Download Key Concept: Synthetic Division

Algebra 2A, Chapter 5 Notes, Part 1
5.1 Polynomial Functions
Terms
Monomial:
Degree of a Monomial:
Polynomial (in one variable):
Degree of a Polynomial:
Polynomial Function:
1
Algebra 2A, Chapter 5 Notes, Part 1
2
You can classify a polynomial l by its degree or its number of terms.
Polynomials degree 0-5 have specific names.
Degree:
0
constant
1
linear
2
quadratic
3
cubic
4
quartic
5
quintic
Number of terms:
1
monomial
2
binomial
3
trinomial
N
Polynomial of “n” terms
Examples:
Example 1: Classifying Polynomials
Write each polynomial in standard form.
What is its classification by degree? By number of terms?
A.
3𝑥 3 − 𝑥 + 5𝑥 4
D.
𝑥 2 − 4𝑥 2
B.
3 − 𝑥 5 + 2𝑥 2 + 10
E.
3 − 𝑥 + 10
C. 3𝑥 3 − 5 + 9𝑥 3
F. 3𝑥 − 𝑥 4 + 5𝑥 2 + 1
Algebra 2A, Chapter 5 Notes, Part 1
3
Degree of a Polynomial Function


Affects the shape of the graph
Affects the end behavior
o End behavior:
4 Types of End Behavior (for functions of degree one or greater)
You can determine the end behavior of a
polynomial function of degree n from the
leading term axn of standard form.
Example 2: Describing the End Behavior of Polynomial Functions
Consider the leading term of each polynomial function. What is its end behavior?
A.
𝑦 = 4𝑥 3 − 8𝑥
B.
𝑦 = −2𝑥 2 − 3𝑥 − 5
C.
𝑦 = −6𝑥 4
D.
𝑦 = − 2 𝑥3 − 7
E.
𝑦 = 𝑥 + 2𝑥 6
F.
𝑦 = 1 + 𝑥 + 3𝑥 3
1
Algebra 2A, Chapter 5 Notes, Part 1
4

y

Example 3: Graphing a Cubic Function


Graph: y  2 x3  3x 2  1





x


























5.2 Polynomials, Linear Factors and Zeros
If P(x) is a polynomial function, the solutions of the related polynomial
equation P(x) = 0 are the zeros of the function.
Example:
Polynomial Function
𝒇(𝒙) = 𝒙𝟐 + 𝟐𝒙 − 𝟖
Polynomial Equation: 𝒙𝟐 + 𝟐𝒙 − 𝟖 = 𝟎
Zeros:
Finding the zeros of a polynomial function will help you factor the polynomial, graph the function, and solve the
related polynomial equation.


Algebra 2A, Chapter 5 Notes, Part 1
5
Key Concept: Roots, Zeros and X-Intercepts
The following are equivalent statements about a real number b and a polynomial
𝑃(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎𝑜
 𝑥 − 𝑏 is a linear factor of the polynomial P(x)
 b is a zero of the polynomial function 𝑦 = 𝑃(𝑥)
 b is a root (or solution) of the polynomial equation 𝑃(𝑥) = 0
 b is an x-intercept of the graph of 𝑦 = 𝑃(𝑥)
Example:
Let 𝑃(𝑥) = 𝑥 3 − 2𝑥 − 4 and let x = 2.
Then P(2) = 0
Therefore,
 _________ is a factor of 𝑥 3 − 2𝑥 − 4

_________ is a zero for the polynomial function P(x)

_________ is a solution to the equation 𝑥 3 − 2𝑥 − 4 = 0

The point (____,____) is an x-intercept of the graph of
𝑃(𝑥) = 𝑥 3 − 2𝑥 − 4
Review: Factoring
Factor the following polynomials completely.
A.
x3  2 x 2  15 x
B.
x3  49 x
C.
6 x3  5 x 2  x
Algebra 2A, Chapter 5 Notes, Part 1
Example 1: Finding Zeros of a Polynomial Function
A) What are the zeros of 𝒚 = 𝒙(𝒙 − 𝟑)(𝒙 + 𝟓)? Sketch the graph of the function.
B) What are the zeros of 𝒚 = (𝒙 − 𝟏)(𝒙 + 𝟒)(𝒙 − 𝟐)? Sketch the graph of the function.
Theorem: Factor Theorem
The expression 𝑥 − 𝑎 is a factor of the polynomial if and only if the value a is a zero of the related function.
6
Algebra 2A, Chapter 5 Notes, Part 1
7
Example 2: Writing a Polynomial From Its Zeros
A.
What is a polynomial function with zeros 4 and -4?
Write a polynomial function with the given zeros.
B.
3, 3, and -4
C.
2, 1, and -1
Key Concept: Multiplicity
𝑦 = (𝑥 − 3)(𝑥 + 2)(𝑥 + 2) can be written as 𝑦 = (𝑥 − 3)(𝑥 + 2)2 .
Because the linear factor (𝑥 + 2) appears twice, -2 is a zero of multiplicity 2.
In general, a is a zero of multiplicity n means that (𝑥 − 𝑎) appears n times as a factor of the polynomial.
Example 3: Finding the Multiplicity of a Zero
Find the zeros of the function. State the multiplicity of any multiple zeros.
A) 𝑦 = (𝑥 − 4)3 (𝑥 + 1)2 (𝑥 − 2)
C)
𝑦 = 𝑥(𝑥 + 6)5 (𝑥 − 2)
B)
𝑦 = 𝑥 2 (𝑥 − 1)4 (𝑥 − 7)
D)
𝑦 = (𝑥 − 1)2 (𝑥 + 1)2
Algebra 2A, Chapter 5 Notes, Part 1
8
5.3 Solving Polynomial Equations
Recall: The expression 𝑥 − 𝑎 is a factor of the polynomial if and only if the value a is a zero of the related
function.
To solve a polynomial equation by factoring,
1. Write the equation in the form P(x) = 0 for the polynomial function P.
2. Factor P(x). Use the Zero Product Property to find the roots.
Practice:
Factor:
x3  64
Factor:
x 3  125
Algebra 2A, Chapter 5 Notes, Part 1
9
Solving Polynomial Equations by Factoring
What are the real or imaginary solutions of each polynomial equation?
1.
𝑥 2 − 5𝑥 + 4 = 0
2.
(𝑥 2 − 1)(𝑥 2 + 4) = 0
3.
𝑥 4 − 13𝑥 2 − 48 = 0
4.
𝑥3 − 1 = 0
5.
𝑥3 + 8 = 0
6.
𝑥 3 + 2𝑥 2 − 8𝑥 = 0
Algebra 2A, Chapter 5 Notes, Part 1
10
7.
𝑥 4 + 7𝑥 3 + 12𝑥 2 = 0
8.
2𝑥 3 − 5𝑥 2 − 3𝑥 = 0
5.4 Dividing Polynomials
You can divide polynomials using steps similar to the long division steps used to divide whole numbers.
Example 1: Using Polynomial Long Division
A)
Divide 3𝑥 2 − 29𝑥 + 56 by 𝑥 − 7.
Algebra 2A, Chapter 5 Notes, Part 1
11
B)
Divide 2𝑥 3 + 3𝑥 2 + 𝑥 + 6 by 𝑥 + 3.
C)
Divide 4 x 4  3 x 3  10 x  2 by x  4
Algebra 2A, Chapter 5 Notes, Part 1
12
Key Concept: Synthetic Division
Synthetic division simplifies the long division process for dividing by a linear expression 𝑥 − 𝑎.
To use synthetic division,
1. Write the coefficients (including zeros) of the polynomial in standard form.
2. For the divisor, reverse the sign (use a).
Example 2: Using Synthetic Division
A) Use synthetic division to divide 𝑥 3 + 2𝑥 2 − 5𝑥 − 6 by 𝑥 + 1.
B) Use synthetic division to divide 𝑥 3 − 57𝑥 + 56 by 𝑥 − 7.
C) Use synthetic division to divide 𝑥 3 + 5𝑥 2 − 3𝑥 − 4 by 𝑥 + 6.
Step 1: Set divisor = 0, solve,
put number in “the box”
Step 2: Write coefficients of
polynomial next to this.
Include zeros for missing
terms.
Step 3: Bring down first
coefficient.
Step 4: Multiply. Add.
Repeat.
Step 5: Write your
polynomial answer. Degree
of answer is one less than
degree of original. Last
number is remainder.
Algebra 2A, Chapter 5 Notes, Part 1
13
Example 3: Checking Factors
(𝑥 − 𝑎) is factor of the polynomial function 𝑃(𝑥) if and only if 𝑃(𝑥) ÷ (𝑥 − 𝑎) has a remainder of zero.
A) Is 𝑥 − 3 a factor of 𝑃(𝑥) = 𝑥 3 − 7𝑥 2 + 11𝑥 + 3? Why or why not?
B) Is 𝑥 + 5 a factor of 𝑃(𝑥) = 4𝑥 3 + 21𝑥 2 − 𝑥 − 24? Why or why not?
Key Concept: Remainder Theorem
If you divide a polynomial 𝑃(𝑥) of degree 𝑛 ≥ 1 by 𝑥 − 𝑎, then the remainder is 𝑃(𝑎).
Example 4: Evaluating a Polynomial Using Synthetic Division
Sometimes this is called “Synthetic Substitution”
Use synthetic division and the remainder theorem to find the following:
A)
𝑃(−4) for 𝑃(𝑥) = 𝑥 5 − 3𝑥 4 − 28𝑥 3 + 5𝑥 + 20
B)
𝑃(3) for 𝑃(𝑥) = 2𝑥 4 − 𝑥 2 − 2
C)
𝑃(5) for 𝑃(𝑥) = 𝑥 3 − 5𝑥 2 − 7𝑥 + 25