
Every polynomial P(x) of degree n ⬎ 0 can be expressed as the product
of n linear factors. Hence, P(x) has exactly n zeros—not necessarily
distinct.
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3 POLYNOMIAL AND RATIONAL FUNCTIONS
Theorems 2 and 3 were first proved in 1797 by Carl Friedrich Gauss, one of
the greatest mathematicians of all time, at the age of 20.
If P(x) is represented as the product of linear factors and x ⫺ r occurs m times,
then r is called a zero of multiplicity m. For example, if
P(x) ⫽ 4(x ⫺ 5)3(x ⫹ 1)2(x ⫺ i)(x ⫹ i)
then this seventh-degree polynomial has seven zeros, not all distinct. That is, 5 is
a zero of multiplicity 3, or a triple zero; ⫺1 is a zero of multiplicity 2, or a double zero; and i and ⫺i are zeros of multiplicity 1, or simple zeros. Thus, this seventh-degree polynomial has exactly seven zeros if we count 5 and ⫺1 with their
respective multiplicities.
Graph each of the following polynomials one at a time for ⫺10 ⱕ x ⱕ 10
and an appropriate interval of y values:
2
y1 ⫽ (x ⫹ 2)2(3 ⫺ x)
y2 ⫽ (x ⫹ 2)(3 ⫺ x)2
y3 ⫽ (x ⫹ 2)2(3 ⫺ x)2
y4 ⫽ (x ⫹ 2)2(3 ⫺ x)3
y5 ⫽ (x ⫹ 2)3(3 ⫺ x)2
y6 ⫽ (x ⫹ 2)3(3 ⫺ x)3
Discuss the various possibilities for the behavior of the graph of a polynomial at a zero of odd multiplicity and at a zero of even multiplicity.
EXAMPLE
2
Solution
Factoring a Polynomial
If ⫺2 is a double zero of P(x) ⫽ x4 ⫺ 7x2 ⫹ 4x ⫹ 20, write P(x) as a product of first-degree factors.
Since ⫺2 is a double zero of P(x), we can write
P(x) ⫽ (x ⫹ 2)2Q(x)
⫽ (x2 ⫹ 4x ⫹ 4)Q(x)
and find Q(x) by dividing P(x) by x2 ⫹ 4x ⫹ 4. Carrying out the algebraic long
division, we obtain
Q(x) ⫽ x2 ⫺ 4x ⫹ 5
The zeros of Q(x) are found, using the quadratic formula, to be 2 ⫺ i and 2 ⫹ i.
Thus, P(x) written as a product of linear factors is
P(x) ⫽ (x ⫹ 2)2[x ⫺ (2 ⫺ i)][x ⫺ (2 ⫹ i)]
3-2 Finding Rational Zeros of Polynomials
199
[Note: Any time Q(x) is a quadratic polynomial, its zeros can be found using the
quadratic formula.]
MATCHED PROBLEM
2
FIGURE 2
If 3 is a double zero of P(x) ⫽ x4 ⫺ 12x3 ⫹ 55x2 ⫺ 114x ⫹ 90, write P(x) as a
product of first-degree factors.
The graph of y ⫽ P(x) from Example 2 is shown in Figure 2. Notice that P(x)
has four zeros, taking multiplicity into account, but only one x intercept—the
double zero at x ⫽ ⫺2. Since an n th degree polynomial has exactly n zeros (Theorem 3), it can have a maximum of n x intercepts, but may have fewer, due to
imaginary zeros and multiple real zeros. Also note that the two imaginary zeros
of P(x) are not visible on the graph in Figure 2. The real zeros of a polynomial
with real coefficients can be approximated graphically on a graphing utility, but
the imaginary zeros cannot.
Approximating real zeros of even multiplicity with a built-in root approximation routine can result in an error message. The maximum and minimum routines on a graphing utility can be used to approximate real zeros
of even multiplicity if the root approximation routine fails.
Imaginary Zeros
Something interesting happens if we restrict the coefficients of a polynomial to
real numbers. Let’s use the quadratic formula to find the zeros of the polynomial
P(x) ⫽ x2 ⫺ 6x ⫹ 13
To find the zeros of P(x), we solve P(x) ⫽ 0:
x2 ⫺ 6x ⫹ 13 ⫽ 0
x⫽
⫽
6 ⫾ 兹36 ⫺ 52
2
6 ⫾ 兹⫺16 6 ⫾ 4i
⫽
⫽ 3 ⫾ 2i
2
2
The zeros of P(x) are 3 ⫺ 2i and 3 ⫹ 2i, conjugate imaginary numbers (see Section 2-4). Also observe that the imaginary zeros in Example 2 are the conjugate
imaginary numbers 2 ⫺ i and 2 ⫹ i.
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3 POLYNOMIAL AND RATIONAL FUNCTIONS
This is generalized in the following theorem:
4
IMAGINARY ZEROS THEOREM
Imaginary zeros of polynomials with real coefficients, if they exist, occur
in conjugate pairs.
As a consequence of Theorems 3 and 4, we also know (think this through)
the following:
5
3
REAL ZEROS AND ODD-DEGREE POLYNOMIALS
A polynomial of odd degree with real coefficients always has at least one
real zero.
(A) Let P(x) be a third-degree polynomial with real coefficients. Indicate
which of the following statements are true and which are false. Justify your conclusions.
(i) P(x) has at least one real zero.
(ii) P(x) has three zeros.
(iii) P(x) can have two real zeros and one imaginary zero.
(B) Let P(x) be a fourth-degree polynomial with real coefficients.
Indicate which of the following statements are true and which are
false. Justify your conclusions.
(i) P(x) has four zeros.
(ii) P(x) has at least two real zeros.
(iii) If we know P(x) has three real zeros, then the fourth zero must
be real.
Rational Zeros
First note that a polynomial with rational coefficients can always be written as a
constant times a polynomial with integer coefficients. For example,
P(x) ⫽ 12x3 ⫺ 23x2 ⫹ 74x ⫹ 5
1
(6x3 ⫺ 8x2 ⫹ 21x ⫹ 60)
⫽ 12
Thus, it is sufficient to confine our attention to polynomials with integer
coefficients.
We introduce the rational zero theorem by examining the following quadratic
polynomial whose zeros can be found easily by factoring:
3-2 Finding Rational Zeros of Polynomials
201
P(x) ⫽ 6x2 ⫺ 13x ⫺ 5 ⫽ (2x ⫺ 5)(3x ⫹ 1)
5
2
Zeros of P(x):
and
1 ⫺1
⫺ ⫽
3
3
Notice that the numerators, 5 and ⫺1, of the zeros are both integer factors of ⫺5,
the constant term in P(x). The denominators 2 and 3 of the zeros are both integer factors of 6, the coefficient of the highest-degree term in P(x). These observations are generalized in Theorem 6.
RATIONAL ZERO THEOREM
If the rational number b/c, in lowest terms, is a zero of the polynomial
6
P(x) ⫽ anxn ⫹ an⫺1xn⫺1 ⫹ ⭈ ⭈ ⭈ ⫹ a1x ⫹ a0
an ⫽ 0
with integer coefficients, then b must be an integer factor of a0 and c
must be an integer factor of an.
The proof of Theorem 6 is not difficult and is instructive, so we sketch it here.
Proof
Since b/c is a zero of P(x),
冢冣
an
b
c
n
冢冣
⫹ an⫺1
b
c
n⫺1
⫹ ⭈ ⭈ ⭈ ⫹ a1
冢bc 冣 ⫹ a ⫽ 0
0
(1)
If we multiply both sides of equation (1) by cn, we obtain
anbn ⫹ an⫺1bn⫺1c ⫹ ⭈ ⭈ ⭈ ⫹ a1bcn⫺1 ⫹ a0cn ⫽ 0
(2)
which can be written in the form
anbn ⫽ c(⫺an⫺1bn⫺1 ⫺ ⭈ ⭈ ⭈ ⫺ a0cn⫺1)
(3)
Since both sides of equation (3) are integers, c must be a factor of anbn. And since
the rational number b/c is given to be in lowest terms, b and c can have no common factors other than ⫾1. That is, b and c are relatively prime. This implies
that bn and c also are relatively prime. Hence, c must be a factor of an.
Now, if we solve equation (2) for a0cn and factor b out of the right side, we
have
a0cn ⫽ b(⫺anbn⫺1 ⫺ ⭈ ⭈ ⭈ ⫺ a1cn⫺1)
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3 POLYNOMIAL AND RATIONAL FUNCTIONS
We see that b is a factor of a0cn and, hence, a factor of a0, since b and c are
relatively prime.
4
Let P(x) ⫽ a3x3 ⫹ a2x2 ⫹ a1x ⫹ a0, where a3, a2, a1, and a0 are integers.
1. If P(2) ⫽ 0, there is one coefficient that must be an even integer.
Identify this coefficient and explain why it must be even.
2. If P(12) ⫽ 0, there is one coefficient that must be an even integer.
Identify this coefficient and explain why it must be even.
3. If a3 ⫽ a0 ⫽ 1, P(⫺1) ⫽ 0, and P(1) ⫽ 0, does P(x) have any
rational zeros? Support your conclusion with verbal arguments and/or
examples.
It is important to understand that Theorem 6 does not say that a
polynomial P(x) with integer coefficients must have rational zeros.
It simply states that if P(x) does have a rational zero, then the numerator of
the zero must be an integer factor of a0 and the denominator of the zero must be
an integer factor of an. Since every integer has a finite number of integer factors,
Theorem 6 enables us to construct a finite list of possible rational zeros. Finding
any rational zeros then becomes a routine, although sometimes tedious, process
of elimination.
As the next example illustrates, a graphing utility can greatly reduce the effort
required to locate rational zeros.
EXAMPLE
3
Solution
Finding Rational Zeros
Find all the rational zeros for P(x) ⫽ 2x3 ⫹ 9x2 ⫹ 7x ⫺ 6.
If b/c in lowest terms is a rational zero of P(x), then b must be a factor of ⫺6
and c must be a factor of 2.
Possible values of b are the integer factors of ⫺6: ⫾1, ⫾2, ⫾3, ⫾6
(4)
Possible values of c are the integer factors of 2: ⫾1, ⫾2
(5)
Writing all possible fractions b/c where b is from (4) and c is from (5), we have
Possible rational zeros for P(x): ⫾1, ⫾2, ⫾3, ⫾6, ⫾21, ⫾23
(6)
[Note that all fractions are in lowest terms and duplicates like ⫾6/⫾2 ⫽ ⫾3 are
not repeated.] If P(x) has any rational zeros, they must be in list (6). We can test
each number r in this list simply by evaluating P(r). However, exploring the graph
of y ⫽ P(x) first will usually indicate which numbers in the list are the most likely
candidates for zeros. Examining a graph of P(x), we see that there are zeros near
⫺3, near ⫺2, and between 1 and 2, so we begin by evaluating P(x) at ⫺3, ⫺2,
and 12 (Fig. 3).
3-2 Finding Rational Zeros of Polynomials
203
FIGURE 3
(a)
(b)
(c)
Thus ⫺3, ⫺2, and are rational zeros of P(x). Since a third-degree polynomial
can have at most three zeros, we have found all the rational zeros. There is no
need to test the remaining candidates in list (6).
1
2
MATCHED PROBLEM
Find all rational zeros for P(x) ⫽ 2x3 ⫹ x2 ⫺ 11x ⫺ 10.
3
As we saw in the solution of Example 3, rational zeros can be located by simply evaluating the polynomial. However, if we want to find multiple zeros, imaginary zeros, or exact values of irrational zeros, we need to consider reduced polynomials. If r is a zero of a polynomial P(x), then we can write
P(x) ⫽ (x ⫺ r)Q(x)
where Q(x) is a polynomial of degree one less than the degree of P(x). The quotient polynomial Q(x) is called the reduced polynomial for P(x). In Example 3,
after determining that ⫺3 is a zero of P(x), we can write
P(x) ⫽ 2x3 ⫹ 9x2 ⫹ 7x ⫺ 6
⫽ (x ⫹ 3)(2x2 ⫹ 3x ⫺ 2)
⫽ (x ⫹ 3)Q(x)
Since the reduced polynomial Q(x) ⫽ 2x2 ⫹ 3x ⫺ 2 is a quadratic, we can find
its zeros by factoring or the quadratic formula. Thus,
P(x) ⫽ (x ⫹ 3)(2x2 ⫹ 3x ⫺ 2) ⫽ (x ⫹ 3)(x ⫹ 2)(2x ⫺ 1)
and we see that the zeros of P(x) are ⫺3, ⫺2, and 12, as before.
EXAMPLE
4
Solution
Finding Rational and Irrational Zeros
Find all zeros exactly for P(x) ⫽ 2x3 ⫺ 7x2 ⫹ 4x ⫹ 3.
First, list the possible rational zeros:
⫾1, ⫾3, ⫾12, ⫾32
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3 POLYNOMIAL AND RATIONAL FUNCTIONS
FIGURE 4
Examining the graph of y ⫽ P(x) (Fig. 4), we see that there is a zero between
⫺1 and 0, another between 1 and 2, and a third between 2 and 3. We test the
only likely candidates, ⫺ 12 and 32:
P(⫺ 12兲 ⫽ ⫺1
and
P(32) ⫽ 0
Thus, 23 is a zero, but ⫺ 12 is not. Using synthetic division (details omitted), we
can write
P(x) ⫽ (x ⫺ 32)(2x2 ⫺ 4x ⫺ 2)
Since the reduced polynomial is quadratic, we can use the quadratic formula to
find the exact values of the remaining zeros:
2x2 ⫺ 4x ⫺ 2 ⫽ 0
x2 ⫺ 2x ⫺ 1 ⫽ 0
x⫽
2 ⫾ 兹4 ⫺ 4(1)(⫺1)
2
⫽
2 ⫾ 2兹2
⫽ 1 ⫾ 兹2
2
Thus, the exact zeros of P(x) are
MATCHED PROBLEM
3
2
and 1 ⫾ 兹2.
Find all zeros exactly for P(x) ⫽ 3x3 ⫺ 10x2 ⫹ 5x ⫹ 4.
4
EXAMPLE
5
Solution
FIGURE 5
Finding Rational and Imaginary Zeros
Find all zeros exactly for P(x) ⫽ x4 ⫺ 6x3 ⫹ 14x2 ⫺ 14x ⫹ 5.
The possible rational zeros are ⫾1 and ⫾5. Examining the graph of P(x)
(Fig. 5), we see that 1 is a zero. Since the graph of P(x) does not appear to change
sign at 1, this may be a multiple root (see Explore/Discuss 2). Using synthetic
division (details omitted), we find that
P(x) ⫽ (x ⫺ 1)(x3 ⫺ 5x2 ⫹ 9x ⫺ 5)
The possible rational zeros of the reduced polynomial
Q(x) ⫽ x3 ⫺ 5x2 ⫹ 9x ⫺ 5
are ⫾1 and ⫾5. Examining the graph of Q(x) (Fig. 6), we see that 1 is a rational
zero. After a division, we have a quadratic reduced polynomial:
Q(x) ⫽ (x ⫺ 1)Q1(x) ⫽ (x ⫺ 1)(x2 ⫺ 4x ⫹ 5)
3-2 Finding Rational Zeros of Polynomials
205
We use the quadratic formula to find the zeros of Q1(x):
FIGURE 6
x2 ⫺ 4x ⫹ 5 ⫽ 0
x⫽
4 ⫾ 兹16 ⫺ 4(1)(5)
2
⫽
4 ⫾ 兹⫺4
⫽2⫾i
2
Thus, the exact zeros of P(x) are 1 (multiplicity 2), 2 ⫺ i, and 2 ⫹ i.
MATCHED PROBLEM
Find all zeros exactly for P(x) ⫽ x4 ⫹ 4x3 ⫹ 10x2 ⫹ 12x ⫹ 5.
5
Remark
FIGURE 7
P(x) ⫽ x3 ⫹ 6x ⫺ 2.
We were successful in finding all the zeros of the polynomials in Examples 4 and
5 because we could find sufficient rational zeros to reduce the original polynomial to a quadratic. This is not always possible. For example, the polynomial
P(x) ⫽ x3 ⫹ 6x ⫺ 2
has no rational zeros, but does have an irrational zero at x ⬇ 0.32748 (Fig. 7).
The other two zeros are imaginary. The techniques we have developed will not
find the exact value of these roots.
5
There is a technique for finding the exact zeros of cubic polynomials,
usually referred to as Cardano’s formula.* This formula shows that the
exact value of the irrational zero of P(x) ⫽ x3 ⫹ 6x ⫺ 2 (see Fig. 7) is
3
3
x ⫽ 兹4 ⫺ 兹2
(A) Verify that this is correct by expanding and simplifying
3
3
P(兹
4⫺兹
2).
(B) Cardano’s formula also shows that the two imaginary zeros are
3
3
3
3
⫺ 2 (兹
4⫺兹
2) ⫾ 21 i 兹3 (兹
4⫺兹
2)
1
If you like algebraic manipulation, you can also verify that these are
correct.
*Girolamo Cardano (1501–1576), an Italian mathematician and physician, was the first to publish a formula for the solution to cubic equations of the form x3 ⫹ ax ⫹ b ⫽ 0 and the first to realize that this technique could be used to solve other cubic equations. Having predicted that he would live to the age of 75,
Cardano committed suicide in 1576.
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3 POLYNOMIAL AND RATIONAL FUNCTIONS
(C) Find a reference for Cardano’s formula in a library or on the
Internet. Use this formula to find the exact value of the irrational
zero of
5
P(x) ⫽ x3 ⫹ 9x ⫺ 6
continued
Check your answer by comparing it with the approximate value
obtained on a graphing utility.
Answers to Matched Problems
1. (A) P(1) ⫽ 154 ⫺ 1 ⫽ 0 implies that x ⫽ 1 is a factor of P(x).
(B) ⫺3, ⫺7, 8, ⫺1
(C) ⫺2i, 2i
(D) No x intercepts
2. P(x) ⫽ (x ⫺ 3)2[x ⫺ (3 ⫺ i)][x ⫺ (3 ⫹ i)]
3. ⫺2, ⫺1, 52
4. 43, 1 ⫺ 兹2, 1 ⫹ 兹2
5. ⫺1 (multiplicity 2), ⫺1 ⫺ 2i, ⫺1 ⫹ 2i
11.
EXERCISE 3-2
A
Write the zeros of each polynomial in Problems 1–4, and
indicate the multiplicity of each if over 1. What is the degree of
each polynomial?
1. P(x) ⫽ (x ⫹ 8)3(x ⫺ 6)2
2. P(x) ⫽ (x ⫺ 5)(x ⫹ 7)2
3. P(x) ⫽ 3(x ⫹ 4)3(x ⫺ 3)2(x ⫹ 1)
4. P(x) ⫽ 5(x ⫺ 2)3(x ⫹ 3)2(x ⫺ 1)
12.
In Problems 5–10, find a polynomial P(x) of lowest degree,
with leading coefficient 1, that has the indicated set of zeros.
Leave the answer in a factored form. Indicate the degree of the
polynomial.
5. 3 (multiplicity 2) and ⫺4
6. ⫺2 (multiplicity 3) and 1 (multiplicity 2)
7. ⫺7 (multiplicity 3), ⫺3 ⫹ 兹2, ⫺3 ⫺ 兹2
8.
1
3
(multiplicity 2), 5 ⫹ 兹7, 5 ⫺ 兹7
9. (2 ⫺ 3i), (2 ⫹ 3i), ⫺4 (multiplicity 2)
10. i兹3 (multiplicity 2), ⫺i兹3 (multiplicity 2), and
4 (multiplicity 3)
In Problems 11–16, find a polynomial of lowest degree, with
leading coefficient 1, that has the indicated graph. Assume all
zeros are integers. Leave the answer in a factored form.
Indicate the degree of each polynomial.
13.
3-2 Finding Rational Zeros of Polynomials
207
25. P(x) ⫽ 12x3 ⫺ 16x2 ⫺ 5x ⫹ 3
14.
26. P(x) ⫽ 2x3 ⫺ 9x2 ⫹ 14x ⫺ 5
In Problems 27–32, write P(x) as a product of linear terms.
27. P(x) ⫽ x3 ⫹ 9x2 ⫹ 24x ⫹ 16; ⫺1 is a zero
28. P(x) ⫽ x3 ⫺ 4x2 ⫺ 3x ⫹ 18; 3 is a double zero
29. P(x) ⫽ x4 ⫺ 1; 1 and ⫺1 are zeros
30. P(x) ⫽ x4 ⫹ 2x2 ⫹ 1; i is a double zero
15.
31. P(x) ⫽ 2x3 ⫺ 17x2 ⫹ 90x ⫺ 41; 12 is a zero
32. P(x) ⫽ 3x3 ⫺ 10x2 ⫹ 31x ⫹ 26; ⫺ 23 is a zero
In Problems 33–40, find all roots exactly (rational, irrational,
and imaginary) for each polynomial equation.
33. 2x3 ⫺ 5x2 ⫹ 1 ⫽ 0
34. 2x3 ⫺ 10x2 ⫹ 12x ⫺ 4 ⫽ 0
35. x4 ⫹ 4x3 ⫺ x2 ⫺ 20x ⫺ 20 ⫽ 0
36. x4 ⫺ 4x2 ⫺ 4x ⫺ 1 ⫽ 0
16.
37. x4 ⫺ 2x3 ⫺ 5x2 ⫹ 8x ⫹ 4 ⫽ 0
38. x4 ⫺ 2x2 ⫺ 16x ⫺ 15 ⫽ 0
39. 2x5 ⫺ 3x4 ⫺ 2x ⫹ 3 ⫽ 0
40. 2x5 ⫹ x4 ⫺ 6x3 ⫺ 3x2 ⫺ 8x ⫺ 4 ⫽ 0
In Problems 41–48, find all zeros exactly (rational, irrational,
and imaginary) for each polynomial.
41. P(x) ⫽ x3 ⫺ 19x ⫹ 30
In Problems 17–20, determine whether the second polynomial
is a factor of the first polynomial without dividing or using
synthetic division. [Hint: Evaluate directly and use the factor
theorem.]
17. x ⫺ 1; x ⫺ 1
18
18. x18 ⫺ 1; x ⫹ 1
19. 3x ⫺ 7x ⫺ 8x ⫹ 2; x ⫹ 1
3
2
20. 3x ⫺ 2x ⫹ 5x ⫺ 6; x ⫺ 1
4
42. P(x) ⫽ x3 ⫺ 7x2 ⫹ 36
2
3
43. P(x) ⫽ x 4 ⫺ 21
10 x ⫹ 5 x
44. P(x) ⫽ x 4 ⫹ 76 x3 ⫺ 73 x2 ⫺ 52 x
2
45. P(x) ⫽ x 4 ⫺ 5x3 ⫹ 15
2 x ⫺ 2x ⫺ 2
5
1
2
46. P(x) ⫽ x 4 ⫺ 13
4 x ⫺ 2x ⫺ 4
47. P(x) ⫽ 3x5 ⫺ 5x4 ⫺ 8x3 ⫹ 16x2 ⫹ 21x ⫹ 5
3
B
For each polynomial in problems 21–26, list all possible
rational zeros (Theorem 6).
48. P(x) ⫽ 2x5 ⫺ 3x4 ⫺ 6x3 ⫹ 23x2 ⫺ 26x ⫹ 10
In Problems 49–54, write each polynomial as a product of
linear factors.
49. P(x) ⫽ 6x3 ⫹ 13x2 ⫺ 4
21. P(x) ⫽ x3 ⫺ 2x2 ⫺ 5x ⫹ 6
50. P(x) ⫽ 6x3 ⫺ 17x2 ⫺ 4x ⫹ 3
22. P(x) ⫽ x3 ⫹ 3x2 ⫺ 6x ⫺ 8
51. P(x) ⫽ x3 ⫹ 2x2 ⫺ 9x ⫺ 4
23. P(x) ⫽ 3x3 ⫺ 11x2 ⫹ 8x ⫹ 4
52. P(x) ⫽ x3 ⫺ 8x2 ⫹ 17x ⫺ 4
24. P(x) ⫽ 2x3 ⫹ x2 ⫺ 4x ⫺ 3
53. P(x) ⫽ 4x4 ⫺ 4x3 ⫺ 9x2 ⫹ x ⫹ 2
208
3 POLYNOMIAL AND RATIONAL FUNCTIONS
54. P(x) ⫽ 2x4 ⫹ 3x3 ⫺ 4x2 ⫺ 3x ⫹ 2
(A) How many cube roots of 8 are there?
(B) 2 is obviously a cube root of 8; find all others.
In Problems 55–60, multiply.
56. [x ⫺ (2 ⫺ 3i)][x ⫺ (2 ⫹ 3i)]
79. If P is a polynomial function with real coefficients of
degree n, with n odd, then what is the maximum number
of times the graph of y ⫽ P(x) can cross the x axis? What
is the minimum number of times?
57. [x ⫺ (3 ⫹ 4i)][x ⫺ (3 ⫺ 4i)]
80. Answer the questions in Problem 79 for n even.
55. [x ⫺ (4 ⫺ 5i)][x ⫺ (4 ⫹ 5i)]
58. [x ⫺ (5 ⫹ 2i)][x ⫺ (5 ⫺ 2i)]
81. Given P(x) ⫽ x2 ⫹ 2ix ⫺ 5 with 2 ⫺ i a zero, show that
2 ⫹ i is not a zero of P(x). Does this contradict Theorem
4? Explain.
59. [x ⫺ (a ⫹ bi)][x ⫺ (a ⫺ bi)]
60. (x ⫺ bi)(x ⫹ bi)
82. If P(x) and Q(x) are two polynomials of degree n, and if
P(x) ⫽ Q(x) for more than n values of x, then how are P(x)
and Q(x) related?
C
In Problems 61–66, find all other zeros of P(x), given the
indicated zero.
61. P(x) ⫽ x3 ⫺ 5x2 ⫹ 4x ⫹ 10; 3 ⫺ i is one zero
APPLICATIONS
62. P(x) ⫽ x3 ⫹ x2 ⫺ 4x ⫹ 6; 1 ⫹ i is one zero
Find all rational solutions exactly, and find irrational solutions
to two decimal places.
63. P(x) ⫽ x3 ⫺ 3x2 ⫹ 25x ⫺ 75; ⫺ 5i is one zero
64. P(x) ⫽ x3 ⫹ 2x2 ⫹ 16x ⫹ 32; 4i is one zero
83. Storage. A rectangular storage unit has dimensions 1 by 2
by 3 feet. If each dimension is increased by the same
amount, how much should this amount be to create a new
storage unit with volume 10 times the old?
65. P(x) ⫽ x4 ⫺ 4x3 ⫹ 3x2 ⫹ 8x ⫺ 10; 2 ⫹ i is one zero
66. P(x) ⫽ x4 ⫺ 2x3 ⫹ 7x2 ⫺ 18x ⫺ 18; ⫺3i is one zero
84. Construction. A rectangular box has dimensions 1 by 1
by 2 feet. If each dimension is increased by the same
amount, how much should this amount be to create a new
box with volume six times the old?
Prove that each of the real numbers in Problems 67–70 is not
rational by writing an appropriate polynomial and making use
of Theorem 6.
67. 兹6
3
69. 兹5
68. 兹12
★
85. Packaging. An open box is to be made from a rectangular
piece of cardboard that measures 8 by 5 inches, by cutting
out squares of the same size from each corner and bending
up the sides (see the figure). If the volume of the box is to
be 14 cubic inches, how large a square should be cut from
each corner? [Hint: Determine the domain of x from physical considerations before starting.]
★
86. Fabrication. An open metal chemical tank is to be made
from a rectangular piece of stainless steel that measures
10 by 8 feet, by cutting out squares of the same size from
each corner and bending up the sides (see the figure). If
the volume of the tank is to be 48 cubic feet, how large a
square should be cut from each corner?
5
70. 兹
8
In Problems 71–76, find all zeros (rational, irrational, and
imaginary) exactly.
71. P(x) ⫽ 3x3 ⫺ 37x2 ⫹ 84x ⫺ 24
72. P(x) ⫽ 2x3 ⫺ 9x2 ⫺ 2x ⫹ 30
73. P(x) ⫽ 4x4 ⫹ 4x3 ⫹ 49x2 ⫹ 64x ⫺ 240
74. P(x) ⫽ 6x4 ⫹ 35x3 ⫹ 2x2 ⫺ 233x ⫺ 360
75. P(x) ⫽ 4x4 ⫺ 44x3 ⫹ 145x2 ⫺ 192x ⫹ 90
76. P(x) ⫽ x5 ⫺ 6x4 ⫹ 6x3 ⫹ 28x2 ⫺ 72x ⫹ 48
77. The solutions to the equation x3 ⫺ 1 ⫽ 0 are all the cube
roots of 1.
(A) How many cube roots of 1 are there?
(B) 1 is obviously a cube root of 1; find all others.
78. The solutions to the equation x3 ⫺ 8 ⫽ 0 are all the cube
roots of 8.