101
Sect 3.3 – Zeros of Polynomials Functions
Objective 1:
Using the Factor Theorem.
Recall that a zero of a polynomial P(x) is a number c such that P(c) = 0. If
the remainder of P(x) ÷ (x – c) is zero, then c is a zero and x – c is factor of
P(x).
Factor Theorem
A number c is a zero of a polynomial P if and only if x – c is a factor of P.
Proof:
(⇒) Suppose c is a zero of P. Using the Division Algorithm, we can rewrite
P as: P(x) = Q(x)•(x – c) + R(x)
But, by the Remainder Theorem, R(x) = P(c) = 0 since c is a zero of P. This
implies that P(x) = Q(x)•(x – c) which means that x – c is a factor of P.
(⇐) Suppose x – c is a factor of P. Then P(x) = Q(x)•(x – c) for some
polynomial Q(x). If we evaluate P at x = c, we get:
P(c) = Q(c)•(c – c) = 0
This implies that c is a zero of P.
Show that the given values of c are zeros of P(x) and find all the other
zeros of P(x).
Ex. 1
P(x) = 2x4 + 9x3 – 9x2 – 46x + 24; c = – 3, 2
Solution:
First, evaluate P at x = – 3 and at x = 2:
P(– 3) = 2(– 3)4 + 9(– 3)3 – 9(– 3)2 – 46(– 3) + 24
= 162 – 243 – 81 + 138 + 24 = 0
P(2) = 2(2)4 + 9(2)3 – 9(2)2 – 46(2) + 24
= 32 + 72 – 36 – 92 + 24 = 0
Thus, – 3 and 2 are zeros of P so by the factor theorem, x + 3 and
x – 2 are factors of P. We will now use synthetic division to break
down P:
2 2
9
– 9 – 46
24
4
26
34 – 24
2 13
17 – 12
0
Thus, P(x) = (x – 2)(2x3 + 13x2 + 17x – 12)
–3
2 13
17 – 12
– 6 – 21
12
2
7
–4
0
102
So, P(x) = (x – 2)(2x3 + 13x2 + 17x – 12) = (x – 2)(x + 3)(2x2 + 7x – 4)
Now, we can factor 2x2 + 7x – 4 using trial and error:
2x2 + 7x – 4 = (2x – 1)(x + 4)
Therefore, P(x) = (x – 2)(x + 3)(2x – 1)(x + 4) and the zeros of P are
{– 4, – 3, ½, 2}
Show that the given values of c are zeros of P(x) and find all the other
zeros of P(x).
Ex. 2
P(x) = 4x4 – 19x3 – 21x2 + 76x + 20; c = – 2, 2
Solution:
First, evaluate P at x = – 2 and at x = 2:
P(– 2) = 4(– 2)4 – 19(– 2)3 – 21(– 2)2 + 76(– 2) + 20
= 64 + 152 – 84 – 152 + 20 = 0
P(2) = 4(2)4 – 19(2)3 – 21(2)2 + 76(2) + 20
= 64 – 152 – 84 + 152 + 20 = 0
Thus, – 2 and 2 are zeros of P so by the factor theorem, x + 2 and
x – 2 are factors of P. We will now use synthetic division to break
down P:
2 4 – 19 – 21
76
20
8 – 22 – 86 – 20
4 – 11 – 43 – 10
0
Thus, P(x) = (x – 2)(4x3 – 11x2 – 43x – 10)
– 2 4 – 11 – 43 – 10
–8
38
10
4 – 19
–5
0
So, P(x) = (x – 2)(4x3 – 11x2 – 43x – 10) = (x – 2)(x + 2)(4x2 – 19x – 5)
Now, we can factor 4x2 – 19x – 5 using trial and error:
4x2 – 19x – 5 = (4x + 1)(x – 5)
Therefore, P(x) = (x – 2)(x + 2)(4x + 1)(x – 5) and the zeros of P are
{– 2, – ¼, 2, 5}
Objective 2: Finding rational zeros of functions.
In the last example, P(x) = 4x4 – 19x3 – 21x2 + 76x + 20, the constant term
was 20 and the leading coefficient was 4. The factors p of the constant
term are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20 and the factors q of the leading
coefficient are ± 1, ± 2, ± 4. If we form all possible ratios of the factors p of
the constant term to the factors q of the leading coefficient, we get:
103
± 1, ± 2, ± 4, ± 5, ± 10, ± 20, ±
1
,
2
±
5
,
2
±
1
,
4
and ±
5
4
. This gives us all the
possible rational zeros of P(x). Notice that the zeros – ¼, – 2, 2, 5 are
contained in this list. This leads to the following theorem.
€ €
€
€
Rational Zeros Theorem
Let P(x) = anxn + an – 1xn – 1 + … + a1x + a0 be a polynomial of degree n with
p
p
integer coefficients. Then every rational zero of P is in the form
where
q
q
is in lowest terms and p is a factor of the constant term a0 and q is a factor
of the leading coefficient an.
Proof:
€
€
p
p
Let
be a zero of a P(x) and
be in lowest terms. Then,
q
q
P(
an
€
n
p
q
) = an(
pn
n
q
p n
)
q
+ an – 1
n–1
+ an – 1 (
pn −1
€
n −1
q
p n–1
)
q
+ … + a1
p
q
+ … + a1 (
+ a0 = 0
p
q
) + a0 = 0
(simplify)
(multiply both sides by qn)
n
n
an – 1p q +€… + a1pqn – 1 + a
0q = 0 (subtract anp from both sides)
€ an p + €
€
an – 1pn – 1q + … + a1pqn – 1 + a0qn = – anpn
(factor out q)
n–1
n–2
n–1
n
q(an – 1p
+ … + a1pq
+ a0 q ) = – an p
€
€This implies
€ that q is a factor
of – anpn. But since p and q are in lowest
terms, then q has to be a factor of an. We can use a similar argument to
show that p is a factor of a0.
Find all the rational zeros of the P(x) and then factor P(x) into linear
factors:
7
1
1
Ex. 3
P(x) = x4 + x3 – 2x2 – x +
6
2
3
Solution:
In order to use the Rational Zeros Theorem, all the coefficients have
to be integers. Since the LCD of the coefficients is 6, we will first
€ 1/6 from all€the terms:
€
factor out
7
1
1
P(x) = x4 + x3 – 2x2 – x +
(Rewrite with a denominator of 6)
=
=
€
€
6
2
3
6 4
7 3
12 2
3
2
x + x –
x – x+
6
6
6
6
6
1
(6x4 + 7x3 – 12x2 – 3x + 2)
6
€
€
€
(Factor out
1
)
6
Now, find the possible rational zeros of 6x4 + 7x3 – 12x2 – 3x + 2
€
€
€
€of p are ± 1, ± 2, €
The possible
values
and the possible values of q are
± 1, ± 2, ± 3, ± 6. Thus, the possible combinations of p over q are
104
±
1
,
1
±
1
,
2
1
,
3
±
±
1
,
6
±
2
,
1
2
2
±
,±
2
,
3
±
2
.
6
Simplifying and eliminating
duplication gives us the possible rational zeros of
1
1
1
2
± 1, ± , ± , ± , ± 2, ± .
2
€
3
6
3
use synthetic
to factor 6x4 + 7x3 – 12x2 – 3x + 2
€ Now,
€ we€will €
€
€ division
€
Let's try x = 1:
1
6
7 – 12
–3 2
€ € €
€6
13
1 –2
6 13
1 –2 0
Yes
Thus, P(x) =
1
(x
6
– 1)(6x3 + 13x2 + x – 2)
Let's try x = 1 again:
1
6 13
1
6 19
€
6 19 20
–2
20
18
No
Now, let's try x = – 1:
–1
6 13
1
–6 –7
6
7 –6
–2
6
4
No
1 –2
50 102
51 100
No
Let's try x = 2:
2
6 13
12
6 25
Now, let's try x = – 2:
–2
6 13
1
– 12 – 2
6
1 –1
Hence, P(x) =
1
(x
6
–2
2
0
Yes
– 1)(x + 2)(6x2 + x – 1)
Now, factor 6x2 + x – 1 using trial and error to get (3x – 1)(2x + 1).
1
Then, our answer is P(x) = (x – 1)(x + 2)(3x – 1)(2x + 1).
6
€
This suggests a procedure for finding the rational zeros of a polynomial.
€
105
Steps to Finding the Rational Zeros
1)
Use the Rational Zeros Theorem to list all possible rational
zeros.
2)
Use synthetic division evaluate each possible rational zero in
the polynomial. If the reminder is zero, write down the quotient
you found.
3)
Repeat the process for the quotient you found in part 2 until you
obtain a quotient that is degree two or easily factors. Factor or
use the quadratic formula to find the remaining zeros.
Find all the zeros of the P(x):
Ex. 4
P(x) = 2x4 – x3 – 67x2 – 159x – 63
Solution:
The possible values of p are ± 1, ± 3, ± 7, ± 9, ± 21, ± 63 and the
possible values of q are ± 1, ± 2. Thus, the possible combinations of
p over q are:
1
3
7
9
21
63
± 1, ± , ± 3, ± , ± 7, ± , ± 9, ± , ± 21, ±
, ± 63, ±
.
2
2
2
2
Now, we will use synthetic division
Let's try x = 1:
1
2 – 1 – 67 – 159 – 63
€
€
€2
€
1 € – 66 – 225
2
1 – 66 – 225 – 288
Try x = – 1
–1
2
2
Try x = 3:
3
2
2
Try x = – 3
–3
2
2
– 63
95
32
2
2
€
No
–1
–2
–3
– 67 – 159
3
64
– 64 – 95
–1
6
5
– 67 – 159 – 63
15 – 156 – 945
– 52 – 315 – 1008
No
–1
–6
–7
– 67 – 159
21
138
– 46 – 21
Yes
– 63
63
0
No
106
Try x = – 3 again.
–3
2 –7
–6
2 – 13
– 46
39
–7
– 21
21
0
Yes
Now, factor 2x2 – 13x – 7 using trial and error:
2x2 – 13x – 7 = (2x + 1)(x – 7).
Solving (2x + 1)(x – 7) = 0 yields x = – ½ and x = 7.
Thus, the zeros are {– 3, – ½, 7}.
If we had to write P(x) in the last example as a product of linear factors, our
answer would have been P(x) = (x + 3)2(2x + 1)(x – 7). Notice that the
factors (2x + 1) and (x – 7) occur only once. In such a case, we say that the
zeros – ½ and 7 have multiplicity of one. Since the factor (x + 3) occurs
twice, the zero – 3 has multiplicity of two.
Find all the real solutions to the following:
Ex. 5
f(x) = x4 + 2x3 + 3x2 – 2x – 4 = 0
Solution:
The possible values of p are ± 1, ± 2, ± 4 and the possible values of q
are ± 1. Thus, the possible combinations of p over q are
± 1, ± 2, ± 4.
Now, we will use synthetic division
Let's try x = 1:
1
1
2
3
–2
–4
1
3
6
4
1
3
6
4
0
Yes
Try x = – 1
–1
1
1
3
–1
2
6
–2
4
4
–4
0
Yes
Using the discriminant on x2 + 2x + 4 = 0, we get:
b2 – 4ac = (2)2 – 4(1)(4) = 4 – 16 = – 12
which tells us that x2 + 2x + 4 = 0 has no real solutions. Thus, the
only real solutions are {– 1, 1}. Each zero has multiplicity of one.
Suppose in the last example, if we were asked to find all solutions (real and
complex), we would have to use the quadratic formula to get the to
complex solutions:
x2 + 2x + 4 = 0
107
x=
−b± b2 −4ac
2a
=
−(2) ± (2)2 −4(1)(4)
2(1)
=
−2±
−12
2
=–1±
3i
3 i, – 1 + 3 i }. In factored
form, f(x) would be (x + 1)(x – 1)(x + 1 + 3 i)(x + 1 – 3 i). Notice that
€ real numbers, it
even
€ though f(x)€does not factor into€linear factors in the
does factor into linear factors in the€complex numbers.
Also, the complex
€
zeros are what we call complex conjugates.
€
€
So, all the solutions would be {– 1, 1, – 1 –
Complex Conjugates
If r = a + bi is a complex number, then r = a – bi is the complex conjugate.
Complex conjugate pairs have the following property:
r•r = (a + bi )(a – bi ) = a2 + b2
Thus, (5 + 3i )(5 – 3i ) = 25 – 15i + 15i – 9i 2 = 25 – 9(– 1) = 25 + 9 = 34.
r and r are called conjugate pairs.
Objective 3: Finding complex zeros of polynomials.
The depressed equation x2 + 2x + 4 = 0 from the last example had no real
solutions. Such quadratic equations are called irreducible if they cannot be
factored over the real numbers. Any irreducible quadratic equation in the
real numbers can be factored in the complex numbers. This suggests that
every complex non-constant polynomial has at least one complex zero.
This is known as the Fundamental Theorem of Algebra.
Fundamental Theorem of Algebra
Every complex polynomial function with degree n ≥ 1 has at least one
complex zero.
Using this theorem in conjunction with the Factor Theorem leads to the
following theorems:
Theorem
Every complex polynomial function f of degree n ≥ 1 can be factored into n
linear factors (not necessarily distinct) of the form:
f(x) = an(x – r1)(x – r2)•…•(x – rn)
where an, r1, r2, …, rn are complex numbers. In other words, every nonconstant polynomial has exact n complex zeros including repeating zeros.
Conjugate Pairs Theorem
Let f be a polynomial function whose coefficients are real numbers. If
r = a + bi is a zero of f, then the complex conjugate r = a – bi is also a
zero of f so long as b ≠ 0.
108
Find all the complex zeros of the P(x). Then write P in factored form:
Ex. 6
P(x) = x4 + 2x3 + 22x2 + 50x – 75
Solution:
The possible values of p are ± 1, ± 3, ± 5, ± 15, ± 25, ± 75 and the
possible values of q are ± 1. Thus, the possible combinations of
p over q are ± 1, ± 3, ± 5, ± 15, ± 25, ± 75 which gives us the possible
rational zeros. Now, we will use synthetic division:
Let's try x = 1:
1
1
2
22
50 – 75
1
3
25
75
1
3
25
75
0
Yes
Try x = – 1
–1
1
1
Try x = 3:
3
1
1
Try x = – 3
–3
1
1
3
–1
2
25
–2
23
75
– 23
52
No
3
3
6
25
18
43
75
129
204
No
3
–3
0
25
0
25
75
– 75
0
Yes
Now, factor x2 + 25 in the complex numbers:
x2 + 25 = (x + 5i )(x – 5i )
Solving (x + 5i )(x – 5i ) = 0 yields x = ± 5i .
Thus, the zeros are {– 5i , 5i , – 3, 1}.
As a product of linear factors, we get:
P(x) = (x + 5i )(x – 5i )(x + 3)(x – 1)
Solve the following:
Ex. 7
f(x) = 2x4 + 5x3 + 9x2 + 2x – 4 = 0
Solution:
The possible values of p are ± 1, ± 2, ± 4 and the possible values of q
are ± 1, ± 2. Thus, the possible combinations of p over q are:
109
± 1, ±
1
,
2
± 2, ± 4.
Now, we will use synthetic division
Let's try x = 1:
€
1
2
5
9
2
2
7
16
2
7
16
18
Try x = – 1
–1
2
5
–2
3
9
–3
6
2
–6
–4
Try x = – 1 again
–1
2
3
–2
2
1
6
–1
5
–4
–5
–9
2
Try x =
1
2
€
–4
18
14
No
–4
4
0
Yes
No
1
2
2
3
6
–4
2
1
4
2
8
4
0
Yes
€ Working with the depressed equation 2x2 + 4x + 8 = 0, we can factor
out a 2 to get 2(x2 + 2x + 4) and then we will need to use the
quadratic formula on x2 + 2x + 4:
x=
−b± b2 −4ac
2a
=
−(2) ± (2)2 −4(1)(4)
2(1)
So, the solution is {– 1 –
3i,–1+
=
−2±
−12
2
3 i , – 1,
=–1±
1
}.
2
3i
If we had to write f(x) in factored form, we would get:
€
1
€
f(x) = 2(x –€(– 1 – 3 i ))(x – (–€1 + 3 i ))(x + 1)(x – )
2
€
€
€
Given the zeros of a polynomial, we can construct a polynomial that would
have those zeros. Let's start with the answer from example #2 and work
€ a polynomial. €
backwards to find
€
110
Find a 4th degree polynomial with the given zeros:
Ex. 8
– ¼, – 2, 2, 5
Solution:
Since – ¼, – 2, 2, & 5 are zeros, then (x – (– ¼)) = (x + ¼) ,
(x – (– 2)) = (x + 2), (x – 2), & (x – 5) are factors. But saying (x + ¼) is
a factor is also equivalent to saying (4x + 1) is a factor since they
have the "same zero." Thus, P is the product of these factors:
P(x) = (4x + 1)(x + 2)(x – 2)(x – 5)
But by FOIL, (4x + 1)(x + 2) = 4x2 + 9x + 2 and
(x – 2)(x – 5) = x2 – 7x + 10
Thus,
P(x) = (4x + 1)(x + 2)(x – 2)(x – 5) = (4x2 + 9x + 2)(x2 – 7x + 10)
Multiplying the two trinomials, we get:
4x2 + 9x + 2
x2 – 7x + 10
40x2 + 90x + 20
– 28x3 – 63x2 – 14x
4x4 + 9x3 + 2x2
4x4 – 19x3 – 21x2 + 76x + 20
Hence, P(x) = 4x4 – 19x3 – 21x2 + 76x + 20.
Note that if we multiply 4x4 – 19x3 – 21x2 + 76x + 20 by any non-zero
constant, we will get another 4th degree polynomial that has the same
zeros. This means that there are an infinite number of answers for the last
example; all such answers are constant multiples of one and other.
Find a 4th degree polynomial f satisfying the following conditions:
1
Ex. 9
Zeros: 4 (multiplicity of 2) , – (multiplicity of 1), and
2
1
3
(multiplicity of 1). f(– 2) = – 189
Solution:
Since 4, –
€ & (x –
1
),
3
1
,
2
&
1
3
€ then (x – 4), (x – (–
are zeros,
are factors. But saying (x +
1
))
2
= (x +
1
),
2
1
) is a factor is also equivalent
2
1
) is a factor is equivalent to
3 €
€
to saying (2x + 1) is a factor and (x –
€
€
saying (3x – 1) is a factor since they have the "same zero." Since the
2, the factor (x – 4) will occur twice.
€ zero of 4 has a multiplicity of €
Thus, f is the product of these factors:
€ – 1)
f(x) = (x – 4)2(2x + 1)(3x
111
But by FOIL, (x – 4)(x – 4) = x2 – 8x + 16 and
(2x + 1)(3x – 1) = 6x2 + x – 1
Thus, f(x) = (x – 4)2(2x + 1)(3x – 1) = (x2 – 8x + 16)(6x2 + x – 1)
Multiplying the two trinomials, we get:
x2 – 8x + 16
6x2 + x – 1
– x2 + 8x – 16
x3 – 8x2 + 16x
6x4 – 48x3 + 96x2
6x4 – 47x3 + 87x2 + 24x – 16
So, f(x) = a(6x4 – 47x3 + 87x2 + 24x – 16). But, f(– 2) = – 189, which
says: f(– 2) = a(6(– 2)4 – 47(– 2)3 + 87(– 2)2 + 24(– 2) – 16) = – 189
a(96 + 376 + 348 – 48 – 16) = – 189
756a = – 189 or a = – ¼
Hence, f(x) = – ¼(6x4 – 47x3 + 87x2 + 24x – 16)
3
47 3
87 2
= – x4 +
x –
x – 6x + 4
2
4
4
Find a polynomial function with the given information:
€
€ Zeros: – 2, 3 – 5i , 4i
Ex. 10 €
Degree
5;
Solution:
Since the polynomial has a degree of 5, then it has five zeros.
The conjugate pair of 4i is – 4i which is also a zero and the
conjugate pair of 3 – 5i is 3 + 5i which is a zero as well. So, the
missing zeros are – 4i and 3 + 5i .
This means that (x – (– 2)), (x – (3 – 5i )), (x – (3 + 5i )), (x – (– 4i )),
& (x – (4i )) are factors. Thus, f can be written as:
f(x) = a(x – (– 2))(x – (3 – 5i ))(x – (3 + 5i ))(x – (– 4i ))(x – (4i ))
where a is any nonzero real number.
f(x) = a(x + 2)[x – (3 – 5i )][x – (3 + 5i )](x + 4i )(x – 4i ) (expand)
f(x) = a(x + 2)[x2 – (3 – 5i + 3 + 5i )x + (3 – 5i )(3 + 5i )](x2 – 16i 2)
f(x) = a(x + 2)[x2 – 6x + 9 – 25i 2](x2 – 16i 2) (replace i 2 by – 1)
f(x) = a(x + 2)[x2 – 6x + 9 + 25](x2 + 16)
f(x) = a(x + 2)(x2 – 6x + 34)(x2 + 16)
f(x) = a(x3 – 4x2 + 22x + 68)(x2 + 16)
f(x) = a(x5 – 4x4 + 38x3 + 4x2 + 352x + 1088)