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Chapter 4
Products and Factors
of Polynomials
Section 4-1
Polynomials
Constant A number
-2, 3/5, 0
Monomial A constant, a variable,
or a product of a
constant and one or
more variables
-7
5u
(1/3)m2
-s2t3
x
Coefficient The constant (or
numerical) factor in a
monomial
3m2
coefficient = 3
u
coefficient = 1
- s2t3
coefficient = -1
Degree of a Variable The number of times the
variable occurs as a factor
in the monomial
For Example – 6xy3
What is the degree of x? y?
Degree of a monomial The sum of the degrees of
the variables in the
monomial. A nonzero
constant has degree 0.
The constant 0 has no
degree.
Examples 6xy3
-s2t3
u
-7
degree = 4
degree = 5
degree = 1
degree = 0
Similar Monomials Monomials that are
identical or that differ only
in their coefficients
Also called like terms
Are - s2t3 and 2s2t3 similar?
Polynomial A monomial or a sum of
monomials.
The monomials in a
polynomial are called the
terms of the polynomial.
Examples x2 + (-4)x + 5
x2 – 4x + 5
What are the terms?
2
x , -4x, and 5
Simplified Polynomial A polynomial in which no
two terms are similar.
The terms are usually
arranged in order of
decreasing degree of one
of the variables
Are they Simplified?
2x3 – 5 + 4x + x3
3x3 + 4x – 5
4x2 – x + 3x4 – 5 + x2
Degree of a Polynomial The greatest of the
degrees of its terms after
it has been simplified
What is the degree?
x4 + 3x
2x3 + 3x – 7
2
x – 5x + 1 7x + 1
x4 – 2x2y3 + 6y -11
Adding Polynomials
To add two or more
polynomials, write their
sum and then simplify by
combining like terms
Add the following-
(x2 + 4x – 3) +
(x3 – 2x2 + 6x – 7)
Subtracting Polynomials
To subtract one
polynomial from another,
add the opposite of each
term of the polynomial
you’re subtracting
3
(x
–
2
5x
+ 2x – 5) –
(2x2 – 3x + 5)
Section 4-2
Using Laws of
Exponents
Laws of Exponents
Let a and b be real
numbers and m and n
be positive integers in
all the following laws
Law 1
am · an = am+n
x2 · x 4 = x 6
3
y
5
y
·
=?
m · m4 = ?
Law 2
(ab)m = ambm
(xy)3 = x3y3
2
(3st)
=?
(xy)5 = ?
Law 3
(am)n = amn
(x3)2 = x6
2
3
4
(x y )
=?
(2mn2)3 = ?
Using Distributive Law
Distribute the variable
using exponent laws
3t2(t3 – 2t2 + t – 4) = ?
– 2x2(x3 – 3x + 4) = ?
Section 4-3
Multiplying
Polynomials
Binomial
A polynomial that has
two terms
2x + 3
3xy – 14
4x – 3y
613 + 39z
Trinomial
A polynomial that has
three terms
2x2 – 3x + 1
14 + 32z – 3x
mn – m2 + n2
Multiplying binomials
When multiplying two
binomials both terms of
each binomial must be
multiplied by the other
two terms
Using the F.O.I.L method
helps you remember the
steps when multiplying
F.O.I.L. Method
F – multiply First terms
O – multiply Outer terms
I – multiply Inner terms
L – multiply Last terms
Add all terms to get
product
Binomial
A polynomial that has
two terms
2x + 3
3xy – 14
4x – 3y
613 + 39z
Trinomial
A polynomial that has
three terms
2x2 – 3x + 1
14 + 32z – 3x
mn – m2 + n2
Multiplying binomials
When multiplying two
binomials both terms of
each binomial must be
multiplied by the other
two terms
Using the F.O.I.L method
helps you remember the
steps when multiplying
F.O.I.L. Method
F – multiply First terms
O – multiply Outer terms
I – multiply Inner terms
L – multiply Last terms
Add all terms to get
product
Example - (2a – b)(3a + 5b)
F – 2a · 3a
O – 2a · 5b
I – (-b) ▪ 3a
L - (-b) ▪ 5b
2
2
6a + 10ab – 3ab – 5b
6a2 + 7ab – 5b2
Example – (x + 6)(x +4)
F – x ▪ x
O – x ▪ 4
I – 6 ▪ x
L – 6 ▪ 4
x2 + 4x + 6x + 24
x2 + 10x + 24
Special Products
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
(a + b)(a – b) =
2
a
-
2
b
Section 4-4
Using Prime
Factorization
Factor
A number over a set of
numbers, you write it as a
product of numbers
chosen from that set
The set is called a factor
set
Example
The number 15 can be
factored in the
following ways
(1)(15)
(5)(3)
(-1)(-15)
(-3)(-5)
Prime Number
An integer greater than 1
whose only positive
integral factors are itself
and 1
Prime Factorization
If the factor set is
restricted to the set of
primes
To find it you write the
integer as a product of
primes
Example
350 = 2 x 175
= 2 x 5 x 35
=2x5x5x7
So the prime factorization
of 350 is 2 x 52 x 7
Greatest Common Factor
The greatest integer that
is a factor of each
number.
To find the GCF, take the
least power of each
common prime factor.
Example
What is the GCF of
100, 120, and 90?
10
Least Common Multiple
The least positive
integer having each as
a factor
To find the LCM, take
the greatest power of
each common prime
factor.
Example
What is the LCM of
100, 120, and 90?
1800
Summary
GCF – take the least
power of each common
prime factor.
LCM – take the greatest
power of each prime
factor
Section 4-5
Factoring
Polynomials
Factor
To factor a polynomial
you express it as a
product of other
polynomials
We will factor using
polynomials with
integral coefficients
Greatest Monomial Factor
The GCF of the terms
What is the GCF of
4
3
2
2x – 4x + 8x ?
2
2x
Now factor:
2x4 – 4x3 + 8x2
Factor out 2x2
2
2
2x (x – 2x + 4)
Perfect Square Trinomials
The polynomials in the
form of a2 + 2ab + b2
2
2
and a – 2ab + b are the
result of squaring a + b
and a – b respectively
Difference of Squares
The polynomial a2 – b2 is
the product of a + b and
a-b
Factor Each Polynomial
z2 + 6z + 9
4s2 – 4 st + t2
2
2
25x – 16a
Factored Form:
(z + 3)2
(2s – t)2
(5x + 4a)(5x – 4a)
Sum and Difference of Cubes
a3 + b 3 =
(a + b)(a2 - ab + b2)
3
a
3
b
–
=
(a – b)(a2 + ab + b2)
Factor Each Polynomial
y3 - 1
8u3 + v3
Factor by Grouping
Factor each polynomial by
grouping terms that have
a common factor
Then factor out the
common factor and write
the polynomial as a
product of two factors
Factor each Polynomial
3xy - 4 - 6x + 2y
xy + 3y + 2x + 6
Section 4-6
Factoring Quadratic
Polynomials
Quadratic Polynomials
Polynomials of the
2
form ax + bx + c
Also called seconddegree polynomials
Terms
- quadratic term
bx - linear term
c - constant term
2
ax
Quadratic Trinomial
A quadratic
polynomial for which
a, b, and c are all
nonzero integers
Factoring Quadratic
Trinomials
ax2 + bx + c can be
rewritten in another
format and then
factored by grouping.
Factors
Multiply ac
Find factors ac that
add to b
Rewrite the trinomial
Factor the Polynomial
x2 + 2x – 15
a = 1, c= – 15
Factors of ac that add
to b are 5 and – 3
x2 + 5x – 3x – 15
Factor the Polynomial
x2 + 5x – 3x – 15
Factor by grouping
2
(x + 5x) + (– 3x – 15)
x(x + 5) –3(x + 5)
(x+5)(x – 3)
Factor the Polynomials
2
15t
- 16t + 4
3 - 2z 2
x
2
z
+ 4x - 3
Irreducible
If a polynomial has more
than one term and cannot
be expressed as a
product of polynomials of
lower degree taken from
a given factor set, it is
irreducible
2
x + 4x - 3 is irreducible
Factored Completely
A polynomial is factored
completely when it is
written as a product of
factors and each factor is
either a monomial, a
prime polynomial, or a
power of a prime
polynomial
Greatest Common Factor
The GCF of two or more
polynomials is the
common factor having
the greatest degree and
the greatest constant
factor
Least Common Multiple
The LCM of two or more
polynomials is the
common multiple
having the least degree
and least positive
constant factor
Section 4-7
Solving Polynomial
Equations
Polynomial Equation
An equation that is
equivalent to one with a
polynomial as one side
and 0 as the other
2
x
= 5x + 24
Root
The value of a variable
that satisfies the
equation
Also called the solution
Solving a polynomial
Equation
You can factor the
polynomial to solve the
equation
Steps to Solving a
polynomial Equation
Write the equation with 0
as one side
Factor the other side of
the equation
Solve the equation
obtained by setting each
factor equal to 0
Example 1
Solve (x – 5)(x + 2) = 0
Step 1: already = 0
Step 2: already factored
Step 3: set each factor = 0
x-5=0
x+2=0
x=5
x = -2
Example 2
Solve x2 = x + 30
1: x2 - x – 30 = 0
2: (x – 6)(x + 5) = 0
3: x – 6 = 0
x+5=0
x=6
x = -5
The solution set is {6, -5}
Zeros
A number r is a zero of a
function f if f(r) = 0
You can find zeros using
the same method that is
used to solve polynomial
equations
Example
Find the zeros of
f(x) = (x – 4)3 – 4(3x – 16)
1: simplify
2: factor
3: set each factor = 0
Double Zero
A number that occurs
as a zero of a function
twice
Double Root
A number that occurs
twice as a root of a
polynomial equation
Solve
x2 + 25 = 10x
12 + 4m = m2
Section 4-8
Problem Solving Using
Polynomial Equations
Example 1
A graphic artist is designing a
poster that consists of a
rectangular print with a
uniform border. The print is
to be twice as tall as it is
wide, and the border is to be
3 in. wide. If the area of the
poster is to be 680 in2, find
the dimensions of the print.
Solution
1.Draw a diagram
2.Let w = width and
2w = height
1.The dimensions are 6 in.
greater than the print,
so they are w + 6 and
2w + 6
Solution
4. The area is
represented by
(w + 6)(2w + 6) = 680
5. Solve the equation.
Example 2
The sum of two
numbers is 9. The sum
of their squares is 101.
Find the numbers.
Solution
1.Let x = one number
2.Then 9 – x = the other
number
3.x2 + (9 – x)2 = 101
Section 4-9
Solving Polynomial
Inequalities
Polynomial Inequality
An inequality that is
equivalent to an
inequality with a
polynomial as one side
and 0 as the other side.
x2 > x + 6
Solve by factoring
The product is positive if
both factors are positive,
or both factors are
negative
The product is negative if
the factors have opposite
signs
Example 1
Solve and graph
x2 – 1 > x + 5
x2 – x – 6 > 0
Both factors must be
positive or negative
Example 2
Solve and graph
3t < 4 – t2
t2 + 3t – 4 < 0
The factors must have
opposite signs