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Chapter 5
5.1 Polynomials and Functions
A polynomial function is a function of the form
f(x) = an x nn + an – 1 x nn – 11 +· · ·+ a 1 x + aa00
Where ann  00 and the exponents are all whole numbers.
For this polynomial function, aan is the leading coefficient,
coefficient
n
aa00 is the constant
constant term,
term and n is the degree.
degree
A polynomial function is in standard form if its terms are
descending order
order of
of exponents
exponents from
from left
left to
to right.
right.
written in descending
You are already familiar with some types of polynomial
functions. Here is a summary of common types of
polynomial functions.
Degree
Type
Standard Form
0
Constant
f (x) = a 0
1
Linear
f (x) = a1x + a 0
2
Quadratic
f (x) = a 2 x 2 + a 1 x + a 0
3
Cubic
f (x) = a 3 x 3 + a 2 x 2 + a 1 x + a 0
4
Quartic
f (x) = a4 x 4 + a 3 x 3 + a 2 x 2 + a 1 x + a 0
Identifying Polynomial Functions
Decide whether the function is a polynomial function. If it is,
write the function in standard form and state its degree, type
and leading coefficient.
f (x) = 1 x 2 – 3x4 – 7
2
SOLUTION
The function is a polynomial function.
4
Its standard form is f (x) = – 3x +
1 2
x – 7.
2
It has degree 4, so it is a quartic function.
The leading coefficient is – 3.
Identifying Polynomial Functions
Decide whether the function is a polynomial function. If it is,
write the function in standard form and state its degree, type
and leading coefficient.
f (x) = x 3 + 3 x
SOLUTION
The function is not a polynomial function because the
x
term 3 does not have a variable base and an exponent
that is a whole number.
Identifying Polynomial Functions
Decide whether the function is a polynomial function. If it is,
write the function in standard form and state its degree, type
and leading coefficient.
–
f (x) = 6x 2 + 2x 1 + x
SOLUTION
The function is not a polynomial function because the term
2x –1 has an exponent that is not a whole number.
Identifying Polynomial Functions
Decide whether the function is a polynomial function. If it is,
write the function in standard form and state its degree, type
and leading coefficient.
f (x) = – 0.5x +  x 2 –
2
SOLUTION
The function is a polynomial function.
Its standard form is f (x) =  x – 0.5x –
2
It has degree 2, so it is a quadratic function.
The leading coefficient is .
2.
Identifying Polynomial Functions
Polynomial function?
f (x) = 12 x 2 – 3x 4 – 7
f (x) = x 3 + 3x
f (x) = 6x2 + 2x– 1 + x
f (x) = – 0.5x +  x2 –
2
Decide whether the function is a polynomial
function. If it is, write the function in
standard form and state the degree and
leading coefficient.
Goal 1: To evaluate polynomial functions
Goal 2: To simplify polynomial functions
Chapter 5
5.2 Addition and Subtraction of
Polynomials
Goal 1: To add polynomial functions
Goal 2: To subtract polynomial functions
The additive inverse of a polynomial.
The additive inverse of a polynomial can be found by
replacing each term by its additive inverse.
The sum of a polynomial and its additive inverse is O.
The additive inverse of a polynomial.
The additive inverse of a polynomial can be found by
replacing each term by its additive inverse.
The sum of a polynomial and its additive inverse is O.
Thus, to subtract one polynomial from another, we add
its additive inverse.
Simplify the polynomial
HW #5.1-2
Pg 208-209 1-29 Odd, 30-36
Pg 212-213 1-31 Odd, 33-35
HW Quiz HW #5.1-2
Wednesday, May 24, 2017
Pg 208 32
Pg 208 34
Pg 212 31
Pg 212 34
Pg 208 30
Pg 208 32
Pg 212 34
Pg 212 35
Chapter 5
5.3 Multiplication of Polynomials
Find the product.
Simplify.
Based on your answers to parts to the above, write a general
formula. Use “2n” to represent a general even integer and let “2n
+ 1” represent a general odd integer, and use “…” for missing
terms.
Answers to challenge
HW #5.3
Pg 217-218 1-39 Odd, 40-49
HW Quiz HW #5.3
Wednesday, May 24, 2017
Missing Parts
5.4 Factoring
Do Examples from Regular book la205bad
HW 5.4 Pg 222-223 3-60 Every Third, 61-76
5.5 More Factoring
HW Handout Factoring
5.6 Factoring A General Strategy
Do bonus problems from Great Factoring Problems WS
HW Pg 231 1-37 Odd, 38-47
HW Quiz HW #5.6
Wednesday, May 24, 2017
Row 1, 3, 5
Factor Completely
1. x2  18x  72
2. x4  18x2  81
3. 4 x3  108
4
2
4. x  9 x  81
Row 2, 4, 6
Factor Completely
1. x2  6 x  72
4
2. x  16
3. 4 x3  108
4
4. x  64
5.7 Solving by Factoring
HW #5.7
Pg 233 1-42 Left Column, 43-46
Pg 228 89-99 Odd
5.8 Using Polynomial Equations
A candy factory needs a box that has a volume of 30
cubic inches. The width should be 2 inches less than
the height and the length should be 5 inches greater
than the height. What should the dimensions of the
box be?
For the city park commission, you are designing a marble planter
in which to plant flowers. You want the length of the planter to be
six times the height and the width to be three times the height.
The sides should be one foot thick. Since the planter will be on the
sidewalk, it does not need a bottom. What should the outer
dimensions of the planter be if it is to hold 4 cubic feet of dirt?
Suppose you have 250 cubic inches of clay with which to make
a rectangular prism for a sculpture. If you want the height and
width each to be 5 inches less than the length, what should the
dimensions of the prism be?
HW #5.8a
Pg 235-236 1-17 Odd, 18-20
5
15
17
18
7
13
17
20
Test Review
Geometry Express the area A of a rectangle as a function of the
length x if the length of the rectangle is twice its width.
Geometry Express the area A of an isosceles right triangle as a
function of the length x of one of the two equal sides.
1. Find the degree of a polynomial
2. Find the degree of a monomial
3. Find the function value of a polynomial
4. Standard form of a polynomial
5. Find volume of a box
6. Properties of Exponents
7. Additive Inverse
8. Polynomial arithmetic
9. Factoring
10. Solving Polynomial equations
11. Study all the challenge problems in the book
If h and k areconstantsand x 2  kx  6 factors into
( x  6)( x  h)findk.
Evaluate: (30)2 + 2(30)(22) + (22)2
x
2
 7 xy  10 y
Find ( x  5 y )
2
  15and ( x  2 y)  5,
Factor
1.
2.
3.
4.
5.
6.
a9 b9c9d2 - d2
x2n + xn
yn – yn-1 + yn-2
72x2n + 120xn + 50
50x4 – 72y6
x2 – w2 –16 + 8w
Factor
1.
2.
3.
4.
5.
6.
7.
x4 + 9x2 + 81
x4 + x2y2 + 25y4
x4 + 64
7y2a + b – 5ya + b + 3ya + 2b
4x2a – 4xa – 3
a6 – 64b6
9a3 +9a2b – 4ab2 – 4b3
HW #R-4
Pg 241 1-42