Transcript Zeros of Polynomial Functions

Zeros of Polynomial
Functions
Objectives:
1.Use the Fundamental Theorem of Algebra to determine the
number of zeros of polynomial functions
2.Find rational zeros of polynomial functions
3.Find conjugate pairs of complex zeros
4.Find zeros of polynomials by factoring
5.Use Descartes’s Rule of Signs and the Upper and Lower
Bound Rules to find zeros of polynomials
WHY???
Finding the zeros of a
polynomial function can
help you analyze the
attendance at women’s
college basketball games.
In the complex number system, every nth-degree polynomial has
precisely “n” zeros.
Fundamental Theorem of Algebra
If f(x) is a polynomial of degree n, where n > 0, then f has at least one zero
in the complex number system
Linear Factorization Theorem
If f(x) is a polynomial of degree n, where n > 0, then f has precisely n
linear factors
f (x)  an (x  c1)(x  c2 )...(x  cn )
where
c1,c2,....,cn
are complex numbers
Zeros of Polynomial Functions
Give the degree of the polynomial, tell how many zeros there are, and
find all the zeros
f (x)  x  2
f (x)  x 2  6x  9
f (x)  x 3  4 x
f (x)  x 4 1
Rational Zero Test
To use the Rational Zero Test, you should list all rational numbers
whose numerators are factors of the constant term and whose
denominators are factors of the leading coefficient
factors_of _constan t _ term
Possible _ Rational _ Zeros 
factors_of _ leading _coefficient
Once you have all the possible zeros test them using substitution or
synthetic division to see if they work and indeed are a zero of the
function (Also, use a graph to help determine zeros to test)
It only test for rational numbers
EXAMPLE: Using the Rational Zero Theorem
List all possible rational zeros of f(x)  15x3  14x2  3x – 2.
Solution
The constant term is –2 and the leading coefficient is 15.
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5

1
,

2


1
,3

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
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
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1
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,
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, 
, 2
, 
, 2
, 
,
5
5
3
3
1
5
1
5
Divide 1
and 2
by 1.
Divide 1
and 2
by 3.
Divide 1
and 2
by 5.
Divide 1
and 2
by 15.
There are 16 possible rational zeros. The actual solution set to f(x)  15x3 
14x2  3x – 2 = 0 is {-1, 1/3, 2/5}, which contains 3 of the 16 possible solutions.
Roots & Zeros of Polynomials II
Finding the Roots/Zeros of Polynomials (Degree 3 or higher):
• Graph the polynomial to find your first zero/root
•
Use synthetic division to find a smaller polynomial
• If the polynomial is not a quadratic follow the 2 steps
above using the smaller polynomial until you get a
quadratic.
• Factor or use the quadratic formula to find your
remaining zeros/roots
Example 1:
Find all the zeros of each polynomial function
First, graph the equation to
1
0
x

9
xx

1
9

6
find the first zero
3 2
ZERO
From looking at the graph
you can see that there is a
zero at -2
Example 1 Continued
Second, use the zero you
1
0
x

9
xx

1
9

6
3 2
found from the graph and do
synthetic division to find a
smaller polynomial
-2
10
9
-19
6
-20 22
-6
10 -11
3
The new, smaller polynomial is:
Don’t forget your
remainder should be zero
0
1
01
x

1
x
3
2
Example 1 Continued:
1
01
x

1
x
3
2
Third, factor or use the quadratic
formula to find the remaining
zeros.
This quadratic can be factored into:
(5x – 3)(2x – 1)
are:
1
0
x

9
xx

1
9

6
31
x
2, ,
52
Therefore, the zeros to the problem
3 2
Rational Zeros
Find the rational zeros.
f (x)  x 3  x  1
f (x)  x 4  x 3  x 2  3x  6
f (x)  x  8x  40x  525
3
2
f (x)  2x  3x  8x  3
3
2
Find all the real zeros (Hint: start by finding the rational zeros)
f (x)  10x  15x  16x 12
3
2
f (x)  3x 3 19x 2  33x  9
Writing a Polynomial given the zeros.
To write a polynomial you must write the zeros out in
factored form. Then you multiply the factors together
to get your polynomial.
Factored Form: (x – zero)(x – zero). . .
***If P is a polynomial function and a + bi is a root,
then a – bi is also a root.
***If P is a polynomial function and a bis a root,
then a b is also a root
Example 1:
The zeros of a third-degree polynomial are 2
(multiplicity 2) and -5. Write a polynomial.
(x – 2)(x – 2)(x – (-5)) = (x – 2)(x – 2)(x+5)
Second, multiply the factors out to
find your polynomial
First, write the
zeros in
factored form
Example 1 Continued
(x – 2)(x – 2)(x+5)
(
x

2
)
(
xx

2
)


4
x

4
2
x
4
x4
2
X
5
4x2 4x
2
20x 20
5x 
x
3
First FOIL or box
two of the factors
Second, box your answer from
above with your remaining
factors to get your polynomial:
x

x

1
6
x

2
0
32
ANSWER
So if asked to find a polynomial that has zeros, 2 and
1 – 3i, you would know another root would be 1 + 3i.
Let’s find such a polynomial by putting the roots in
factor form and multiplying them together.
If x = the root then
x - the root is the factor form.










x

2
x

1

3
i
x

1

3
i

 
 
Multiply the last two factors
x

2
x

1

3
i
x

1

3
i
together. All i terms should
disappear when simplified.




x

2
x

x

3
xi

x

1

3
i

3
xi

3
i

9
i
2

2




x

2
x

2
x

10
Now multiply the x – 2 through
2
-1
3 2

x

4
x

14
x

20
Here is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i
Conjugate Pairs
Complex Zeros Occur in Conjugate Pairs = If a + bi is a zero of the
function, the conjugate a – bi is also a zero of the function
(the polynomial function must have real coefficients)
EXAMPLES: Find a polynomial with the given zeros
-1, -1, 3i, -3i
2, 4 + i, 4 – i
STEPS For Finding the Zeros
given a Solution
1) Find a polynomial with the given solutions
(FOIL or BOX)
2) Use long division to divide your polynomial you
found in step 1 with your polynomial from the
problem
3) Factor or use the quadratic formula on the
answer you found from long division.
4) Write all of your answers out
Find Roots/Zeros of a Polynomial
If the known root is imaginary, we can use the Complex
Conjugates Thm.
Ex: Find all the roots of
If one root is 4 - i.
32
f
(
x
)

x

5
x

7
x

51
Because of the Complex Conjugate Thm., we know that
another root must be 4 + i.
Example (con’t)
Ex: Find all the roots of
If one root is 4 - i.
32
f
(
x
)

x

5
x

7
x

51
If one root is 4 - i, then one factor is [x - (4 - i)], and
Another root is 4 + i, & another factor is [x - (4 + i)].
Multiply these factors:
[
x

(
4

i
)
]
[
x

(
4

i
)
]

(
x

4

i
)
(
x

4

i
)
X
x

4i
x 2 4x ix
-4
4x 16 4i
-i
ix 4i i2

x

8
x

1
7
2
Example (con’t)
32
(
x
)

x

5
x

7
x

51
Ex: Find all the roots of f
If one root is 4 - i.
2
If the product of the two non-real factors is x

8
x

17
then the third factor (that gives us the real root) is the
2
x

8
x

17
quotient of P(x) divided by
x
3
2
3
2
x

8
x
17
x

5
x

7
x
51 The third root
3
2
x

5
x

7
x
51 is x = -3
0

So, all of the zeros are: 4 – i, 4 + i, and -3
FIND ALL THE ZEROS
f (x)  x 4  3x 3  6x 2  2x  60
(Given that 1 + 3i is a zero of f)
f (x)  x 3  7x 2  x  87
(Given that 5 + 2i is a zero of f)
More Finding of Zeros
f (x)  x  x  2x 12x  8
5
3
2
f (x)  3x 3  4 x 2  8x  8

Descartes’s Rule of Signs
n
n1
2
f
(x)

a
x

a
x

...
a
x
 a1 x  a0 be a polynomial
Let
n
n1
2
with real coefficients and a0  0
The number of positive real zeros of f is either equal to the number of
variations in sign of f(x) or less than that number by an even integer
The number of negative real zeros of f is either equal to the number of
variations in 
sign of f(-x) or less than that number by an even integer
Variation in sign = two consecutive coefficients have opposite signs
EXAMPLE:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f(x)  x3  2x2  5x + 4.
Solution
1. To find possibilities for positive real zeros, count the number of sign
changes in the equation for f(x). Because all the terms are positive, there
are no variations in sign. Thus, there are no positive real zeros.
2. To find possibilities for negative real zeros, count the number of sign
changes in the equation for f(x). We obtain this equation by replacing x
with x in the given function.
f(x)  x3  2x2 5x + 4
Replace x with x.
f(x)  (x)3  2(x)2 5x4
 x3  2x2  5x + 4
This is the given polynomial function.
EXAMPLE:
Using Descartes’ Rule of Signs
Determine the possible number of positive and negative real zeros of
f(x)  x3  2x2  5x + 4.
Solution
Now count the sign changes.
f(x)  x3  2x2  5x + 4
1
2
3
There are three variations in sign.
# of negative real zeros of f is either equal to 3, or is less than this number by
an even integer.
This means that there are either 3 negative real zeros
or 3  2  1 negative real zero.
Descartes’s Rule of Signs
EXAMPLES: describe the possible real zeros
f (x)  3x 3  5x 2  6x  4
f (x)  3x 3  2x 2  x  3
Upper & Lower Bound Rules
Let f(x) be a polynomial with real coefficients and a positive leading
coefficient. Suppose f(x) is divided by x – c, using synthetic didvision
If c > 0 and each number in the last row is either positive or zero, c is an
upper bound for the real zeros of f
If c < 0 and the numbers in the last row are alternately positive and
negative (zero entries count as positive or negative), c is a lower bound
for the real zeros of f
EXAMPLE: find the real zeros

f (x)  6x  4 x  3x  2
3
2
h(x) = x4 + 6x3 + 10x2 + 6x + 9
3,
9
Factors
of
9 1,

1
Factors
of
1
1
1
6
10
2
6
9
4
6
0
1
4
6
0
9
Signs are all positive, therefore 1 is an upper
bound.
EXAMPLE
You are designing candle-making kits. Each kit contains 25 cubic
inches of candle wax and a model for making a pyramid-shaped
candle. You want the height of the candle to be 2 inches less than the
length of each side of the candle’s square base. What should the
dimensions of your candle mold be?