Zeros of Polynomial Functions

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Zeros of Polynomial Functions

• Pre-Calculus
• Day 17

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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)

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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).
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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).
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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.
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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
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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.

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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.

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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.

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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.

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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.

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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

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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

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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
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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
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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
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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.

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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

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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

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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,

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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

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Use factoring to solve 5x3 12x2 4x 0.
x(5x2 12x 4) 0
x(5x 2)(x 2) 0
x 0
, x 2
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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.
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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.
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But what if you dont know a zero to get you
started?
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Rational Root Theorem

• p is a factor of the constant term of P

• q is a factor of the leading coefficient of P

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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
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• We can eliminate any duplicate ratios

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Example 1
Find all rational roots of 8x3 10x2 11x 2
0.
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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
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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
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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.
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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.
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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.
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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
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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.

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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.

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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.)

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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

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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.)

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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.

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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.

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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

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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.