Zeros of Polynomial Functions - PowerPoint PPT Presentation

1 / 47
About This Presentation
Title:

Zeros of Polynomial Functions

Description:

Zeros of Polynomial Functions Pre-Calculus Day 17 Example 1 Find all rational roots of 8x3 + 10x2 11x + 2 = 0. Step 3: Use substitution or synthetic division to ... – PowerPoint PPT presentation

Number of Views:1277
Avg rating:3.0/5.0
Slides: 48
Provided by: Jer886
Category:

less

Transcript and Presenter's Notes

Title: Zeros of Polynomial Functions


1
Zeros of Polynomial Functions
  • Pre-Calculus
  • Day 17

2
Plan for the day
  • Quiz polynomials
  • Review Homework
  • Review of ideas learned so far about polynomials
  • A couple of Theorems
  • Factor Theorem
  • Remainder Theorem
  • Finding the zeros of polynomials
  • Rational Zero test
  • Using Division
  • Descartes Rule of Signs
  • Homework

3
What you should learn
  • Find rational zeros of polynomial functions.
  • Find conjugate pairs of complex zeros.
  • Find zeros of polynomials by factoring.
  • Use Descartes Rule of Signs

4
Analyzing a Quadratic Equation
  • Useful formats of quadratic equations
  • f (x) x2 6x 2
  • g(x) 2(x 1)2 3
  • m(x) (x 2)(x 1)

5
Quadratic Function in Standard Form
Example Graph the parabola f (x) 2x2 4x 1
and find the axis and vertex.
f (x) 2x2 4x 1
f (x) 2x2 4x 1 original equation
f (x) 2( x2 2x) 1 factor out 2
f (x) 2( x2 2x 1) 1 2 complete the
square
f (x) 2( x 1)2 3 standard form
a gt 0 ? parabola opens upward like y 2x2.
h 1, k 3 ? axis x 1, vertex (1, 3).
6
Vertex of a Parabola
Vertex of a Parabola
Example Find the vertex of the graph of f (x)
x2 10x 22.
f (x) x2 10x 22 original equation
a 1, b 10, c 22
So, the vertex is (5, -3).
7
Example Basketball
Example A basketball is thrown from the free
throw line from a height of six feet. What is the
maximum height of the ball if the path of the
ball is
The path is a parabola opening downward. The
maximum height occurs at the vertex.
So, the vertex is (9, 15).
The maximum height of the ball is 15 feet.
8
Polynomials
  • What shape does it have?
  • What are some key attributes?
  • Given key attributes, what is the polynomial?

9
Exploring End Behavior using the degree (n) and
leading coefficient (a)
Rise or Fall???
a gt 0 a gt 0 a lt 0 a lt 0
left right left right
n is even
n is odd
10
Review
  • Previously, we learned that the maximum number of
    real zeros of a polynomial is equal to its
    degree.
  • The maximum number of turns a polynomial can have
    will be one less than the degree.
  • A polynomial can have roots that are repeated,
    known root multiplicity.

11
The Fundamental Theorem of Algebra
  • The Fundamental Theorem of Algebra states that
    every polynomial has exactly as many complex
    roots as its degree, counting root multiplicity.
  • Remember, the set of complex numbers includes all
    numbers real and imaginary.
  • Any roots of a polynomial that arent real are
    imaginary.

12
Example
  • If a polynomial has a degree of 9, then it will
    have a total of nine complex solutions.
  • Possible examples
  • 9 real
  • 5 real, 4 imaginary
  • 3 real (each with a multiplicity of 2), 1 real
    (multiplicity of 1), 2 imaginary
  • If asked, you should be able to find all the
    roots of a polynomial.
  • If you can factor it down to quadratic factors,
    the Quadratic Formula will do the rest.

13
The Linear Factorization Theorem
  • The Linear Factorization Theorem is related to
    (and derived from) the Fundamental Theorem of
    Algebra. It states
  • A polynomial of degree n (n gt 0) will have
    precisely n linear factorswhere c1, c2, cn
    are complex numbers.
  • In other words, every root has an associated
    factor, whether the root is real or imaginary.

14
More on complex roots
  • If the polynomial have real coefficients, then
    you always must have an even number of imaginary
    roots.
  • Remember imaginary solutions always travel in
    pairs.
  • If a complex number is a root of a polynomial,
    the complex conjugate of that number will also be
    a root.
  • Even though they are zeros of the polynomial,
    complex solutions will not be x-intercepts.

15
Find the equation
  • of a polynomial with the following roots. Make
    sure the polynomial has integer coefficients.
  • Roots 3, 2i.
  • Complex roots always travel in pairs, so -2i is
    also a root.
  • We can write the equation in factored form
  • y (x 3)(x 2i)(x 2i)
  • y (x 3)(x2 4)
  • y x3 3x2 4x 12

16
Find the equation
  • of a polynomial with the following roots. Make
    sure the polynomial has integer coefficients.
  • Roots ?2, 1 i.
  • Complex roots always travel in pairs, so 1 i is
    also a root.
  • Since the coefficients are integers, that means
    radical roots also travel in pairs, so -?2 is
    also a root.
  • We can write the equation in factored form

17
Continued
  • Lets talk about how to multiply this out.

Lets distribute those negatives. We could
multiply the trinomials together.
This is the difference of squares factoring
pattern. It becomes
18
Continued
  • Lets talk about how to multiply this out.

Lets distribute those negatives. We could
multiply the trinomials together.
This is the difference of squares factoring
pattern. It becomes
19
Continued
  • Lets talk about how to multiply this out.
  • Now you multiply the binomial by the trinomial.

(
(
)
)
Lets distribute those negatives. OR, we could
group a bit differently. Do you recognize that
pattern? Difference of squares.
This is the difference of squares factoring
pattern. It becomes
20
Dividing Polynomials
  • Long Division
  • Synthetic Division
  • Linear divisors only
  • If when dividing one polynomial by another, if
    the remainder is zero, then the divisor is a
    factor of the numerator.

21
Remainder Theorem
  • If the polynomial f(x) is divided by (x k),
    the remainder r is
  • r f(k)
  • In other words, the remainder from division is
    equal to the function evaluated at the zero of
    the divisor

22
Look at this
  • f(x) x2 3x 4
  • (x2 3x 4) (x 1)
  • Here the remainder is 8.
  • f(1) (1)2 3(1) 4 8

23
Factor Theorem
  • A polynomial f(x) has a factor (x k) if and
    only if f(k) 0.
  • In other words, if the remainder is zero then the
    divisor (k) is a zero and (x k) is a factor,

24
Finding Key Attributes
  • We know left and right
  • We can find the y-intercept
  • How can we find the x- intercepts (or zeros)
  • Factoring
  • Division

25
Use factoring to solve 5x3 12x2 4x 0.
x(5x2 12x 4) 0
x(5x 2)(x 2) 0
x 0
, x 2
26
Use a graph, synthetic division, and factoring to
find all roots of x3 3x2 4 0.
First, graph the polynomial function to
approximate the roots.
Then use synthetic division to test your choices.
1 3 0 -4
1
4
4
1
4
4
0
Since the remainder is 0, x 1 is a factor of x3
3x2 - 4.
27
Example
Use a graph, synthetic division, and factoring to
find all roots of x3 3x2 4 0.
x3 3x2 4 0
(x 1)(x2 4x 4) 0
(x 1)(x 2)(x 2) 0
x 1
x -2
x -2
The roots of x3 3x2 4 are 1 and -2, with the
root -2 occurring twice.
28
But what if you dont know a zero to get you
started?
29
Rational Root Theorem
  • p is a factor of the constant term of P
  • q is a factor of the leading coefficient of P

30
Example 1
Find all rational roots of 8x3 10x2 11x 2
0.
Step 1 Make an organized list of all possible
quotients.
factors of 2
factors of 8
31
  • We can eliminate any duplicate ratios

32
Example 1
Find all rational roots of 8x3 10x2 11x 2
0.
33
Example 1
Find all rational roots of 8x3 10x2 11x 2
0.
Step 2 Use substitution or synthetic division
to test all possible roots.
8 10 -11 2
-16
12
-2
8
-6
1
0
roots -2
34
Example 1
Find all rational roots of 8x3 10x2 11x 2
0.
Step 3 Use substitution or synthetic division
to test all possible roots. (Or convert back to
polynomial form and try to factor.)
8 10 -11 2
-16
12
-2
8
-6
1
0
8x2 6x 1 (4x 1)(2x 1)
roots -2
, ¼, ½
35
Example 2
Find all of the zeros of Q(x) x3 4x2 6x -
12.
First, use the Rational Root Theorem to determine
some possibilities.
Then use synthetic division to test your choices.
1 4 -6 -12
2
12
12
1
6
6
0
Since the remainder is 0, x 2 is a factor of x3
4x2 6x - 12.
36
Example 2
Find all of the zeros of Q(x) x3 4x2 6x -
12.
x3 4x2 6x 12 0
(x 2)(x2 6x 6) 0
x 2
or
Since the remainder is 0, x 2 is a factor of x3
4x2 6x - 12.
37
Example 3
Find all of the zeros of P(x) -4x3 2x2 x
3.
First, use the Rational Root Theorem and a graph
of the polynomial function (if available) to
determine some possibilities.
Then use synthetic division to test your choices.
-4 2 -1 3
-4
-2
-3
-4
-2
-3
0
Since the remainder is 0, x 1 is a factor of
-4x3 2x2 x 3.
38
Example
Find all of the zeros of P(x) -4x3 2x2 x
3.
-4x3 2x2 x 3 0
(x 1)(-4x2 - 2x - 3) 0
x 1
or
No real solution
39
Descartes Rule of Signs
  • Remember, the Rational Roots Test only gives
    potential (rational) roots it does not tell you
    what the actual roots will be.
  • There is another tool that can help us narrow
    down our possibilities Descartes Rule of
    Signs.
  • Descartes Rule of Signs will tell you the
    maximum number of positive and negative roots you
    can expect.
  • For this, you need to be aware of the changes in
    signs among the coefficients of the polynomial.

40
Descartes Rule of Signs
  • Example
  • I want to note the changes in sign among the
    coefficients.
  • There are four sign changes, which means I will
    have a maximum of four positive roots.
  • However, this number could decrease by multiples
    of two, so I could also have two or zero positive
    roots.

41
Descartes Rule of Signs
  • For the negative roots, I need to evaluate f(-x).
    When I do this, all the signs of the
    coefficients with odd powers will change.
  • Now do the sign change thing again.
  • We will have one negative root. (Normally, wed
    count down by twos again, but we cant have fewer
    than zero roots.)

42
Descartes Rule of Signs
  • This can be useful in finding roots. For
    example, if you were using the Rational Roots
    Test in the previous problem, and had already
    found a negative root, then you know that any
    other root will be positive.
  • Conversely, if you had already found a positive
    root, then you can expect one or three more.
  • Use Descartes Rule of Signs to find the possible
    number positive and negative roots of the
    following polynomial

43
Upper and Lower Bound Rules
  • One more test to help us narrow down the list of
    possible roots involves the upper and lower bound
    rules.
  • These rules involve observing the results of our
    synthetic division, and they can help us exclude
    certain possible rational roots.
  • Upper Bound Rule
  • Lets say were using synthetic division on
    polynomial f(x), using c as our possible root.
  • If c gt 0, and the bottom row of the synthetic
    division is all non-negative (zero or positive
    numbers), then c is an upper bound of the roots.
  • Lower Bound Rule
  • If c lt 0, and the bottom row of the synthetic
    division has alternating signs, then c is a lower
    bound of the roots. (Zero can count as either
    sign in this case.)

44
Example
  • Find real zeros of f(x) 6x3 - 4x2 3x 2.
  • By the rational roots test, our possibilities are
  • Well test 1 with synthetic division
  • It didnt work, but since 1 is positive and the
    bottom row is all positive, 1 is an upper bound.
  • All roots will be less than 1, which eliminates 2
    as a possibility.

45
Example
  • Find real zeros of f(x) 6x3 - 4x2 3x 2.
  • Now well test -1
  • Since our test was a negative number, and the
    bottom row alternates signs, -1 is a lower bound.

46
Finding the Zeros of a Polynomial
  • Get an overall picture
  • Know the shape and possible number of zeros
  • Estimate the zeros
  • Graphing calculator
  • Rational Root Theorem
  • Descartes Rule of Signs
  • Upper and Lower Bounds
  • Test and confirm
  • Division
  • Remainder Theorem and Factor Theorem
  • Find the factors and zeros
  • Factor, complete the square, quadratic formula

47
Homework 10
  • Section 2.5 Page 160 11 23 odd, 37-41 odd,
    83-89 odd.
  • Demonstrate your work with division and verify
    with your calculator.
Write a Comment
User Comments (0)
About PowerShow.com