Download 3.5 More on Zeros of Polynomial Functions

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3.5
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More on Zeros of Polynomial Function
 Rational Zero Theorem
 Descartes’ Rule of Sign
 Boundedness Theorem
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Introduction
Recall that the fundamental theorem of algebra states
that a polynomial function of degree 𝑛 has exactly 𝑛
zeros (counting multiplicity).
The rational zero theorem will enable us to make a list
of all possible rational zeros. While, Descartes rule of
signs will enable us to determine the number of
positive and negative real zeros.
The boundedness theorem maybe used to provide
upper, 𝑈, and lower, 𝐿, bounds on the zeros of a given
polynomial 𝑃 𝑥 of degree ≥ 1.
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Rational Zero (or Root) Theorem
For a rational number
𝑃
𝑞
to be a zero of a polynomial
function
𝑓 𝑥 = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0
with integer coefficients 𝑎𝑛 , 𝑎𝑛−1 , … , 𝑎1 , 𝑎0 and 𝑎0 ≠
0, we must have that 𝑃 is a factor of 𝑎0 and 𝑞 is a
factor of 𝑎𝑛 .
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Rational Zero (or Root) Theorem
This theorem enables us to make a list of all possible
real zeros using the following steps:
Step 1: list all the factors of 𝑎0 , say
𝑠 = ±𝑃1 , ±𝑃2 , … , ±𝑃𝑛
Step 2: list all positive factors of 𝑎𝑛 , say
𝑅 = 𝑞1 , 𝑞2 , … , 𝑞𝑠
Step 3: make list of all possible rational zeros of 𝑓 𝑥
±𝑃1
±𝑃1 ±𝑃2
±𝑃2
±𝑃𝑟
±𝑃𝑟
𝑧=
,…,
,
,…,
,…,
,…,
𝑞1
𝑞𝑠 𝑞1
𝑞𝑠
𝑞1
𝑞𝑠
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Conjugate Zero Theorem
Each zero 𝑧𝑗 of the polynomial function
𝑃 𝑥 = 𝑎𝑛 𝑥 − 𝑧1 𝑚1 𝑥 − 𝑧2 𝑚2 … 𝑥 − 𝑧𝑟 𝑚𝑟
may be real or complex, and if 𝑧𝑗 = 𝛼 + 𝛽𝑖 is a nonreal complex zero of
𝑃 𝑥 = 𝑎𝑛 𝑥 𝑛 + ⋯ + 𝑎1 𝑥 + 𝑎0
then 𝑧𝑗 = 𝛼 − 𝛽𝑖 is a non-real complex zero of 𝑃 𝑥
provided that 𝑎𝑛 , 𝑎𝑛−1 , … , 𝑎𝑛 are all real numbers.
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Example 1
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Make a list of all possible rational
zeros of 𝑓 𝑥 = −𝑥 5 + 4𝑥 4 + 5𝑥 − 12
and
𝑓 𝑥 = −6𝑥 8 + 5𝑥 4 + 3𝑥 − 20
(a)
𝑠 = ±1, ±2, ±3, ±4, ±6, ±12
𝑅 = 1 , and
𝑧 = ±1, ±2, ±3, ±4, ±6, ±12
(b)
𝑠 = ±1, ±2, ±4, ±5, ±10, ±20
𝑅 = 1,2,3,4 , and
1
5
𝑧 = ±1, ±2, ±4, ±5, ±10, ±20, ± , ± ,
2
2
1
2
4
5
10
20
1
5
± ,± ,± ,± ,± ,± ,± ,± }
3
3
3
3
3
3
6
6
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Find all the zeros of
𝑓 𝑥 = 𝑥 4 − 5𝑥 3 + 20𝑥 − 16
Example 2
First, we make the list z of all possible zeros of 𝑓 𝑥
𝑧 = ±1, ±2, ±4, ±8, ±16
Using synthetic division we find
1
1
1
−5
1
−4
0
−4
−4
20
−4
16
− 16
16
0
Therefore, 1 is a zero of 𝑓 𝑥 , and 𝑓 𝑥 factors into
𝑓 𝑥 = 𝑥 − 1 𝑥 3 − 4𝑥 2 − 4𝑥 + 16 .
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Example 2
Find all the zeros of
𝑓 𝑥 = 𝑥 4 − 5𝑥 3 + 20𝑥 − 16
Therefore, the rest of the zeros of 𝑓 𝑥 are those of
𝑔 𝑥 = 𝑥 3 − 4𝑥 2 − 4𝑥 + 16
Going through the list of z again, we find that
2
1
−4
2
−2
−4
16
− 4 − 16
−8
0
1
2 is a zero of 𝑔 𝑥 .
Furthermore, it follows that
𝑓 𝑥 = 𝑥 − 1 𝑥 − 2 𝑥 2 − 2𝑥 − 8 .
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Example 2
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Find all the zeros of
𝑓 𝑥 = 𝑥 4 − 5𝑥 3 + 20𝑥 − 16
Thus
ℎ 𝑥 = 𝑥 2 − 2𝑥 − 8
which is a quadratic polynomial, whose zeros can be
found by factoring or using the quadratic formula. We
find that
ℎ 𝑥 = 𝑥−4 𝑥+2 .
And therefore,
𝑓 𝑥 = 𝑥−1 𝑥−2 𝑥−4 𝑥+2
And the zeros of 𝑓 𝑥 are −2, 1, 2 and 4.
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Descartes’ Rule of Signs
This theorem applies to any polynomial 𝑓 𝑥 with real
coefficients, and states that:
Part 1: The number of positive zeros of 𝑓 𝑥 is equal
to the number of sign changes in 𝑓 𝑥 or less by an
even integer,
Part 2: The number of negative zeros of 𝑓 𝑥 is equal
to the number of sign changes in 𝑓 −𝑥 or less by an
even integer.
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Example 3
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Use the Descartes’ rule of signs to
determine the number of positive
and negative zeros of
𝑓 𝑥 = 𝑥4 + 𝑥2 + 1
𝑓 𝑥 = 𝑥 4 + 𝑥 2 + 1 has no positive zeros
(since 𝑓 𝑥 has no changes in signs)
Also,
𝑓 −𝑥 = 𝑥 4 + 𝑥 2 + 1
𝑓 −𝑥 has no changes in signs, then 𝑓 𝑥 does not
have any negative zero either. Therefore, 𝑓 𝑥 doesn’t
have any real zeros.
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Example 4
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Use the Descartes’ rule of signs to
determine the number of positive
and negative zeros of
𝑓 𝑥 = 𝑥 5 − 5𝑥 3 + 3𝑥 2 − 𝑥 − 10
𝑓 𝑥 = 𝑥 5 − 5𝑥 3 + 3𝑥 2 − 𝑥 − 10
has four sign changes. Therefore, 𝑓 𝑥 either has 4, 2
or 0 positive zeros.
Also
𝑓 −𝑥 = −𝑥 5 + 5𝑥 3 + 3𝑥 2 + 𝑥 + 10
has only one sign change, 𝑓 𝑥 must have exactly one
negative zero.
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Boundedness Theorem
1. If when dividing 𝑓 𝑥 by 𝑥 − 𝑐1 for 𝑐1 > 0 using
synthetic division, the sign of the terms in the bottom
row are all non-negative, then 𝑐1 is an upper bound
for all the real zeros of 𝑓 𝑥 . That is, for any real zero
z of 𝑓 𝑥 we have 𝑧 ≤ 𝑐1 .
2. If when dividing 𝑓 𝑥 by 𝑥 − 𝑐2 , for 𝑐2 < 0 using
synthetic division the signs in the bottom row
alternate between non-negative and non-positive then
𝑐2 is a lower bound for all the real zeros of 𝑓 𝑥 . That
is, for any real zero z of 𝑓 𝑥 we have: 𝑐2 ≤ 𝑧
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Show that the real zeros of
Example 5
𝑓 𝑥 = −𝑥 5 + 5𝑥 3 + 10𝑥 2 − 12𝑥 + 20
are all between −4 and 4.
We first make the leading coefficient positive. Consider
𝑔 𝑥 = −𝑓 𝑥 = 𝑥 5 − 5𝑥 3 − 10𝑥 2 + 12𝑥 − 20.
We divide 𝑔 𝑥 by 𝑥 − 4 using synthetic division
4
1
0
− 5 − 10
12 − 20
4
16
44 136
592
1
4
11
34 148
572
Whose bottom row consists of non-negative numbers.
Therefore, any real zero z of 𝑔 𝑥 must satisfy:
𝑧 ≤ 4.
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Example 5 continue
We next divide 𝑔 𝑥 by 𝑥 + 4 using synthetic division
−4
1
0 − 5 − 10
12 − 20
−4
16 − 44 216 − 912
1 −4
11 − 54 228 − 932
Whose bottom row consists of alternating numbers
(from non-negative to non-positive or vice versa).
Therefore, any real zero z of 𝑔 𝑥 must satisfy:
𝑧 ≥ −4
Therefore, we conclude that all the zeros of 𝑓 𝑥 lies
within the interval −4, 4
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