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Solved by verified expert:Pick one concept presented this week. Discuss one real life scenario in which this concept is applicable, In other words, state one way in which you can use this concept in every day life.No need to make a big post. Keep it simple using simple use. I am attaching this week resource. (please don’t write on projectile motion or any other common examples….found on internet)Let me know if you will have any question?
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3
Polynomialand
Equations
Inequalities
and
Rational
Functions
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
1
3.3
Zeros of Polynomial Functions
• Factor Theorem
Rational Zeros Theorem
• Number of Zeros
• Conjugate Zeros Theorem
• Finding Zeros of a Polynomial Function
• Descartes’ Rule of Signs
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
2
Factor Theorem
For any polynomial function f(x), x – k
is a factor of the polynomial if and only
if (k) = 0.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
3
DECIDING WHETHER x – k IS A
FACTOR
Determine whether x – 1 is a factor of each polynomial.
Example 1
(a)
Solution By the factor theorem, x – 1 will be a
factor if (1) = 0. Use synthetic division and the
remainder theorem to decide.
Use a zero
coefficient for
the missing
term.
(1) = 7
The remainder is 7 and not 0, so
x – 1 is not a factor of (x).
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
4
DECIDING WHETHER x – k IS A
FACTOR OF
Determine whether x – 1 is a factor of each polynomial.
Example 1
(b)
Solution
(1) = 0
Because the remainder is 0, x – 1 is a factor.
Additionally, we can determine from the coefficients in
the bottom row that the other factor is
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
5
DECIDING WHETHER x – k IS A
FACTOR OF
Determine whether x – 1 is a factor of each polynomial.
Example 1
(b)
Solution
(1) = 0
Thus,
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
6
Example 2
FACTORING A POLYNOMIAL
GIVEN A ZERO
Factor
into linear
factors if – 3 is a zero of .
Solution Since – 3 is a zero of ,
x – (– 3) = x + 3 is a factor.
Use synthetic division to
divide (x) by x + 3.
The quotient is 6×2 + x – 1, which is the
factor that accompanies x + 3.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
7
Example 2
FACTORING A POLYNOMIAL
GIVEN A ZERO
Factor the following into linear factors if – 3 is
a zero of .
Solution
Factor 6×2 + x – 1.
These factors are all linear.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
8
Rational Zeros Theorem
If
is a rational number written in
lowest terms, and if is a zero of , a
polynomial function with integer
coefficients, then p is a factor of the
constant term and q is a factor of the
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
9
Proof of the Rational Zeros Theorem
Multiply by
qn and
subtract
a 0 q n.
Factor
out p.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
10
Proof of the Rational Zeros Theorem
This result shows that – a0qn equals the
product of the two factors p and
(anpn–1 +  + a1qn–1).
For this reason, p must be a factor of – a0qn.
Since it was assumed that is written in
lowest terms, p and q have no common
factor other than 1, so p is not a factor of qn.
Thus, p must be a factor of a0. In a similar
way, it can be shown that q is a factor of an.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
11
Example 3
USING THE RATIONAL ZEROS
THEOREM
Consider the polynomial function.
(a) List all possible rational zeros.
Solution For a rational number to be a
zero, p must be a factor of a0 = 2 and q
must be a factor of a4 = 6. Thus, p can be
1 or 2, and q can be 1, 2, 3, or 6.
The possible rational zeros, are
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
12
Example 3
USING THE RATIONAL ZEROS
THEOREM
Consider the polynomial function.
(b) Find all rational zeros and factor (x) into
linear factors.
Solution Use the remainder theorem to show
that 1 is a zero.
Use “trial and
error” to find
zeros.
(1) = 0
The 0 remainder shows that 1 is a zero. The quotient is
6×3 +13×2 + x – 2, so (x) = (x – 1)(6×3 +13×2 + x – 2).
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
13
Example 3
USING THE RATIONAL ZEROS
THEOREM
Consider the polynomial function.
(b) Find all rational zeros and factor (x) into
linear equations.
Solution Now, use the quotient polynomial
and synthetic division to find that – 2 is a zero.
(– 2 ) = 0
The new quotient polynomial is 6×2 + x – 1.
Therefore, (x) can now be completely factored.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
14
Example 3
USING THE RATIONAL ZEROS
THEOREM
Consider the polynomial function.
(b) Find all rational zeros and factor (x) into
linear equations.
Solution
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
15
Example 3
USING THE RATIONAL ZEROS
THEOREM
Consider the polynomial function.
(b) Find all rational zeros and factor (x) into
linear equations.
Solution Setting 3x – 1 = 0 and 2x + 1 = 0
yields the zeros ⅓ and – ½. In summary the
rational zeros are 1, – 2, ⅓, and – ½. The linear
factorization of (x) is
Check by
multiplying
these factors.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
16
Note In Example 3, once we obtained
the quadratic factor 6×2 + x – 1, we were
able to complete the work by factoring it
directly. Had it not been easily factorable,
we could have used the quadratic formula
to find the other two zeros (and factors).
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
17
Caution The rational zeros theorem gives only
possible rational zeros. It does not tell us whether
these rational numbers are actual zeros. We must rely
on other methods to determine whether or not they are
indeed zeros. Furthermore, the function must have
integer coefficients. To apply the rational zeros theorem
to a polynomial with fractional coefficients, multiply
through by the least common denominator of all the
fractions. For example, any rational zeros of p(x) defined
below will also be rational zeros of q(x).
Multiply the terms of
p(x) by 6.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
18
Fundamental Theorem of
Algebra
Every function defined by a polynomial
of degree 1 or more has at least one
complex zero.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
19
Fundamental Theorem of Algebra
From the fundamental theorem, if (x) is of
degree 1 or more, then there is some
number k1 such that f(k1) = 0. By the factor
theorem,
for some polynomial q1(x).
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
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Fundamental Theorem of Algebra
If q1(x) is of degree 1 or more, the
fundamental theorem and the factor
theorem can be used to factor q1(x) in the
same way. There is some number k2 such
that q1(k2) = 0, so
and
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
21
Fundamental Theorem of Algebra
Assuming that (x) has degree n and
repeating this process n times gives
where a is the leading coefficient of (x).
Each of these factors leads to a zero of (x),
so (x) has the same n zeros k1, k2, k3,…, kn.
This result suggests the number of zeros
theorem.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
22
Number of Zeros Theorem
A function defined by a polynomial of
degree n has at most n distinct zeros.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
23
Example 4
FINDING A POLYNOMIAL FUNCTION
THAT SATISFIES GIVEN CONDITIONS
(REAL ZEROS)
Find a polynomial function (x) of degree 3 with real
coefficients that satisfies the given conditions.
(a) Zeros of – 1, 2, and 4; (1) = 3
Solution These three zeros give
x – (– 1) = x + 1, x – 2, and x – 4 as factors of
(x). Since (x) is to be of degree 3, these
are the only possible factors by the number
of zeros theorem. Therefore, (x) has the
form
for some real number a.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
24
Example 4
FINDING A POLYNOMIAL FUNCTION
THAT SATISFIES GIVEN CONDITIONS
(REAL ZEROS)
Find a polynomial function (x) of degree 3 with real
coefficients that satisfies the given conditions.
(a) Zeros of – 1, 2, and 4; (1) = 3
Solution To find a, use the fact that (1) = 3.
Let x = 1.
(1) = 3
Multiply.
Divide by 6.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
25
Example 4
FINDING A POLYNOMIAL FUNCTION
THAT SATISFIES GIVEN CONDITIONS
(REAL ZEROS)
Find a polynomial function (x) of degree 3 with real
coefficients that satisfies the given conditions.
(a) Zeros of – 1, 2, and 4; (1) = 3
Solution Thus,
or
Multiply.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
26
Example 4
FINDING A POLYNOMIAL FUNCTION
THAT SATISFIES GIVEN CONDITIONS
(REAL ZEROS)
Find a polynomial function (x) of degree 3 with real
coefficients that satisfies the given conditions.
(b) – 2 is a zero of multiplicity 3; (– 1) = 4
Solution The polynomial function defined
by (x) has the following form.
Factor theorem
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
27
FINDING A POLYNOMIAL FUNCTION
THAT SATISFIES GIVEN CONDITIONS
(REAL ZEROS)
Example 4
Find a polynomial function (x) of degree 3 with real
coefficients that satisfies the given conditions.
(b) – 2 is a zero of multiplicity 3; (– 1) = 4
Solution To find a, use the fact that (–1) = 4.
Remember:
(x + 2)3 ≠ x3 + 23
Thus
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
28
Note In Example 4a, we cannot clear
the denominators in (x) by multiplying
each side by 2 because the result would
equal 2 • (x), not (x).
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
29
Properties of Conjugates
For any complex numbers c and d, the
following properties hold.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
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Conjugate Zeros Theorem
If (x) defines a polynomial function
having only real coefficients and if
z = a + bi is a zero of (x), where a
and b are real numbers, then
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
31
Proof of the Conjugate Zeros Theorem
where all coefficients are real numbers. If
the complex number z is a zero of (x), then
Take the conjugate of both sides of this
equation.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
32
Proof of the Conjugate Zeros Theorem
Using generalizations of the properties
Now use the property
that for any real number a,
and the fact
Hence is also a zero of (x),
which completes the proof.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
33
Caution When the conjugate zeros
theorem is applied, it is essential that
the polynomial have only real
coefficients. For example,
(x) = x – (1 + i)
has 1 + i as a zero, but the conjugate 1 – i
is not a zero.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
34
Example 5
FINDING A POLYNOMIAL FUNCTION THAT
SATISFIES GIVEN CONDITIONS (COMPLEX
ZEROS)
Find a polynomial function f(x) of least
degree having only real coefficients and
zeros 3 and 2 + i.
Solution The complex number 2 – i must
also be a zero, so the polynomial has at
least three zeros: 3, 2 + i, and 2 – i. For the
polynomial to be of least degree, these must
be the only zeros. By the factor theorem
there must be three factors, x – 3, x – (2 + i),
and x – (2 – i).
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
35
Example 5
FINDING A POLYNOMIAL FUNCTION THAT
SATISFIES GIVEN CONDITIONS (COMPLEX
ZEROS)
Find a polynomial function f(x) of least
degree having only real coefficients and
zeros 3 and 2 + i.
Solution
Remember:
i2 = – 1
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
36
Example 5
FINDING A POLYNOMIAL FUNCTION THAT
SATISFIES GIVEN CONDITIONS (COMPLEX
ZEROS)
Find a polynomial function of least degree
having only real coefficients and zeros 3 and
2 + i.
Solution
Any nonzero multiple of
x3 – 7×2 + 17x – 15 also satisfies the given
conditions on zeros. The information on
zeros given in the problem is not sufficient to
give a specific value for the leading
coefficient.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
37
Example 6
FINDING ALL ZEROS GIVEN ONE
ZERO
Find all zeros of (x) = x4 – 7×3 + 18×2 – 22x + 12,
given that 1 – i is a zero.
Solution Since the polynomial function has
only real coefficients and since 1 – i is a
zero, by the conjugate zeros theorem 1 + i is
also a zero. To find the remaining zeros,
first use synthetic division to divide the
original polynomial by x – (1 – i).
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
38
Example 6
FINDING ALL ZEROS GIVEN ONE
ZERO
Find all zeros of (x) = x4 – 7×3 + 18×2 – 22x + 12,
given that 1 – i is a zero.
Solution
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
39
Example 6
FINDING ALL ZEROS GIVEN ONE
ZERO
Find all zeros of (x) = x4 – 7×3 + 18×2 – 22x + 12,
given that 1 – i is a zero.
Solution By the factor theorem, since
x = 1 – i is a zero of (x), x – (1 – i) is a
factor, and (x) can be written as follows.
We know that x = 1 + i is also a zero of (x).
Continue to use synthetic division and divide the
quotient polynomial above by x – (1 + i).
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
40
Example 6
FINDING ALL ZEROS GIVEN ONE
ZERO
Find all zeros of (x) = x4 – 7×3 + 18×2 – 22x + 12,
given that 1 – i is a zero.
Solution
Now f(x) can be written in the following factored
form.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
41
Example 6
FINDING ALL ZEROS GIVEN ONE
ZERO
Find all zeros of (x) = x4 – 7×3 + 18×2 – 22x + 12,
given that 1 – i is a zero.
Solution
The remaining zeros are 2 and 3. The four
zeros are 1 – i, 1 + i, 2, and 3.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
42
Descartes’ Rule of Signs
Let (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 of  either equals the
number of variations in sign occurring in the coefficients of
(x), or is less than the number of variations by a positive
even integer.
(b) The number of negative real zeros of  either equals
the number of variations in sign occurring in the coefficients
of (– x), or is less than the number of variations by a
positive even integer.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
43
APPLYING DESCARTES’ RULE OF
SIGNS
Example 7
Determine the different possibilities for the
number of positive, negative, and nonreal
complex zeros of
Solution We first consider the possible
number of positive zeros by observing that 
(x) has three variations in signs:
1
2
3
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
44
Example 7
APPLYING DESCARTES’ RULE OF
SIGNS
Determine the different possibilities for the
number of positive, negative, and nonreal
complex zeros of
Solution Thus, by Descartes’ rule of signs, 
has either 3 or 3 – 2 = 1 positive real zeros.
For negative zeros, consider the variations in
signs for (– x):
1
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
45
Example 7
APPLYING DESCARTES’ RULE OF
SIGNS
Determine the different possibilities for the
number of positive, negative, and nonreal
complex zeros of
Solution
1
Since there is only one variation in sign, (x)
has exactly one negative real zero.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
46
3
Polynomial
and Rational
Functions
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
1
3.2 Synthetic Division
• Synthetic Division
• Remainder Theorem
• Potential Zeros of Polynomial Functions
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
2
Division Algorithm
Let (x) and g(x) be polynomials with g(x) of
lesser degree than (x) and g(x) of degree 1 or
more. There exist unique polynomials q(x) and
r(x) such that
where either r(x) = 0 or the degree of r(x) is less
than the degree of g(x).
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
3
Synthetic Division
Synthetic division provides an efficient
process for dividing a polynomial by a
binomial of the form x – k.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
4
Synthetic Division
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
5
Synthetic Division
Here the division
process is simplified
by omitting all
variables and writing
only coefficients, with
0 used to represent
the coefficient of any
missing terms.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
6
Synthetic Division
The numbers
in color that are
repetitions of
the numbers
directly above
them can be
omitted as
shown here.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
7
Synthetic Division
The entire process can now be condensed vertically.
The top row of numbers can be omitted since it
duplicates the bottom row if the 3 is brought down.
The rest of the bottom row is obtained by subtracting
– 12, – 40, and – 160 from the corresponding terms
above them.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
8
Synthetic Division
inverse
Signs
changed
Remainder
Quotient
To simplify the arithmetic, we replace subtraction in
the second row by addition and compensate by
changing the – 4 at the upper left to its additive
inverse, 4.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
9
Caution To avoid errors, use 0 as
the coefficient for any missing terms,
including a missing constant, when
setting up the division.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
10
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide.
Solution Express x + 2 in the form x – k by
writing it as x – (– 2).
Coefficients
x + 2 leads
to – 2
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
11
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
Solution Bring down the 5, and multiply:
– 2(5) = – 10
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
12
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
Solution Add – 6 and – 10 to obtain – 16.
Multiply – 2(– 16) = 32.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
13
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
Solution Add – 28 and 32, obtaining 4.
Finally, – 2(4) = – 8.
Be careful
with signs.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
14
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
Solution Add – 2 and – 8 to obtain – 10.
Remainder
Quotient
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
15
Example 1
USING SYNTHETIC DIVISION
Since the divisor x – k has degree 1, the
degree of the quotient will always be written
one less than the degree of the polynomial to
be divided. Thus,
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
16
Special Case of the Division
Algorithm
For any polynomial (x) and any
complex number k, there exists a
unique polynomial q(x) and number r
such that the following holds.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
17
For Example
The mathematical statement
illustrates the special case of the division
algorithm. This form of the division algorithm is
useful in developing the remainder theorem.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
18
Remainder Theorem
If the polynomial (x) is divided by
x – k, the remainder is equal to (k).
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
19
Remainder Theorem
In Example 1, when (x) = 5×3 – 6×2 – 28x – 2
was divided by x + 2, or x – (– 2), the remainder
was – 10. Substitute – 2 for x in (x).
Use parentheses
around substituted
values to avoid errors.
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
20
Remainder Theorem
An alternative way to find the value of a
polynomial is to use synthetic division. By the
remainder theorem, instead of replacing x by – 2
to find (– 2), divide (x) by x + 2 using synthetic
division as in Example 1. Then (– 2) is the
remainder, – 10.
(– 2)
Copyright © 2017, 2013, 2009 Pearson Education, Inc.
21
Example 2
APPLYING THE REMAINDER
THEOREM
Let (x) = – x …
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