Transcript General Education

Chapter 6
Polynomials and
Polynomial Functions
In this chapter, you will …
Learn to write and graph polynomial
functions and to solve polynomial
equations.
Learn to use important theorems about
the number of solutions to polynomial
equations.
Learn to solve problems involving
permutations, combinations and
binomial probability.
6-1 Polynomial Functions
What you’ll learn …
To classify polynomials
To model data with polynomial functions
A monomial- “one term” is an expression that is
a real number, a variable or a product of real
numbers and variables.
13, 3x, -57, x², 4y², -2xy, or 520x²y²
A binomial “two terms” is the sum of two
monomials. It has two unlike terms.
3x + 1, x² - 4x, 2x + y, or y - y²
A trinomial “three Terms” is the sum of three
monomials. It has three unlike terms.
x2 + 2x + 1, 3x² - 4x + 10,
2x + 3y + 2
A polynomial “many terms” is the sum of one
or more terms. x2 + 2x, 3x3 + x² + 5x + 6, 4x 6y + 8
• The exponent or sum of exponents of
the variable(s) in a term determines
the degree of that term.
• The terms in the polynomial must be
in descending order by degree.
• This order demonstrates the standard
form of a polynomial.
2
3
Leading
Coefficient
Cubic
Term
Quadratic
Term
Linear
Term
Constant
Term
Degree
Name Using
Degree
Polynomial
Example
Number
of
Terms
Name Using
Number of
Terms
0
Constant
6
1
monomial
1
Linear
x+3
2
binomial
2
Quadratic
3x2
1
monomial
3
Cubic
2x3 -5x2 -2x
3
trinomial
4
Quartic
x4 + 3x2
2
binomial
5
Quintic
-2x5+3x2-x+4
4
polynomial
Example 1 Classifying Polynomials
Write each in standard form and classify it by
degree and number of terms.
-7x + 5x4
x2 – 4x + 3x3 +2x
(4x)(6x + 5)
6-2 Polynomials and Linear Factors
What you’ll learn …
To analyze the factored form of a polynomial
To write a polynomial function from its zeros
Cubic Function
A cubic polynomial is
a polynomial of degree 3
Example 1 Writing a Polynomial in
Standard Form
Write the expression (x+1)(x+2)(x+3) as a
polynomial in standard form.
Write the expression (x+1)(x+1)(x+2) as a
polynomial in standard form.
Example 2 Writing a Polynomial in
Factored Form
Write 2x3 +10x2 + 12x in factored form.
Write 3x3 - 3x2 - 36x in factored form.
Example 4 Finding Zeros
of a Polynomial Function
Find the zeros of
y= (x-2)(x+1)(x+3).
Then graph the function.
Find the zeros of
y= (x-7)(x-5)(x-3).
Then graph the function.
Example 5 Writing a Polynomial Function
From Its Zeros
Write a polynomial function in standard form with
zeros at -2, 3, and 3.
Write a polynomial function in standard form with
zeros at -4, -2, and 1.
If a factor of a polynomial is repeated,
then the zero is repeated. A repeated
zero is called a multiple zero. A
multiple zero has a multiplicity equal
to the number of times the zero
occurs.
Example 6 Finding the Multiplicity
of a Zero
Find any multiple zeros of f(x)=x4 +6x3+8x2 and
state the multiplicity.
Find any multiple zeros of f(x)=x3 - 4x2+4x and state
the multiplicity.
Equivalent Statements
about Polynomials
1. -4 is a solution of x2 +3x -4 =0.
2. -4 is an x-intercept of the graph of
y= x2 +3x -4.
3. -4 is a zero of y= x2 +3x -4.
4. x+4 is a factor of x2 +3x -4.
6-3 Dividing Polynomials
What you’ll learn …
To divide polynomials using long division
To divide polynomials using synthetic division
You can use polynomial division to help
find all the factors (then, zeros) of a
polynomial function. Division of
polynomials is similar to numerical long
division.
2 56
21 65465
x
2
2x
Example 1a Polynomial Long Division
Divide x2 +3x – 12 by x - 3
Example 1a Polynomial Long Division
Divide x3 +3x2 +7x – 12 by x - 3
Example 1a Polynomial Long Division
Divide 6x3 +3x – 12 by 3x - 6
Example 2 Checking Factors
Determine whether x+4 is a factor of each polynomial
x2 + 6x + 8
x3 + 3x2 -6x - 7
To check (divisor) (quotient) + r = dividend
Synthetic division is a simplified
process in which you omit all variables
and exponents. By reversing the sign
of the divisor, you can add throughout
the process instead of subtracting.
Example 3a Using Synthetic Division
Use synthetic division to divide
3x3 – 4x2 +2x – 1 by x +1
Example 3b Using Synthetic Division
Use synthetic division to divide
x3 + 4x2 + x – 6 by x +1
Check for Understanding
Use synthetic division to divide
x3 - 2x2 - 5x + 6 by x + 2
Remainder Theorem
If a polynomial P(x) of degree n>1 is
divided by (x-a). Where a is a constant,
then the remainder is P(a).
Example 5a Evaluating a Polynomial by
Synthetic Division
Use synthetic division to find P(-4)
for P(x) = x4 - 5x2 + 4x + 12
Example 5b Evaluating a Polynomial by
Synthetic Division
Use synthetic division to find P(-1)
for P(x) = 2x4 + 6x3 – x2 - 60
6-4 Solving Polynomial Equations
What you’ll learn …
To solve polynomials equations by graphing
To solve polynomials equations by factoring
Example 1 Solving by Graphing
Solve x3 + 3x2 = x + 3
Solve x3 - 19x = -2x2 + 20
Sometimes you can solve polynomial
equations by factoring the polynomial
and using the factor Theorem. Recall
that a quadratic expression that is the
difference of squares has a special
factoring pattern. Similarly, a cubic
expression may be the sum of cubes or
the difference of cubes.
Sum and Differences of Cubes
a3 + b3 = (a + b)(a2 – ab + b2)
a3 - b3 = (a - b)(a2 + ab + b2)
Example 3 Factoring a Sum of Cubes
Factor 8x3 + 1
Factor 64x3 + 27
Example 3 Factoring a Difference of Cubes
Factor 8x3 - 27
Factor 125x3 - 64
Example 4 Solving a Polynomial Equation
Factor x3 + 8 = 0
Factor 27x3 + 1
Example 5 Factoring by Using a Quadratic Form
Factor x4 - 2x2 – 8 = 0
Factor x4 + 7x2 + 6 = 0
6-5 Theorems About Roots of
Polynomial Equations
What you’ll learn …
To solve equations using the Rational Root Theorem
To use the Irrational Root Theorem and the
Imaginary Root Theorem
1.02 Define and compute with complex
numbers.
1.03 Operate with algebraic expressions
(polynomial, rational, complex fractions) to
solve problems.
Consider the equivalent
equations….
x3 – 5x2 -2x +24 = 0 and (x+2)(x-3)(x-4) =0
-2, 3 and 4 are the roots of the equation.
• The product of -2,3 and 4 is 24.
• Notice that all the roots are factors of the
constant term 24.
• In general, if the coefficients in a polynomial
equation are integers, then any integer root of
the equation is a factor of the constant term.
Both the constant and the leading coefficient of a
polynomial can play a key role in identifying the rational
roots of the related polynomial equation. The role is
expressed in the Rational Root Theorem.
p
If q is in simplest form and is a
rational root of the polynomial equation
with integer coefficients, then p must
be a factor of the constant term and q
must be a factor of the leading
coefficient.
Example 1a Finding Rational Roots
x3
-
4x2
- 2x + 8 = 0
1.
2.
Steps
List the possible rational
roots of the leading
coefficient and the
constant.
Test each possible root.
Example 1b Finding Rational Roots
2x3
-
x2
+2x - 1 = 0
1.
2.
Steps
List the possible rational
roots of the leading
coefficient and the
constant.
Test each possible root.
Example 1c Finding Rational Roots
x3
-
2x2
- 5x + 10 = 0
1.
2.
Steps
List the possible rational
roots of the leading
coefficient and the
constant.
Test each possible root.
Example 1d Finding Rational Roots
3x3
+
x2
-x+1=0
1.
2.
Steps
List the possible rational
roots of the leading
coefficient and the
constant.
Test each possible root.
In Chapter 5 you learned to find irrational
solutions to quadratic equations. For
example, by the Quadratic Formula,
the solutions of x2 – 4x -1 =0 are
2+√5 and 2 - √5.
Number pairs of the form a+√b and a-√b
are called conjugates.
You can often use conjugates to find the
irrational roots of a polynomial equation.
Irrational Root Theorem
Let a and b be rational numbers and
let √b be an irrational number. If
a+ √b is a root of a polynomial
equation with rational coefficients, then
the conjugate a- √b also is a root.
Example 3 Finding Irrational Roots
A polynomial equation with integer coefficients has
the following roots. Find two additional roots..
1 + √3 and -√11
2 - √7 and √5
Number pairs of the form a+bi and a-bi are
complex conjugates. You can use complex
conjugates to find an equation’s imaginary roots.
Imaginary Root Theorem
If the imaginary number a+bi is a root
of a polynomial equation with real
coefficients then the conjugate a-bi
also is a root.
Example 4 Finding Imaginary Roots
A polynomial equation with integer coefficients has
the following roots. Find two additional roots..
3i and -2 + i
3 - i and 2i
Example 5a Writing a Polynomial
Equation from its Roots
Find a third degree polynomial equation with
rational coefficients that has roots -1 and 2-i.
1.
2.
3.
Steps
Find the other root using
the Imaginary Root
Theorem.
Write the factored form
of the polynomial using
the Factor Theorem.
Multiply the factors.
Example 5b Writing a Polynomial
Equation from its Roots
Find a third degree polynomial equation with
rational coefficients that has roots 3 and 1+i.
1.
2.
3.
Steps
Find the other root using
the Imaginary Root
Theorem.
Write the factored form
of the polynomial using
the Factor Theorem.
Multiply the factors.
Example 5c Writing a Polynomial
Equation from its Roots
Find a fourth degree polynomial equation with
rational coefficients that has roots i and 2i.
1.
2.
3.
Steps
Find the other root using
the Imaginary Root
Theorem.
Write the factored form
of the polynomial using
the Factor Theorem.
Multiply the factors.
6-6 The Fundamental Theorem
of Algebra
What you’ll learn …
To use the Fundamental Theorem of Algebra in
solving polynomial equations with complex roots
1.02 Define and compute with complex
numbers.
1.03 Operate with algebraic expressions
(polynomial, rational, complex fractions) to
solve problems.
You have solved polynomial equations
and found that their roots are included
in the set of complex numbers. That
is, the roots have been integers,
rational numbers, irrational numbers
and imaginary numbers.
But can all polynomial equations be
solved using complex numbers?
In 1799, the German
mathematician Carl Friedrich
Gauss proved that the answer
to this question is yes. The
roots of every polynomial
equation, even those with
imaginary coefficients, are
complex numbers.
Carl Friedrich Gauss
The answer is so important
that his theorem is called the
Fundamental Theorem of
Algebra.
Fundamental Theorem of Algebra
If P(x) is a polynomial of degree n>1
with complex coefficients, then P(x) = 0
has at least one complex root.
Corollary
Including imaginary roots and multiple
roots, an nth degree polynomial
equation has exactly n roots; the
related polynomial function has exactly
n zeros.
Example 1a Using the Fundamental
Theorem of Algebra
Find the number of complex roots, the
possible number of real roots and possible
number of rational roots.
x4 - 3x3 + x2 – x +3 = 0
Example 1b Using the Fundamental
Theorem of Algebra
Find the number of complex roots, the
possible number of real roots and possible
number of rational roots.
x3 - 2x2 + 4x -8 = 0
Example 1c Using the Fundamental
Theorem of Algebra
Find the number of complex roots, the
possible number of real roots and possible
number of rational roots.
x5 + 3x4 - x - 3 = 0
6-8 The Binomial Theorem
What you’ll learn …
To use Pascal’s Triangle
To use the Binomial Theorem

1.03 Operate with algebraic expressions
(polynomial, rational, complex fractions)
to solve problems.
You have learned to multiply binomials
using the FOIL method and the
Distributive Property. If you are raising
a single binomial to a power, you have
another option for finding the product.
Consider the expansion of several
binomials. To expand a binomial
being raised to a power, first multiply;
then write the result as a polynomial in
standard form.
Pascal’s Triangle
(a + b)2
=
(a + b) (a + b)
a2 + 2ab + b2
The coefficients of the product are 1,2 1.
(a + b)3
=
(a + b) (a + b) (a + b)
a3 + 3a2b + 3ab2 + b3
The coefficients of the product are 1,3,3,1.
Pascal’s Triangle
Example 1a Using Pascal’s Triangle
Expand (a+b)8
Example 1b Using Pascal’s Triangle
Expand (x - 2)4
Example 1c Using Pascal’s Triangle
Expand (m + 3)5
Example 1d Using Pascal’s Triangle
Expand (3 – 2x)6
Quadratic
Inequalities
Quadratics
Before we get started let’s review.
A quadratic equation is an equation that can
2
be written in the form ax  bx  c  0 ,
where a, b and c are real numbers and a cannot equal
zero.
In this lesson we are going to discuss quadratic
inequalities.
Quadratic Inequalities
What do they look like?
Here are some examples:
x 2  3x  7  0
3x  4 x  4  0
2
x  16
2
Quadratic Inequalities
When solving inequalities we are trying to
find all possible values of the variable
which will make the inequality true.
Consider the inequality
x2  x  6  0
We are trying to find all the values of x for which the
quadratic is greater than zero or positive.
Solving a quadratic inequality
We can find the values where the quadratic equals zero
2
x
 x6  0
by solving the equation,
x  3x  2  0
x  3  0 or x  2  0
x  3 or x  2
Solving a quadratic inequality
2
x
For the quadratic inequality,  x  6  0
we found zeros 3 and –2 by solving the equation
x 2  x  6  0. Put these values on a number line and
we can see three intervals that we will test in the
inequality. We will test one value from each interval.
-2
3
Solving a quadratic inequality
Interval
 ,2
Test Point
Evaluate in the inequality
True/False
x2  x  6  0
x  3
 32   3  6  9  3  6  6  0 True
x2  x  6  0
 2, 3
 3, 
x0
02  0  6  0  0  6  6  0
False
x2  x  6  0
x4
42  4  6  16  4  6  6  0
True
Example 2:
2
2
x
 3x  1  0
Solve
2
2
x
 3x  1  0
First find the zeros by solving the equation,
2 x 2  3x  1  0
2x 1x 1  0
2 x  1  0 or x  1  0
1
x  or x  1
2
Example 2:
Now consider the intervals around the zeros and test
a value from each interval in the inequality.
The intervals can be seen by putting the zeros on a
number line.
1/2
1
Example 2:
Interval
Test Point
Evaluate in Inequality
True/False
2 x 2  3x  1  0
1

  , 
2

x0
20  30  1  0  0  1  1  0
2
False
2 x 2  3x  1  0
1 
 ,1
2 
9 9
1
3 3
2   3   1    1   0
8 4
8
4 4
2
3
x
4
True
2 x 2  3x  1  0
1, 
x2
22  32  1  8  6  1  3  0
2
False
Summary
In general, when solving quadratic inequalities
1. Find the zeros by solving the equation you get
when you replace the inequality symbol with an
equals.
2. Find the intervals around the zeros using a number
line and test a value from each interval in the
number line.
3. The solution is the interval or intervals which make
the inequality true.
x  5 x  24  0
2
12  x  x  0
2
16 x  1  0
2
x  5x  4  0
2
In this chapter, you should have …
Learned to write and graph polynomial
functions and to solve polynomial
equations.
Learned to use important theorems
about the number of solutions to
polynomial equations.
Learned to solve problems involving
permutations, combinations and
binomial probability.