ALGEBRA 1
SET OF REAL NUMBERS AND RATIONAL EXPRESSIONS
INTEGERS
15 Minute Math
This presentation will review different math skills that will help you with
every day math problems.
Each lesson takes approximately 15 minutes to do. Almost anyone can
find an extra 15 minutes out of his or her day, whether it be during
breakfast or right before bed. In just under 3 weeks, you can review all
the TABE Computation Math sections.
Look for the Professor. He has special hints to help make
working the math problems faster and easier.
This lesson explains how to add and subtract negative
numbers.
Problem: You want to buy a house that costs $100,000. You only have
$5,000. If you buy the house, what will your debt be?
5,000
- 100,000
-95,000
If you spend more money than you have, you go into debt. Debt is a good
example of working with negative integers.
Negative numbers are like a debt.
Positive numbers move to the right on the number line.
Negative numbers move to the left on the number line.
If you have 8 dollars, and then you get 2 dollars, you have 10 dollars.
 8 + 2 = 10
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1
0
2
3
4
5
6
7
8
9
10
3
4
5
6
7
8
9
10
7
8
9
10
If you have 8 dollars, and then you lose 2 dollars, you have only 6 dollars.
 8-2=6
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1
0
2
If you are 8 dollars in debt, and then you get 2 dollars, you are only 6 dollars in debt.
 -8 + 2 = -6
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1
0
2
3
4
5
6
If you are 8 dollars in debt, and then you lose another 2 dollars, you are 10 dollars in debt.
 -8 - 2 = -10
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1
0
2
3
4
5
6
7
8
9
10
Use these hints to help work problems.
If both numbers are negative, add the numbers, then make the answer negative.
-8 – 2  -(8 + 2) = -10
When one sign is positive and one sign is negative, subtract the
numbers and take the sign of the larger.
8 – 2 = 6 (“8” is positive, so the answer is positive)
2 – 8 = -6 (think “8-2” and take the sign of the larger number, “8”)
Examples:
++  9
--  -9
+-  9
-+  -9
+
–
–
+
2
2
2
2
= 11
= -11
= 7
= -7
Try using this chart:
Integer Addition/Subtraction
++  Add
--  Add
+-  Subtract
-+  Subtract
Always use the sign of the largest
number for the sign of the answer
Try it yourself
3+6=9
-3 – 6 = -9 3 – 6 = -3 -3 + 6 = 3
4+7=
-4 – 7 =
4–7=
-4 + 7 =
7+4=
-7 – 4 =
7–4=
-7 + 4 =
5+1=
-5 – 1 =
5–1=
-5 + 1 =
10 + 20 = -10 – 20 = 10 – 20 = -10 + 20 =
Click for answers:
4 + 7 = 11
-4 – 7 = -11
4 – 7 = -3
-4 + 7 = 3
7 + 4 = 11
-7 – 4 = -11
7–4=3
-7 + 4 = -3
5+1=6
-5 – 1 = -6
5–1=4
-5 + 1 = -4
10 + 20 = 30
-10 – 20 = -30 10 – 20 = -10 -10 + 20 = 10
If there are 2 signs next to each other, rewrite the problem with just
one sign.
To subtract a negative number, the 2 negatives make a positive.
8 – (-2)  8 - -2  8 + 2 = 10
Adding a negative number is the same as subtraction.
8 + (-2)  8 + -2  8 – 2 = 6
We don’t usually write the
positive sign, but you could if you
wanted to:
8 + 2  (+8) + (+2)
8 – 2  (+8) – (+2) or (+8) + (-2)
The parentheses are only
there to keep the signs
separate. You can remove
them if you want to.
Try it yourself
3 – (+6)  3 – 6 = -3
-2 – (-4)  -2 + 4 = 2
4 – (-2) =
-8 + (-3) =
-10 + (-3) =
9 + (+2) =
End of Lesson 12
10 – (+2) =
-7 – (-3) =
Click for answers:
4 – (-2) = 6
-8 + (-3) = -11
-10 + (-3) = 7
10 – (+2) = 8
9 + (+2) = 11
-7 – (-3) = -4
This lesson explains how to
multiply and divide negative numbers,
plus the absolute value sign,||.
Problem: You are $30 in debt. If your debt is increased 2 times, how much is
your debt?
-30 x 2 = -60
Multiplying by a negative number is multiplying your debt.
Some simple rules will help you when multiplying negative numbers. The rules
are the same for division.
• If there is one negative number, then the answer is negative.
• If two numbers are negative, then the answer is positive.
8 x 5 = 40
40 ÷ 8 = 5
-8 x 5 = -40
-40 ÷ -8 = 5
8 x -5 = -40
-40 ÷ 8 = -5
-8 x -5 = 40
40 ÷ -8 = -5
If the signs are the same = positive answer.
If the signs are different = negative answer.
Integer Mult./Division
++  Answer is
--  Answer is
+-  Answer is
-+  Answer is
• If you have more than 2 numbers, the rule is:
 An odd number of negative #s = negative answer
 An even number of negative #s = positive answer
(-1 )x (-2) x (-3) x (4) = -24 (3 negative #s)
(-1) x (-2) x (-3) x (-4) = 24 (4 negative #s)
+
+
-
The absolute value symbols “| |” make everything positive.
To work an absolute value problem, first work everything inside the
absolute value symbols.
Then, make it positive.
| -10 | = 10
| 10 | = 10
| 6 x -3 | = 18
|3–7|
= | -4 |
=4
Work inside the symbols | 3 – 7 |= | -4 |
Now, make it positive  | -4 | = +4
Try it yourself
2x3=6
-2 x 3 = -6 2 x -3 = -6 -2 x -3 = 6
6x8=
-6x8=
6 x -8 =
-6 x -8 =
7x4=
-7 x 4 =
7 x -4 =
-7 x -4 =
54  6 =
-54  6 = 54  -6 =
|-8 x 3| = |5 x (-2) – 1| =
2 x |3 – 9| =
-54  -6 =
|-12  -4| =
Click for answers:
6 x 8 = 48
- 6 x 8 = -48
6 x -8 = -48
7 x 4 = 28
-7 x 4 = -28
7 x -4 = -28
54  6 = 9
-54  6 = -9
54  -6 = -9
|-8 x 3| = 24
|5 x (-2) – 1| = 11 2 x |3 – 9| = 12
-6 x -8 = 48
-7 x -4 = 28
-54  -6 = 9
|-12  -4| = 3
Practice for Problem Solving
• Fiona spends $5 per week on bus fare. How much does she
spend in 2 weeks?
• Lucy spends 2 per week on snacks. How much does she
spend in 4 weeks?
• Anton earns $8 each week for baby-sitting. How much does
he earn in 3 weeks?
Practice for Problem Solving
• Lional pays $3 per day for bus transportation. How much
does she pay in a school week?
• Jill has $100 in the bank. She owes 3 of her friends $10
dollars each. What is her net worth?
THE NUMBER LINE
Constructing a Number Line
• A number line is drawn by choosing a starting position, usually 0, and
marking off equal distance from that point
17
• Although only a portion of the number line is shown, the arrowheads
indicate that the line and the set of numbers continue
18
• The set of numbers used on the number line is called the set of
integers
• This set can be written
{…, -3, -2, -1, 0, 1, 2, 3, …}
19
• The number line is made up of 3 sets of numbers
• Natural numbers 1, 2, 3, 4, …
• Whole numbers 0, 1, 3, 4, …
• Integers …, -2, -1, 0, 1, 2, …
20
• We can use a Venn Diagram to
represent thee 3 sets of numbers and
their relationship to one another
21
• You should not include the number 0
• You should not include numbers less than 0 (negative numbers)
• You should not include the number 6 or any higher
22
• Positive numbers move to the right
• Negative numbers move to the left
• + 3 + (-8) = -5
23
In Summary
24
EXPONENTS
The term 27 is called a power. If a number is in exponential
form, the exponent represents how many times the base is
to be used as a factor.
Exponent
Base
2
7
Write in exponential form.
A. 4 • 4 • 4 • 4
4•4•4•4=
44
Identify how many times 4
is a factor.
B. d • d • d • d • d
d • d • d • d • d = d5
Reading Math
Read 44 as “4 to the 4th power.”
Identify how many times d
is a factor.
Write in exponential form.
C. (–6) • (–6) • (–6)
(–6) • (–6) • (–6) = (–6)3
Identify how many times –
6 is a factor.
Remember to keep the – sign inside the ( )! If it is
outside, your answer will be negative even if you have
an even number of – signs.
D. 5 • 5
5 • 5 = 52
Identify how many times 5
is a factor.
Write in exponential form.
A. x • x • x • x • x
x • x • x • x • x=
x5
Identify how many times x
is a factor.
B. d • d • d
d • d • d = d3
Identify how many times d
is a factor.
Evaluate.
Find the product of five 3’s.
A. 35
35 = 3 • 3 • 3 • 3 • 3
= 243
B. (–3)5
(–3)5
Find the product of five –3’s.
= (–3) • (–3) • (–3) • (–3) • (–3)
= –243
Helpful Hint
Always use parentheses to raise a negative number to a power.
Evaluate.
C. 74
74 = 7 • 7 • 7 • 7
= 2401
D. (–9)3
Find the product of four 7’s.
Find the product of three –9’s.
(–9)3 = (–9) • (–9) • (–9)
= –729
Simplifying Expressions
Containing Powers
Simplify (25 – 32 ) + 6(4)
= (32 – 9) + 6(4)
= (23) + 6(4)
Evaluate the exponents.
Subtract inside the parentheses.
= 23 + 24
Multiply from left to right.
= 47
Add from left to right.
Simplify (32 – 82) + 2 • 3
= (9 – 64) + 2 • 3
= (–55) + 2 • 3
= –55 + 6
= –49
Evaluate the exponents.
Subtract inside the parentheses.
Multiply from left to right.
Add from left to right.
Lesson Quiz
Write the product or quotient as one power.
1. n3  n4
109
3.
105
n7
104
2. 8 • 88 89
t9
4.
t7
t2
5. 32 • 33 • 35 310
6. (m2)19 m38
7. (9-8)9 9–72
8. (104)0
1
Exponents
Power
5
exponent
3
base
Example: 125  53 means that 53 is the exponential
form of the number 125.
53 means 3 factors of 5 or 5 x 5 x 5
The Laws of Exponents:
#1: Exponential form: The exponent of a power indicates
how many times the base multiplies itself.
x  x  x  x  x  x  x  x
n
n times
n factors of x
Example: 5  5  5  5
3
#2: Multiplying Powers:
If you are multiplying Powers
with the same base, KEEP the BASE & ADD the EXPONENTS!
x x  x
m
So, I get it!
When you
multiply
Powers, you
add the
exponents!
n
mn
2 6  23  2 6  3  29
 512
#3: Dividing Powers: When dividing Powers with the same base,
KEEP the BASE & SUBTRACT the EXPONENTS!
m
x
m
n
mn

x

x

x
n
x
So, I get it!
When you
divide
Powers, you
subtract the
exponents!
6
2
6 2
4

2

2
2
2
 16
Try these:
12
1. 3  3 
2
2
7.
2. 52  54 
3.
8.
a a 
5
2
4. 2s  4s 
2
7
12 8
9.
5. (3)  (3) 
2
6.
3
s t s t 
2 4
7 3
s

4
s
9
3

5
3
s t

4 4
st
5 8
10.
36a b

4 5
4a b
SOLUTIONS
2
2 2
4
a a  a
5 2
a
1. 3  3  3  3  81
2 4
6
2
4
2. 5  5  5  5
2
3.
5
2
4. 2s  4s  2  4  s
2
7
5. (3)  (3)  (3)
2
6.
3
s t s t 
2 4
7 3
s
7
27
23
 8s
 (3)  243
27 43
t
9
5
s t
9 7
SOLUTIONS
12
7.
8.
9.
10.
s
12 4
8
s

s

4
s
9
3
9 5
4
3

3
 81

5
3
12 8
s t
12 4 8 4
8 4
s t s t

4 4
st
5 8
36a b
5 4 85
3
36

4

a
b

9
ab

4 5
4a b
#4: Power of a Power: If you are raising a Power to an
exponent, you multiply the exponents!
x 
n
m
So, when I
take a Power
to a power, I
multiply the
exponents
x
mn
32
(5 )  5
3
2
5
5
#5: Product Law of Exponents: If the product of the
bases is powered by the same exponent, then the result is a
multiplication of individual factors of the product, each powered
by the given exponent.
 xy 
So, when I take
a Power of a
Product, I apply
the exponent to
all factors of
the product.
n
x y
n
n
(ab)  a b
2
2
2
#6: Quotient Law of Exponents: If the quotient of the
bases is powered by the same exponent, then the result is both
numerator and denominator , each powered by the given exponent.
n
 x
x
   n
y
 y
So, when I take a
Power of a
Quotient, I apply
the exponent to
all parts of the
quotient.
n
4
16
2 2
   4 
81
3 3
4
Try these:
 
5
2 5
1. 3

 
3. 2a  
4. 2 a b  
2. a
3 4
2 3
2
5 3 2
5. (3a ) 
2 2
 
2 4 3
6. s t

s
7.   
t
9 2
3 
8.  5  
3 
2
 st 
9.  4  
 rt 
5 8 2
 36a b 
 
10. 
4 5 
 4a b 
8
SOLUTIONS
 
2 5
1. 3
 
2. a
3 4


2
3

 
3. 2a
10
2 3
a 12
3
 2 a

5 3 2
4. 2 a b
23
 8a
22 52 32
4 10 6
10 6
 2 a b  2 a b  16a b
5. (3a )   3  a
2
2 2
 
2 4 3
6. s t
6
23 43
s t
22
s t
 9a
6 12
4
SOLUTIONS
5
s
7.   
t
5
s
5
t
2
3 
8.  5   34
3 
9
 
2
3
8
2
4 2
2 8
 st 


st
s
t
9.  4   
  2

r
 rt 
 r 
8
 36a b
10 
4 5
 4a b
5 8
2

  9ab3



2
2 32
9 a b
2
 81a b
2 6
#7: Negative Law of Exponents: If the base is powered
by the negative exponent, then the base becomes reciprocal with the
positive exponent.
So, when I have a
Negative Exponent, I
switch the base to its
reciprocal with a
Positive Exponent.
Ha Ha!
If the base with the
negative exponent is in
the denominator, it
moves to the
numerator to lose its
negative sign!
x
m
1
 m
x
1
1
5  3 
5
125
and
3
1
2

3
9
2
3
#8: Zero Law of Exponents: Any base powered by zero
exponent equals one.
x 1
0
So zero
factors of a
base equals 1.
That makes
sense! Every
power has a
coefficient
of 1.
50  1
and
a0  1
and
(5 a ) 0  1
Try these:
1.
2a b
0
2

2.
y 2  y 4 
3.
a 
5 1
2

4. s  4s 

7
 
s t  
2
5. 3x y
6.
3 4
2 4 0
1
2 
7.   
 x 2
 39 
8.  5  
3 
2
2
s t 
9.  4 4  
s t 
2
5
 36a 
10.  4 5  
 4a b 
2 2
SOLUTIONS


0
1. 2a b  1
2
 
1
3. a
 5
a
5
2
7
4. s  4s  4s
5 1

2
5. 3x y
 
2 4 0
6. s t

3 4

4
 3 x y
 1
8
12

8
x

81y12
SOLUTIONS
1
2 
7.  
 x
9 2
3 
8.  5 
3 
2
1
x
4
 x   4
 
 3
2

4 2

1
3  8
3
8

s t 
 2  2 2
4 4
s t
9.  4 4   s t
s t 
10
2
5
b

2

2
10
 36a 
9
a
b

2


10.  4 5  
81a
4
a
b


2 2
Basic Examples
x x  x
2
3
2 3
x 
x
xy
x y
4 3
3
43
3
x
x
3
12
5
Basic Examples
3
x
x
   3
y
y
 
7
74
x
x
3


x
4
x
1
5
x
1
1


7
7 5
2
x
x
x
3
More Examples
2a 3  7 a 4  2  7a 34  14a 7
 5r 2  8r 3  2r 2   5  8  2r 23 2   80r 7
2m n 
2 5 3
3xy
3
2
13
23 53
2 m n
 2 m n  8m n
3
 3 x y  27 x y
3
3
3
3
3
2 2
2
2
a
2
a
4
a
 
   2 2  2
9b
 3b  3 b
4
4 1
8x
8x

 4x 3
2x 2 1
9z3 9 1
1
3

3 2  2
5
53
x
3z
3z
x
6 15
6 15
More Examples
3x 3 y 2 z  7 xyz2  3  7x31 y 21 z1 2  21x 4 y 3 z 3


3x y  2 xy 
3
 8 xy2  3xy   2 xy3   8  3  2x111 y 213  48x 3 y 6
2
3 2
2 2

12


x 22 y 32 212 x12 y 22 
9 x 4 y 6  4 x 2 y 4  9  4x 4 2 y 6 4  36 x 6 y10
3
 5a b  513 a 33b13 53 a 9b 3 125a 9b 3 125a 93 125a 6

  13 13 23  3 3 6 


6 3
3 6
3
2 
3
a
b
27
b
3
a
b
27
a
b
27
b
3
ab


3
POLYNOMIALS
Monomial: A number, a variable or the product of a number and one or more
variables.
Polynomial: A monomial or a sum of monomials.
Binomial: A polynomial with exactly two terms.
Trinomial: A polynomial with exactly three terms.
Coefficient: A numerical factor in a term of an algebraic expression.
Degree of a monomial: The sum of the exponents of all of the variables in the
monomial.
Degree of a polynomial in one variable: The largest exponent of that variable.
Standard form: When the terms of a polynomial are arranged from the largest
exponent to the smallest exponent in decreasing order.
Polynomials and Polynomial Functions
Definitions
Term: a number or a product of a number and variables raised
to a power.
2
2
3, 5x ,  2 x, 9 x y
Coefficient: the numerical factor of each term.
5x 2 ,  2 x, 9 x 2 y
Constant: the term without a variable.
3,  6, 5, 32
Polynomial: a finite sum of terms of the form axn, where a is a
real number and n is a whole number.
15 x  2 x  5
3
2
21y 6  7 y 5  2 y 3  6 y
Polynomials and Polynomial Functions
Definitions
Monomial: a polynomial with exactly one term.
2
ax ,
rt ,
4
2x ,
9m,
2
9x y
Binomial: a polynomial with exactly two terms.
x  8, r  3, 5 x 2  2 x, 2 x  9 x 2 y
Trinomial: a polynomial with exactly three terms.
x 2  x  8,
r 5  3r  3, 5 x 2  2 x  7
Polynomials and Polynomial Functions
Definitions
The Degree of a Term with one variable is the exponent on
the variable.
5x 2  2,
2x 4  4,
9m  1
The Degree of a Term with more than one variable is
the sum of the exponents on the variables.
7x y  3,
2
2x y  6,
4
2
9mn z  10
5 4
The Degree of a Polynomial is the greatest degree of
the terms of the polynomial variables.
2 x  3 x  7  3,
3
2 x y  5x y  6 x  6
4
2
2
3
Polynomials and Polynomial Functions
Practice Problems
Identify the degrees of each term and the degree of the
polynomial.
5x  4 x  5x
2
1
3
3
4𝑎2 𝑏 4 + 3𝑎3 𝑏 5 − 9𝑏 4 + 4
2
8
6
3
8
3
4𝑥 5 𝑦 4 + 5𝑥 4 𝑦 6 − 6𝑥 3 𝑦 3 + 2𝑥𝑦
9
10
6
10
2
0
What is the degree of the monomial?
4
5x b
2
The degree of a monomial is the sum of the exponents of the variables in the
monomial.
The exponents of each variable are 4 and 2. 4+2 = 6.
The degree of the monomial is 6.
The monomial can be referred to as a sixth degree monomial.
A polynomial is a monomial or the sum of monomials
4x
2
3x  8
3
5 x  2 x  14
2
Each monomial in a polynomial is a term of the polynomial.
The number factor of a term is called the coefficient.
The coefficient of the first term in a polynomial is the lead coefficient.
A polynomial with two terms is called a binomial.
A polynomial with three terms is called a trinomial.
The degree of a polynomial in one variable is the largest exponent of that
variable.
2
A constant has no variable. It is a 0 degree polynomial.
4x 1 This is a 1
st
degree polynomial. 1st degree polynomials are linear.
5 x  2 x  14
2
3x  8
3
This is a 2nd degree polynomial. 2nd degree polynomials
are quadratic.
This is a 3rd degree polynomial. 3rd degree polynomials are cubic.
Classify the polynomials by degree and number of terms.
Classify by
number of terms
Polynomial
Degree
Classify by
degree
a.
5
Zero
Constant
Monomial
b.
2x  4
First
Linear
Binomial
c.
3x 2  x
Second
Quadratic
Binomial
Third
Cubic
Trinomial
d.
x  4x 1
3
2
To rewrite a polynomial in standard form, rearrange the terms of the
polynomial starting with the largest degree term and ending with the lowest
degree term.
The leading coefficient, the coefficient of the first term in a polynomial written
in standard form, should be positive.
Write the polynomials in standard form.
5x  4 x 4  x 2  7
2 x3  x 4  7  5x  5x 2
4x  x  5x  7
 x  2x  5x  5x  7
4
2
4
3
2
Remember: The lead
coefficient should be
positive in standard form.
 1( x  2 x  5x  5x  7)
To do this, multiply the
polynomial by –1 using the
distributive property.
x 4  2x 3  5x 2 5x  7
4
3
2
Write the polynomials in standard form and identify the polynomial by degree
and number of terms.
1.
7  3x  2 x
2.
1  3x  2 x
3
2
2
7  3x  2 x
3
2
7  3x 3  2 x 2
 3x  2x  7
3
2

 1  3x 3  2 x 2  7
3x 3  2 x 2  7
This is a 3rd degree, or cubic, trinomial.

1  3x  2 x
2
1  3x 2  2 x
3x  2x  1
2
This is a 2nd degree, or quadratic, trinomial.
Polynomials and Polynomial Functions
Practice Problems
Evaluate each polynomial function
f  x   3 x 2  10
f  1
3  1  10
2
g 3

3 1 10  3 10  7
g  y   6 y 2  11 y  20
6  3  11 3  20  6  9  33  20 
2
54  33  20  87  20  67
Vocabulary
• Monomials - a number, a variable, or a product of a
number and one or more variables. 4x, 20x2yw3, -3,
a2b3, and 3yz are all monomials.
• Polynomials – one or more monomials added or
subtracted
• 4x + 6x2, 20xy - 4, and 3a2 - 5a + 4 are all polynomials.
Like Terms
Like Terms refers to monomials that have the same variable(s) but may have
different coefficients. The variables in the terms must have the same powers.
Which terms are like?
3a2b, 4ab2, 3ab, -5ab2
4ab2 and -5ab2 are like.
Even though the others have the same variables, the exponents are not the same.
3a2b = 3aab, which is different from 4ab2 = 4abb.
Like Terms
Constants are like terms.
Which terms are like?
2x, -3, 5b, 0
-3 and 0 are like.
Which terms are like?
3x, 2x2, 4, x
3x and x are like.
Which terms are like?
2wx, w, 3x, 4xw
2wx and 4xw are like.
Adding Polynomials
Add: (x2 + 3x + 1) + (4x2 +5)
Step 1: Underline like terms:
(x2 + 3x + 1) + (4x2 +5)
Notice: ‘3x’ doesn’t have a like term.
Step 2: Add the coefficients of like terms, do not change the powers of the
variables:
(x2 + 4x2) + 3x + (1 + 5)
5x2 + 3x + 6
Adding Polynomials
Some people prefer to add polynomials by stacking them. If you choose to do
this, be sure to line up the like terms!
(x2 + 3x + 1) + (4x2 +5)
(x2 + 3x + 1)
+ (4x2
+5)
5x2 + 3x + 6
Stack and add these polynomials: (2a2+3ab+4b2) + (7a2+ab+-2b2)
(2a2 + 3ab + 4b2)
(2a2+3ab+4b2) + (7a2+ab+-2b2)
+ (7a2 + ab + -2b2)
9a2 + 4ab + 2b2
Adding Polynomials
• Add the following polynomials; you may stack them if you
prefer:
1) 3x  7x  3x  4x   6x3  3x
3
3
2) 2w  w  5  4w  7w  1 6w  8w  4
2
2
2
3) 2a  3a  5a  a  4a  3 
3
2
3
3a  3a  9a  3
3
2
Subtracting Polynomials
Subtract: (3x2 + 2x + 7) - (x2 + x + 4)
Step 1: Change subtraction to addition (Keep-Change-Change.).
(3x2 + 2x + 7) + (- x2 + - x + - 4)
Step 2: Underline OR line up the like terms and add.
(3x2 + 2x + 7)
+ (- x2 + - x + - 4)
2x2 + x + 3
Subtracting Polynomials
• Subtract the following polynomials by changing to addition
(Keep-Change-Change.), then add:
1) x  x  4 3x  4x  1 2x  3x  5
2
2
2
2) 9y  3y  1 2y  y  9 7y  4y  10
2
2
2
3) 2g  g  9 g  3g  3  g  g  g  12
2
3
2
3
2
Polynomials and Polynomial Functions
Multiplication
Multiplying Monomials by Monomials
Examples:
2
90x
10x  9x 
10

88x
8 x  11x  
3
 5x
7
4
5

x

   5x
Polynomials and Polynomial Functions
Multiplication
Multiplying Monomials by Polynomials
Examples:
4 x  x  4 x  3  4x 16x 12x
3
2
2
8 x  7 x  1  56x 8x
5
4
 5x  3x
3
2
 x  2   15x 5x 10x
5
4
3
Polynomials and Polynomial Functions
Multiplication
Multiplying Two Polynomials
Examples:
2
3
2

5x

10x
x
3x
50x 15
 x  5  x  10 x  3 
2
x 3 15x 2 47x 15
2
4
x
  x  5  3x  4  
12x 3 16x 23x 24x 15x 20 
12x 13x 11x 20
3
2
Multiplying Polynomials
Special Products
Multiplying Two Binomials using FOIL
First terms
Outer terms
Inner terms
2
x
x

3
x

4

4x 3x 12 
 

x 7x 12
2
2
x
x

7
x

4

4x 7x 28 



x 2 3x 28
Last terms
Multiplying Polynomials
Special Products
Multiplying Two Binomials using FOIL
First terms
Outer terms
Inner terms
2
3
2
2
y

3
y

4

2
y

8y
3y 12 

 
2 y 8y 3y 12
3
y
2
 4 
2
2
y
2
 4  y  4  
2
y 4 y 4 y 16  y 8y 16
4
2
2
4
2
Last terms
Multiplying Polynomials
Special Products
Squaring Binomials
2
2
a

b

a

2
ab

b


2
 a  b
2
 a  2ab  b
2
2
2
4
x

5

16x
2  20x  25 


2
16 x  40 x  25
2
 3 x  6 
2
 9x 2 18x  36 
2
9 x 2  36 x  36
Multiplying Polynomials
Special Products
Multiplying the Sum and Difference of Two Binomials
2
2
a

b
a

b

a

b



2
2
x
 81
x
x

9
x

9

9x 9x 81 

 
3x  53x  5  9x
2
15x 15x 25  9 x  25
x  8x  8  x  64
2
5 x  135 x  13  25 x  169
2
2
Multiplying Polynomials
Special Products
Dividing by a Monomial
ab a b
 
c
c c
where c  0
21x8  9 x 6  12 x 4

3
3x
8
21x
3x3
6
4
9x
12 x
 3  3 
3x
3x
7 x 3x 4x
5
3
Extra
Problems
Polynomials and Polynomial Functions
Multiplication
Multiplying Two Polynomials
Examples:
2

x
 10x  5x  50 
x  5x  10
x 2  15x  50
2
12x
4x  53x  4  16 x  15x  20 
12x 2  x  20
Polynomials and Polynomial Functions
Multiplication
Multiplying Two Polynomials
Examples:
 x  3  2 x
2
 5x  4 
2x 5x 4x 6x 15x 12 
3
2
2
2x 3  x 2 11x 12
Polynomials and Polynomial Functions
Multiplication
Multiplying Two Polynomials
Examples:
 x  3 y  2
2

 x  3 y  2 x  3 y  2 
x 3xy 2x 3xy 9 y 6 y 2x 6 y 4 
2
2
x 2 6xy 4x 9 y 2 12 y 4
Dividing Polynomials
The objective is to be able to divide a polynomial by a monomial.
( 6x  3)  3x
2
Step 1
( 6x  3)  3x
2
Divide each term of the polynomial by the monomial.
2
6x
3

3x 3x
Step 2
2
6x
3

3x 3x
Factor each expression.
2  3 x  x
3

3 x
3 x
Step 3
2  3 x  x
3

3 x
3 x
Divide out the common factors in each expression.
2  3 x  x
3

3 x
3 x
The numbers and
variables which are
crossed out divide out
to 1.
Step 4
2  3 x  x
3

3 x
3 x
Write in simplified form.
1
2x 
x
1.
8x
2.
(12 x  2 x  3)  2 x
3
2

 3x  24x
A
A
Dividing a Polynomial
by a Polynomial
The objective is to be able to divide a polynomial by a polynomial by using long division.
Dividend – the number which is being divided.
Divisor – The number that is being divided into the dividend.
Quotient – The result obtained when numbers or expressions are divided.
Remainder – The part that is left over when the divisor no longer goes into the dividend a
whole number of times.
Polynomial Long Division
Divide x  3x  1 by x  2.
2
Step 1: Write it as you would a regular long division problem.
x  2 x  3x  1
2
The x+2 is the divisor and the x2+3x-1 is the dividend.
Step 2
x  2 x  3x  1
2
x
2
x  2 x  3x  1
x  2x
x 1
2
Divide x2 by x to get x. Place this on
top.
Multiply x+2 by x to get x2 +2x.
Subtract the x2+2x from the x2+3x-1.
Step 3
Divide the x by x to get 1.
Multiply x+2 by 1 to get x+2.
Subtract x+2 from the x-1.
x 1
x  2 x 2  3x  1
x  2x
x 1
x2
3
2
Step 4
Write your final answer.
x 2  3x  1 divided by x  2 is
3
x  1
x2
The x+1 is the quotient.
3
The is the remainder.
x+2
1.
x
2.
12x
2

 x  15   x - 2
2

 5x  10   4x  1
A
A
Example: Divide x2 + 3x – 2 by x – 1 and check the answer.
2
x
1. x x   x
x
2. x( x  1)  x 2  x
x + 2
2
x  1 x22  3x  2
x + x
3. ( x 2  3x)  ( x 2  x)  2 x
4. x 2 x  2 x  2
x
5. 2( x  1)  2 x  2
2x – 2
2x + 2
–4
remainder
Answer: x + 2 +
6. (2 x  2)  (2 x  2)  4
–4
x 1
Check:
(x + 2)
(x + 1)
+ (– 4)
= x2 + 3x – 2
quotient divisor remainder
dividend
correct
Example: Divide 4x + 2x3 – 1 by 2x – 2 and check the answer.
x2 + x
+ 3
2x  2 2x  0x  4x  1
3
2
2x3 – 2x2
2x2
Since there is no x2 term in the
dividend, add 0x2 as a placeholder.
+ 4x
2x2 – 2x
6x – 1
6x – 6
5
Answer:
x2
+x+3

Check: (x2 + x + 3)(2x – 2) + 5
= 4x + 2x3 – 1
Write the terms of the dividend in
descending order.
5
2x  2
3
2
x
1.
2. x 2 (2 x  2)  2 x3  2 x 2
 x2
2x
2
2
x
3
3
2
2
3. 2 x  (2 x  2 x )  2 x
4.
x
2x
5. x(2 x  2)  2 x 2  2 x
6. (2 x 2  4 x)  (2 x 2  2 x)  6 x
8. 3(2 x  2)  6 x  6
7. 6 x  3
2x
9. (6 x  1)  (6 x  6)  5  remainder
Example: Divide x2 – 5x + 6 by x – 2.
x – 3
x  2 x22  5 x  6
x – 2x
– 3x + 6
– 3x + 6
0
Answer: x – 3 with no remainder.
Check: (x – 2)(x – 3)
= x2 – 5x + 6
Example: Divide x3 + 3x2 – 2x + 2 by x + 3 and check the answer.
x2 + 0x
– 2
x  3 x 3  3x 2  2 x  2
Note: the first subtraction
eliminated two terms from
the dividend.
Therefore, the quotient
skips a term.
Answer:
x2
–2+
8
x3
x3 + 3x2
0x2
– 2x
+ 2
– 2x – 6
8
Check: (x + 3)(x2 – 2) + 8
= x3 + 3x2 – 2x + 2
BINOMIAL EXPANSION
The binomial theorem provides a useful method for raising any
binomial to a nonnegative integral power.
Consider the patterns formed by expanding (x + y)n.
(x + y)0 = 1
1 term
(x + y)1 = x + y
2 terms
(x + y)2 = x2 + 2xy + y2
3 terms
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
4 terms
5 terms
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
Notice that each expansion has n + 1 terms.
Example: (x + y)10 will have 10 + 1, or 11 terms.
Copyright © by Houghton Mifflin
Company, Inc. All rights reserved.
116
6 terms
Consider the patterns formed by expanding (x + y)n.
(x + y)0 = 1
(x + y)1 = x + y
(x + y)2 = x2 + 2xy + y2
(x + y)3 = x3 + 3x2y + 3xy2 + y3
(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4
(x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
1. The exponents on x decrease from n to 0.
The exponents on y increase from 0 to n.
2. Each term is of degree n.
Example: The 5th term of (x + y)10 is a term with x6y4.”
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The coefficients of the binomial expansion are called binomial
coefficients. The coefficients have symmetry.
(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5
The first and last coefficients are 1.
The coefficients of the second and second to last terms
are equal to n.
Example: What are the last 2 terms of (x + y)10 ? Since n = 10,
the last two terms are 10xy9 + 1y10.
The coefficient of xn–ryr in the expansion of (x + y)n is written  n 
or nCr . So, the last two terms of (x + y)10 can be expressed  r 
as 10C9 xy9 + 10C10 y10 or as 10  xy 9 + 10  y10.
9 
 
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10 
 
The triangular arrangement of numbers below is called Pascal’s
Triangle.
1
0th row
1
1
1+2=3
1
6 + 4 = 10
1
1
2
3
4
1st row
1
3
6
2nd row
1
4
3rd row
1
1 5 10 10 5 1
4th row
5th row
Each number in the interior of the triangle is the sum of the two
numbers immediately above it.
The numbers in the nth row of Pascal’s Triangle are the binomial
coefficients for (x + y)n .
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Example: Use the fifth row of Pascal’s Triangle to generate the
6
6
sixth row and find the binomial coefficients  ,  , 6C4 and 6C2 .
1 5
5th row
1
5
10
10
5
1
6th row
1
6 15 20
15 6
1
6 6
  1
0  
6
 
 2
6
 
 3
 6 6
  5
 4  
6
 
6
6C0
6C2
6C3
6C4
6C6
6C1
6
 =6=
1
6C5
6
  and 6C4 = 15 = 6C2.
5
There is symmetry between binomial coefficients.
nCr = nCn–r
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Example: Use Pascal’s Triangle to expand (2a + b)4.
0th row
1
1
1
1
1
2
3
4
1st row
1
3
6
2nd row
1
3rd row
1
4
1
4th row
(2a + b)4 = 1(2a)4 + 4(2a)3b + 6(2a)2b2 + 4(2a)b3 + 1b4
= 1(16a4) + 4(8a3)b + 6(4a2b2) + 4(2a)b3 + b4
= 16a4 + 32a3b + 24a2b2 + 8ab3 + b4
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The symbol n! (n factorial) denotes the product of the first n
positive integers. 0! is defined to be 1.
1! = 1
4! = 4 • 3 • 2 • 1 = 24
6! = 6 • 5 • 4 • 3 • 2 • 1 = 720
n! = n(n – 1)(n – 2)  3 • 2 • 1
Formula for Binomial Coefficients For all nonnegative
n!
integers n and r,
n
Cr 
(n  r )!r !
7!
7!
7
Example: 7 C3 


(7  3)! • 3! 4! • 3! 4! • 3!
(7 • 6 • 5 • 4) • (3 • 2 • 1) 7 • 6 • 5 • 4


 35
(4 • 3 • 2 • 1) • (3 • 2 • 1) 4 • 3 • 2 • 1
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Example: Use the formula to calculate the binomial coefficients
 12  and  50  .
C
,
C
,
10 5 15 0
 
1
 
 48 
10!
10!
(10 • 9 • 8 • 7 • 6) • 5! 10 • 9 • 8 • 7 • 6



 252
10 C5 
(10  5)! • 5! 5! • 5!
5! • 5!
5 • 4 • 3• 2 •1
10!
10!
1! 1

  1
10 C0 
•
•
(10  0)! 0! 10! 0! 0! 1
50!
(50 • 49) • 48! 50 • 49
 50 
50!



 1225
  
•
•
•
2! 48!
2 1
 48  (50  48)! • 48! 2! 48!
12 
12! 12 • 11! 12
12!


  12
  
•
•
1
1  (12  1)! • 1! 1! 1! 11! 1!
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Binomial Theorem
n 1
( x  y)  x  nx y 
n
n
 nCr x
n r
y 
r
 nxy
n 1
y
n
n!
with nCr 
(n  r )!r !
Example: Use the Binomial Theorem to expand (x4 + 2)3.
(x 4  2)3  3 C0(x 4 )3  3 C1( x 4 ) 2 (2)  3 C2(x 4 )(2) 2  3 C3(2)3
 1 (x 4 )3  3( x 4 ) 2 (2)  3(x 4 )(2) 2  1(2)3
 x12  6 x8  12 x 4  8
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Although the Binomial Theorem is stated for a binomial which
is a sum of terms, it can also be used to expand a difference of
terms.
Simply rewrite
(x + y) n as (x + (– y)) n
and apply the theorem to this sum.
Example: Use the Binomial Theorem to expand (3x – 4)4.
(3x  4) 4  (3x  (4)) 4
 1(3x) 4  4(3x)3 (4)  6(3x) 2 (4) 2  4(3x)(4)3  1(4) 4
 81x 4  4(27 x3 )(4)  6(9 x 2 )(16)  4(3x)(64)  256
 81x 4  432 x 3  864 x 2  768x  256
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Example: Use the Binomial Theorem to write the first three
terms in the expansion of (2a + b)12 .
12 
12 
12 
12
11
(2a  b)   (2a)   (2a) b   (2a)10 b 2  ...
0 
1 
2 
12
12!
12!
12!
12 12
11 11

(2 a ) 
(2 a )b 
(210 a10 )b 2  ...
(12  0)! • 0!
(12  1)! • 1!
(12  2)! • 2!
 (212 a12 )  12(211 a11 )b  (12 • 11)(210 a10 )b 2  ...
 4096 a12  24576 a11b  135168 a10b 2  ...
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8.4 rth Term of a Binomial Expansion
rth Term of the Binomial Expansion
The rth term of the binomial expansion of (x + y)n,
where n > r – 1, is
 n  n ( r 1) r 1
y

x
 r  1
Example: Find the eighth term in the expansion of (x + y)13 .
Think of the first term of the expansion as x13y 0 . The power of
y is 1 less than the number of the term in the expansion.
The eighth term is 13C7 x 6 y7.
13! (13 • 12 • 11 • 10 • 9 • 8) • 7!

13 C7 
6! • 7!
6! • 7!
13 • 12 • 11 • 10 • 9 • 8

 1716
6 • 5 • 4 • 3 • 2 •1
Therefore, the eighth term of (x + y)13 is 1716 x 6 y7.
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Find the 6th term in the expansion of (3a + 2b)12
Using the Binomial Theorem, let x = 3a and y = 2b
and note that in the 6th term, the exponent of y is
m = 5 and the exponent of x is n – m = 12 – 5 = 7.
Consequently, the 6th term of the expansion is:
12 1110  9  8  7!
7
5
3a  2b 
12 C5 x y 
7!5!
7
5
= 55,427,328 a7b5
8.4 Finding a Specific Term of a Binomial
Expansion.
10
(
a

2
b
)
Example Find the fourth term of
.
Solution Using n = 10, r = 4, x = a, y = 2b in the
formula, we find the fourth term is
10  7
3
7
3
7 3
  a (2b)  120a 8b  960a b .
3
FACTORING POLYNOMIALS
Definitions
• Recall: Factors of a number are the numbers that
divide the original number evenly.
• Writing a number as a product of factors is called a
factorization of the number.
• The prime factorization of a number is the
factorization of that number written as a product of
prime numbers.
• Common factors are factors that two or more
numbers have in common.
• The Greatest Common Factor (GCF) is the largest
common factor.
Ex: Find the GCF(24, 40).
Prime factor each number:
24
2 12
2 6
2 3  24 = 2*2*2*3 = 23*3
 GCF(24,40) = 23 = 8
40
2 20
2 10
2 5  40 = 2*2*2*5 = 23*5
The Greatest Common Factor of terms of a polynomial
is the largest factor that the original terms share
• Ex: What is the GCF(7x2, 3x)
7x2 = 7 * x * x
3x = 3 * x
The terms share a factor of x
 GCF(7x2, 3x) = x
Ex: Find the
5
3
2
GCF(6a ,3a ,2a )
• 6a5 = 2*3*a*a*a*a*a
• 3a3 = 3*a*a*a
• 2a2 = 2*a*a
The terms share two
factors of a
 GCF(6a5,3a3,2a2)= a2
Note: The exponent of the variable in the GCF is
the smallest exponent of that variable the terms
Definitions
• To factor an expression means to write an equivalent
expression that is a product
• To factor a polynomial means to write the polynomial
as a product of other polynomials
• A factor that cannot be factored further is said to be a
prime factor (prime polynomial)
• A polynomial is factored completely if it is written as a
product of prime polynomials
To factor a polynomial completely, ask
• Do the terms have a common factor (GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
• Is the polynomial a difference of squares?
• a2 – b 2
• Is the polynomial a sum/difference of cubes?
• a3 + b3 or a3 – b3
• Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two binomials?
• Factored completely?
Ex: Factor 7x2 + 3x
Think of the Distributive Law: a(b+c) = ab + ac
 reverse it ab + ac = a(b + c)
Do the terms share a common factor?
What is the GCF(7x2, 3x)?
Recall: GCF(7x2, 3x) = x
2
 7 x + 3 x = x( + )
x
x
 7x2 + 3x = x(7x + 3)
Factor out
What’s left?
Ex: Factor 6a5 – 3a3 – 2a2
Recall: GCF(6a5,3a3,2a2)= a2
6a5
a2
3
–
1
3
3a –
a2
2a2 = a2( 6a3 - 3a - 2 )
a2
 6a5 – 3a3 – 2a2 = a2(6a3 – 3a – 2)
Your Turn to Try a Few
Ex: Factor x(a + b) – 2(a + b)
Always ask first if there is common factor the terms share . . .
x(a + b) – 2(a + b)
Each term has factor (a + b)
 x(a + b) – 2(a + b) = (a + b)( x – 2 )
(a + b)
(a + b)
 x(a + b) – 2(a + b) = (a + b)(x – 2)
Ex: Factor a(x – 2) + 2(2 – x)
As with the previous example, is there a common factor
among the terms?
Well, kind of . . . x – 2 is close to 2 - x . . . Hum . . .
Recall: (-1)(x – 2) = - x + 2 = 2 – x
 a(x – 2) + 2(2 – x) = a(x – 2) + 2((-1)(x – 2))
= a(x – 2) + (– 2)(x – 2)
= a(x – 2) – 2(x – 2)
 a(x – 2) – 2(x – 2) = (x – 2)( a
(x – 2)
(x – 2)
– 2
)
Ex: Factor b(a – 7) – 3(7 – a)
Common factor among the terms?
Well, kind of . . . a – 7 is close to 7 - a
Recall: (-1)(a – 7) = - a + 7 = 7 – a
 b(a – 7) – 3(7 – a) = b(a – 7) – 3((-1)(a – 7))
= b(a – 7) + 3(a – 7)
= b(a – 7) +3(a – 7)
 b(a – 7) + 3(a – 7) = (a – 7)( b
(a – 7)
(a – 7)
+ 3
)
Your Turn to Try a Few
To factor a polynomial completely, ask
• Do the terms have a common factor (GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
• Is the polynomial a difference of squares?
• a2 – b 2
• Is the polynomial a sum/difference of cubes?
• a3 + b3 or a3 – b3
• Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two binomials?
• Factored completely?
Factor by Grouping
• If the polynomial has four terms, consider factor by grouping
1. Factor out the GCF from the first two terms
2. Factor out the GCF from the second two terms (take the negative
sign if minus separates the first and second groups)
3. If factor by grouping is the correct approach, there should be a
common factor among the groups
4. Factor out that GCF
5. Check by multiplying using FOIL
Ex: Factor 6a3 + 3a2 +4a + 2
Notice 4 terms . . . think two groups: 1st two and 2nd two
Common factor among the 1st two terms?
GCF(6a3, 3a2) = 3a2
2a
1
 6a3 + 3a2 = 3a2( 2a + 1 )
2
3a2 3a
Common factor among the 2nd two terms?
2
1
GCF(4a, 2) = 2
 4a + 2 = 2( 2a + 1
2
2
Now put it all together . . .
)
6a3 + 3a2 +4a + 2 = 3a2(2a + 1) + 2(2a + 1)
Four terms  two terms. Is there a common factor?
Each term has factor (2a + 1)
3a2(2a + 1) + 2(2a + 1) = (2a + 1)( 3a2 + 2 )
(2a + 1)
(2a + 1)
6a3 + 3a2 +4a + 2 = (2a + 1)(3a2 + 2)
Ex: Factor 4x2 + 3xy – 12y – 16x
Notice 4 terms . . . think two groups: 1st two and 2nd two
Common factor among the 1st two terms?
4x
GCF(4x2, 3xy) = x
3y
 4x2 + 3xy = x( 4x + 3y )
x
x
Common factor among the 2nd two terms?
3y
GCF(-12y, - 16x) = -4
4x
 -12y – 16x = - 4( 3y + 4x )
-4
-4
Now put it all together . . .
4x2 + 3xy – 12y – 16x = x(4x + 3y) – 4(4x + 3y)
Four terms  two terms. Is there a common factor?
Each term has factor (4x + 3y)
x(4x + 3y) – 4(4x + 3y) = (4x + 3y)( x – 4 )
(4x + 3y)
(4x + 3y)
4x2 + 3xy – 12y – 16x = (4x + 3y)(x – 4)
Ex: Factor 2ra + a2 – 2r – a
Notice 4 terms . . . think two groups: 1st two and 2nd two
Common factor among the 1st two terms?
GCF(2ra, a2) = a
 2ra + a2 = a( 2r + a )
a
a
Common factor among the 2nd two terms?
GCF(-2r, - a) = -1
 -2r – a = - 1( 2r + a )
-1 -1
Now put it all together . . .
2ra + a2 –2r – a = a(2r + a) – 1(2r + a)
Four terms  two terms. Is there a common factor?
Each term has factor (2r + a)
a(2r + a) – 1(2r + a) = (2r + a)( a – 1 )
(2r + a)
(2r + a)
2ra + a2 – 2r – a = (2r + a)(a – 1)
Your Turn to Try a Few
To factor a polynomial completely, ask
• Do the terms have a common factor (GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
• Is the polynomial a difference of squares?
• a2 – b 2
• Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two binomials?
• Factored completely?
Special Polynomials
Is the polynomial a difference of squares?
• a2 – b2 = (a – b)(a + b)
Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 = (a + b)2
• a2 – 2ab + b2 = (a – b)2
Ex: Factor x2 – 4
Notice the terms are both perfect squares
and we have a difference  difference of squares
x2 = (x)2
4 = (2)2
 x2 – 4 = (x)2 – (2)2 = (x – 2)(x + 2)
a2 – b2 = (a – b)(a + b)
factors as
Ex: Factor 9p2 – 16
Notice the terms are both perfect squares
and we have a difference  difference of squares
9p2 = (3p)2
16 = (4)2
 9a2 – 16 = (3p)2 – (4)2 = (3p – 4)(3p + 4)
a2 – b2 = (a – b)(a + b)
factors as
Ex: Factor y6 – 25
Notice the terms are both perfect squares
and we have a difference  difference of squares
y6 = (y3)2
25 = (5)2
 y6 – 25 = (y3)2 – (5)2 = (y3 – 5)(y3 + 5)
a2 – b2 = (a – b)(a + b)
factors as
Ex: Factor 81 – x2y2
Notice the terms are both perfect squares
and we have a difference  difference of squares
81 = (9)2
x2y2 = (xy)2
 81 – x2y2 = (9)2 – (xy)2 = (9 – xy)(9 + xy)
a2 – b2 = (a – b)(a + b)
factors as
Your Turn to Try a Few
To factor a polynomial completely, ask
• Do the terms have a common factor (GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
• Is the polynomial a difference of squares?
• a2 – b 2
• Is the polynomial a sum/difference of cubes?
• a3 + b3 or a3 – b3
• Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two binomials?
• Factored completely?
FOIL Method of Factoring
• Recall FOIL
• (3x + 4)(4x + 5) = 12x2 + 15x + 16x + 20 = 12x2 + 31x + 20
The product of the two binomials is a trinomial
The constant term is the product of the L terms
The coefficient of x, b, is the sum of the O & I products
The coefficient of x2, a, is the product of the F terms
FOIL Method of Factoring
1. Factor out the GCF, if any
2. For the remaining trinomial, find the F terms (__ x +
= ax2
3. Find the L terms ( x + __ )( x + __ ) = c
4. Look for the outer and inner products to sum to bx
5. Check the factorization by using FOIL to multiply
)(__ x +
)
Ex: Factor b2 + 6b + 5
1. there is no GCF
2. the lead coefficient is 1  (1b
3. Look for factors of 5
)(1b
)
1, 5 & 5, 1
(b + 1)(b + 5) or (b + 5)(b + 1)
4. outer-inner product?
(b + 1)(b + 5)  5b + b = 6b
or (b + 5)(b + 1)  b + 5b = 6b
Either one works  b2 + 6b + 5 = (b + 1)(b + 5)
5. check: (b + 1)(b + 5) = b2 + 5b + b + 5
= b2 + 6b + 5
Ex: Factor y2 + 6y – 55
1. there is no GCF
2. the lead coefficient is 1  (1y
)(1y
)
3. Look for factors of – 55 1, -55 & 5, - 11 & 11, - 5 & 55, - 1
(y + 1)(y – 55) or (y + 5)(y - 11) or ( y + 11)(y – 5) or (y + 55)(y – 1)
4. outer-inner product?
(y + 1)(y - 55)  -55y + y = - 54y (y + 5)(y - 11)  -11y + 5y = -6y
(y + 55)(y - 1)  -y + 55y = 54y
(y + 11)(y - 5)  -5y + 11y = 6y
 y2 + 6y - 55 = (y + 11)(y – 5)
5. check: (y + 11)(y – 5) = y2 – 5y + 11y - 55
= y2 + 6y – 55
Factor completely – 3 Terms
•
Always look for a common factor
•
•
•
•
immediately take it out to the front of the expression all common factors
show what’s left inside ONE set of parenthesis
Identify the number of terms.
If there are three terms, and the leading coefficient is positive:
•
•
find all the factors of the first term, find all the factors of the last term
Within 2 sets of parentheses,
•
•
•
•
place the factors from the first term in the front of the parentheses
place the factors from the last term in the back of the parentheses
NEVER put common factors together in one parenthesis.
check the last sign,
•
•
•
if the sign is plus: use the SAME signs, the sign of the 2nd term
if the sign is minus: use different signs, one plus and one minus
“smile” to make sure you get the middle term
•
multiply the inner most terms together then multiply the outer most
terms together, and add the two products together.
Factor completely:
2
2x
– 5x – 7
•Factors of the first term: 1x & 2x
•Factors of the last term:
-1 & 7 or 1 & -7
•(2x – 7)(x + 1)
Factor Completely.
4x2 + 83x + 60
• Nothing common
• Factors of the first term: 1 & 4 or 2 & 2
• Factors of the last term: 1,6 2,30 3,20 4,15 5,12 6,10
• Since each pair of factors of the last has an even number,
we can not use 2 & 2 from the first term
•
(4x + 3)(1x + 20 )
Sign Pattern for the Binomials
Trinomial Sign Pattern
Binomial Sign Pattern
+
+
(
+
)(
+
)
-
+
(
-
)(
-
)
-
-
1 plus and 1 minus
+
-
1 plus and 1 minus
But as you can tell from the previous example, the
FOIL method of factoring requires a lot of trial and
error (and hence luck!) . . . Better way?
Your Turn to Try a Few
ac Method for factoring ax2 + bx + c
1.
2.
3.
4.
5.
6.
Factor out the GCF, if any
For the remaining trinomial, multiply ac
Look for factors of ac that sum to b
Rewrite the bx term as a sum using the factors found in step 3
Factor by grouping
Check by multiplying using FOIL
Ex: Factor 3x
3 2 – 4x
4 – 15
1. Is there a GCF? No
2. Multiply ac  a = 3 and c = – 15 3(-15) = - 45
3. Factors of -45 that sum to – 4
1
– 45  – 44
3
5
– 15  – 12
–9 –4
Note: although there are more
factors of – 45, we don’t have to
check them since we found what
we were looking for!
4. Rewrite the middle term
3x2 – 4x – 15 = 3x2 – 9x + 5x – 15
Four-term polynomial . . . How should we
proceed to factor?
Factor by grouping . . . 3x2 – 9x + 5x – 15
Common factor among the 1st two terms?
3x
3
 3 x 2 – 9x = 3x( x – 3 )
3x
3x
Common factor among the 2nd two terms?
5
3
 5 x – 15 = 5( x – 3 )
5
5
 3x2 – 9x + 5x – 15 = 3x(x – 3) + 5(x – 3)
= (x – 3)( 3x + 5 )
Ex: Factor 2t
2 2 + 5t
5 – 12
1. Is there a GCF? No
2. Multiply ac  a = 2 and c = – 12 2(-12) = - 24
3. Factors of -24 that sum to 5
1
– 24  – 23
2
3
– 12  – 10
–8 –5
Close but wrong sign so reverse it
-3 8 5
4. Rewrite the middle term
2t2 + 5t – 12 = 2t2 – 3t + 8t – 12
Four-term polynomial . . . Factor by grouping . . .
2t2 – 3t + 8t – 12
Common factor among the 1st two terms?
t
3
 2 t 2 – 3t = t( 2t –
t
t
3)
Common factor among the 2nd two terms?
2
4
3
 8 t – 12 = 4( 2t – 3 )
4
4
 2t2 – 3t + 8t – 12 = t(2t – 3) + 4(2t – 3)
= (2t – 3)( t + 4 )
Ex: Factor 9x
9 4 + 18x
18 2 + 88
1. Is there a GCF? No
2. Multiply ac  a = 9 and c = 8
9(8) = 72
3. Factors of 72 that sum to 18
1 72  73
Bit big  think bigger factors
3 24  27
6 12  18 
4. Rewrite the middle term
9x4 + 18x2 + 8 = 9x4 + 6x2 + 12x2 + 8
Four-term polynomial . . . Factor by grouping . . .
9x4 + 6x2 + 12x2 + 8
Common factor among the 1st two terms?
3x2
3x2
2
 9x4 + 6x2 = 3x2(3x2 + 2 )
3x2 3x2
Common factor among the 2nd two terms?
3
4
3
 12x2 + 8 = 4( 3x2 + 2 )
4
4
 9x4 + 6x2 + 12x2 + 8 = 3x2(3x2 + 2) + 4(3x2 + 2)
= (3x2 + 2)( 3x2 + 4 )
y + 6y
Ex: Factor 12x
12 2 – 17
17xy
6y22 Pick one to be the variable
1. Is there a GCF? No, but notice two variables
2. Multiply ac  a = 12x2 and c = 6y2 12x2(6y2) = 72y2
3. Factors of 72x2y2 that sum to
- 17xy
-1xy -72xy  -73xy Each factor need a y, both need to
be negative
-6xy -12xy  -18xy
Too big, think bigger factors
-8xy -9xy  -17xy 
4. Rewrite the middle term
12x2 – 17xy + 6y2 = 12x2 – 8xy – 9xy + 6y2
Four-term polynomial . . . Factor by grouping . . .
12x2 – 8xy – 9xy + 6y2
Common factor among the 1st two terms?
3x
4x
2y
 12x2 – 8xy = 4x( 3x – 2y )
4x
4x
Common factor among the 2nd two terms?
3
- 3y
-2y
 – 9xy + 6y2 = - 3y( 3x – 2y )
-3y
-3y
 12x2 – 8xy – 9xy + 6y2 = 4x(3x – 2y) – 3y(3x – 2y)
= (3x – 2)( 4x – 3y )
Your Turn to Try a Few
To factor a polynomial completely, ask
• Do the terms have a common factor (GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
• Is the polynomial a difference of squares?
• a2 – b 2
• Is the polynomial a sum/difference of cubes?
• a3 + b3 or a3 – b3
• Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two binomials?
• Factored completely?
Ex: Factor x3 + 3x2 – 4x – 12
1. Is there a GCF? No
2. Notice four terms  grouping
Common factor among the 1st two terms? x2
x
 x3 + 3x2 = x2( x
x2
+ 3 )
x2
Common factor among the 2nd two terms?
3
 – 4x – 12 = – 4( x
-4
+3 )
-4
 x3 + 3x2 - 4x – 12 = x2(x + 3) – 4(x + 3)
= (x + 3)( x2 – 4 )
-4
Cont: we have (x + 3)(x2 – 4)
But are we done? No. We have to make sure
we factor completely.
Is (x + 3) prime?  can x + 3 be factored further?
No . . . It is prime
What about (x2 – 4)? Recognize it?
Difference of Squares
x2 = (x)2
4 = (2)2
 x2 – 4 = (x)2 – (2)2 = (x – 2)(x + 2)
Therefore x3 + 3x2 – 4x – 12 = (x + 3)(x2 – 4)
= (x + 3)(x – 2)(x + 2)
Your Turn to Try a Few
To factor a polynomial completely, ask
• Do the terms have a common factor (GCF)?
• Does the polynomial have four terms?
• Is the polynomial a special one?
• Is the polynomial a difference of squares?
• a2 – b 2
• Is the polynomial a sum/difference of cubes?
• a3 + b3 or a3 – b3
• Is the trinomial a perfect-square trinomial?
• a2 + 2ab + b2 or a2 – 2ab + b2
• Is the trinomial a product of two binomials?
• Factored completely?
Special Polynomials
Is the polynomial a sum/difference of cubes?
• a3 + b3 = (a + b)(a2 - ab + b2)
• a3 – b3 = (a - b)(a2 + ab + b2)
Ex: Factor 8p3 – q3
Notice the terms are both perfect cubes
and we have a difference  difference of cubes
8p3 = (2p)3
q3 = (q)3
 8p3 – q3 = (2p)3 – (q)3 = (2p – q)((2p)2 + (2p)(q) + (q)2)
a3 – b3 = (a – b)(a2 + ab + b2)
factors as
= (2p – q)(4p2 + 2pq + q2)
Ex: Factor x3 + 27y9
Notice the terms are both perfect cubes
and we have a sum  sum of cubes
x3 = (x)3
27y9 = (3y3)3
 x3 + 27y9 = (x)3 + (3y3)3 = (x + 3y3)((x)2 - (x)(3y3) + (3y3)2)
a3 + b3 = (a + b)(a2 - ab + b2)
factors as
= (x + 3y3)(x2 – 3xy3 + 9y6)
SERIES AND SEQUENCES
Arithmetic Sequences and Series
An introduction…………
1, 4, 7, 10, 13
35
2, 4, 8, 16, 32
62
9, 1,  7,  15
12
9,  3, 1,  1/ 3
20 / 3
6.2, 6.6, 7, 7.4
27.2
,   3,   6
3  9
1, 1/ 4, 1/16, 1/ 64 85 / 64
9.75
, 2.5, 6.25
Arithmetic Sequences
Geometric Sequences
ADD
To get next term
MULTIPLY
To get next term
Arithmetic Series
Sum of Terms
Geometric Series
Sum of Terms
Find the next four terms of –9, -2, 5, …
Arithmetic Sequence
2  9  5  2  7
7 is referred to as the common difference (d)
Common Difference (d) – what we ADD to get next term
Next four terms……12, 19, 26, 33
Find the next four terms of 0, 7, 14, …
Arithmetic Sequence, d = 7
21, 28, 35, 42
Find the next four terms of x, 2x, 3x, …
Arithmetic Sequence, d = x
4x, 5x, 6x, 7x
Find the next four terms of 5k, -k, -7k, …
Arithmetic Sequence, d = -6k
-13k, -19k, -25k, -32k
Vocabulary of Sequences (Universal)
a1  First term
an  nth term
n  number of terms
Sn  sum of n terms
d  common difference
nth term of arithmetic sequence  an  a1  n  1 d
sum of n terms of arithmetic sequence  Sn 
n
 a1  an 
2
Given an arithmetic sequence with a15  38 and d  3, find a1.
x
a1  First term
38 an  nth term
15
n  number of terms
NA Sn  sum of n terms
-3
d  common difference
an  a1  n  1 d
38  x  15  1 3 
X = 80
Find S63 of  19,  13, 7,...
-19 a1  First term
353
??
an  nth term
n  number of terms
63
x
Sn  sum of n terms
6
d  common difference
an  a1  n  1 d
??  19   63  1 6 
??  353
n
 a1  an 
2
63

 19  353 
2
Sn 
S63
S63  10521
Try this one: Find a16 if a1  1.5 and d  0.5
1.5 a1  First term
x
16
an  nth term
n  number of terms
NA Sn  sum of n terms
0.5
d  common difference
an  a1  n  1 d
a16  1.5  16  1 0.5
a16  9
Find n if an  633, a1  9, and d  24
9
a1  First term
633 an  nth term
x
n  number of terms
NA Sn  sum of n terms
24
d  common difference
an  a1  n  1 d
633  9   x  1 24
633  9  24x  24
X = 27
Find d if a1  6 and a29  20
-6
a1  First term
20 an  nth term
29
n  number of terms
NA Sn  sum of n terms
x
d  common difference
an  a1  n  1 d
20  6   29  1 x
26  28x
13
x
14
Find two arithmetic means between –4 and 5
-4, ____, ____, 5
-4
a1  First term
5
an  nth term
n  number of terms
4
NA
x
Sn  sum of n terms
d  common difference
an  a1  n  1 d
5  4   4  1 x 
x3
The two arithmetic means are –1 and 2, since –4, -1, 2, 5
forms an arithmetic sequence
Find three arithmetic means between 1 and 4
1, ____, ____, ____, 4
1
a1  First term
4
an  nth term
5
NA
x
n  number of terms
Sn  sum of n terms
d  common difference
an  a1  n  1 d
4  1   5  1 x 
3
x
4
The three arithmetic means are 7/4, 10/4, and 13/4
since 1, 7/4, 10/4, 13/4, 4 forms an arithmetic sequence
Find n for the series in which a1  5, d  3, Sn  440
5
a1  First term
y
an  nth term
x
n  number of terms
440 Sn  sum of n terms
3
d  common difference
an  a1  n  1 d
y  5   x  1 3
x
440   5  5   x  1 3 
2
x  7  3x 
440 
2
880  x  7  3x 
0  3x 2  7x  880
Graph on positive window
X = 16
n
Sn   a1  an 
2
x
440   5  y 
2
The sum of the first n terms of an infinite sequence
is called the nth partial sum.
Sn  n (a1  an)
2
Example 6. Find the 150th partial sum of the arithmetic sequence, 5,
16, 27, 38, 49, …
a1  5
d  11
 c  5  11  6
an  11n  6  a150  11150  6  1644
S150
150

 5  1644   75 1649   123,675
2
Example 7. An auditorium has 20 rows of seats. There are 20 seats in
the first row, 21 seats in the second row, 22 seats in the third row, and
so on. How many seats are there in all 20 rows?
d 1
c  20  1  19
an  a1   n 1 d  a20  20  19 1  39
20
S 20   20  39   10  59   590
2
Example 8. A small business sells $10,000 worth of sports memorabilia
during its first year. The owner of the business has set a goal of
increasing annual sales by $7500 each year for 19 years. Assuming that
the goal is met, find the total sales during the first 20 years this business
is in operation.
a1  10,000
d  7500
c  10,000  7500  2500
an  a1   n 1 d  a20  10,000  19  7500  152,500
20
S20  10,000  152,500   10 162,500   1,625,000
2
So the total sales for the first 2o years is $1,625,000
Geometric Sequences and Series
1, 4, 7, 10, 13
35
2, 4, 8, 16, 32
62
9, 1,  7,  15
12
9,  3, 1,  1/ 3
20 / 3
6.2, 6.6, 7, 7.4
27.2
,   3,   6
3  9
1, 1/ 4, 1/16, 1/ 64 85 / 64
9.75
, 2.5, 6.25
Arithmetic Sequences
Geometric Sequences
ADD
To get next term
MULTIPLY
To get next term
Arithmetic Series
Sum of Terms
Geometric Series
Sum of Terms
Vocabulary of Sequences (Universal)
a1  First term
an  nth term
n  number of terms
Sn  sum of n terms
r  common ratio
nth term of geometric sequence  an  a1r n1


a1 r n  1 

sum of n terms of geometric sequence  Sn  
r 1
Find the next three terms of 2, 3, 9/2, ___, ___, ___
3 – 2 vs. 9/2 – 3… not arithmetic
3 9/2
3

 1.5  geometric  r 
2
3
2
9 9 3 9 3 3 9 3 3 3
2, 3, ,  ,   ,   
2 2 2 2 2 2 2 2 2 2
9 27 81 243
2, 3, , , ,
2 4 8 16
1
2
If a1  , r  , find a9 .
2
3
a1  First term
1/2
an  nth term
x
n  number of terms
9
Sn  sum of n terms
NA
r  common ratio
2/3
an  a1r n1
 1  2 
x    
 2  3 
9 1
28
27
128
x
8 
8 
23
3
6561
Find two geometric means between –2 and 54
-2, ____, ____, 54
a1  First term
-2
an  nth term
54
n  number of terms
4
Sn  sum of n terms
NA
r  common ratio
x
an  a1r n1
54   2  x 
41
27  x 3
3  x
The two geometric means are 6 and -18, since –2, 6, -18, 54
forms an geometric sequence
Find a2  a 4 if a1  3 and r 
2
3
-3, ____, ____, ____
2
Since r  ...
3
3,  2,
4 8
,
3 9
 8  10
a 2  a 4  2  


9
 9 
Find a9 of 2, 2, 2 2,...
a1  First term
2
an  nth term
x
n  number of terms
Sn  sum of n terms
r  common ratio
9
NA
r
an  a1r n1
 2
2  2
x 2
x
x  16 2
9 1
8
2
2 2

 2
2
2
If a5  32 2 and r   2, find a 2
____, ____, ____,____,32 2
a1  First term
x
an  nth term
32 2
n  number of terms
Sn  sum of n terms
r  common ratio
an  a1r n1
NA
 2
 
2  x  2 
32 2  x  2
32
5
32 2  4x
8 2x
5 1
4
*** Insert one geometric mean between ¼ and 4***
*** denotes trick question
1
,____,4
4
a1  First term
an  nth term
n  number of terms
Sn  sum of n terms
r  common ratio
an  a1r n1
1 2
1 31
4  r  4  r  16  r 2  4  r
4
4
1/4
4
3
NA
x
1
, 1, 4
4
1
,  1, 4
4
1 1 1
Find S7 of    ...
2 4 8
a1  First term
1/2
an  nth term
NA
n  number of terms
Sn  sum of n terms
7
x
r  common ratio


 a1 r n  1 

Sn  
r 1
 1   1 7

     1 
 2   2 
 
x

1
1
2
 1   1 7

     1 
 2   2 
  63

1
64

2
1 1
1
r 4  8 
1 1 2
2 4
INFINITE SERIES
1, 4, 7, 10, 13, ….
Infinite Arithmetic
3, 7, 11, …, 51
Finite Arithmetic
1, 2, 4, …, 64
Finite Geometric
1, 2, 4, 8, …
Infinite Geometric
r>1
r < -1
No Sum
1 1 1
3,1, , , ...
3 9 27
Infinite Geometric
-1 < r < 1
a1
S
1 r
No Sum
n
Sn   a1  an 
2
a1 r n  1
Sn 
r 1
1 1 1
Find the sum, if possible: 1     ...
2 4 8
1 1
1
2
4
r  
 1  r  1  Yes
1 1 2
2
a1
S

1 r
1
1
1
2
2
Find the sum, if possible: 2 2  8  16 2  ...
8
16 2
r

 2 2  1  r  1  No
8
2 2
NO SUM
2 1 1 1
 ...
Find the sum, if possible:   
3 3 6 12
1 1
1
3
6
r  
 1  r  1  Yes
2 1 2
3 3
a1
S

1 r
2
3
4

1 3
1
2
2 4 8
   ...
Find the sum, if possible:
7 7 7
4 8
r  7  7  2  1  r  1  No
2 4
7 7
NO SUM
5
Find the sum, if possible: 10  5   ...
2
5
5
1
2
r
 
 1  r  1  Yes
10 5 2
a1
10
S

 20
1
1 r
1
2
The Bouncing Ball Problem – Version A
A ball is dropped from a height of 50 feet. It rebounds 4/5 of
it’s height, and continues this pattern until it stops. How far
does the ball travel?
50
40
40
32
32
32/5
32/5
S
50
40

 450
4
4
1
1
5
5
The Bouncing Ball Problem – Version B
A ball is thrown 100 feet into the air. It rebounds 3/4 of
it’s height, and continues this pattern until it stops. How far
does the ball travel?
100
100
75
75
225/4
225/4
S
100
100

 800
3
3
1
1
4
4
SIGMA NOTATION
UPPER BOUND
(NUMBER)
B
SIGMA
(SUM OF TERMS)
a
n A
n
NTH TERM
(SEQUENCE)
LOWER BOUND
(NUMBER)
4
  j  2  1  2   2  2   3  2    4  2  18
j1
7
  2a    2  4    2 5    2  6   2 7  44
a4

4
n 0
 
 
 
 
 
4
3
2

0.5

2

0.5

2

0.5

2
0.5  2  0.5  2  0.5  2
n
 33.5
0
1


n
0
2
1
3
3
3
3
6

6

6

5
 5   6 
 5   ...

 
 
b 0 
5
a1
S

1 r
6
3
1
5
 15
  2x  1   2 7  1   2 8  1   2 9  1  ...   2  23  1
23
x 7
n
23  7  1
Sn   a1  an  
15  47   527
2
2
  4b  3    4  4  3   4 5  3   4  6  3   ...   4 19  3
19
b 4
Sn 
n
19  4  1
a

a

 1 n
19  79   784
2
2
Rewrite using sigma notation: 3 + 6 + 9 + 12
Arithmetic, d= 3
an  a1  n  1 d
an  3  n  1 3
an  3n
4
 3n
n1
Rewrite using sigma notation: 16 + 8 + 4 + 2 + 1
Geometric, r = ½
an  a1r n1
n1
 1
an  16  
2
 1
16  

2
n1
5
n1
Rewrite using sigma notation: 19 + 18 + 16 + 12 + 4
Not Arithmetic, Not Geometric
19 + 18 + 16 + 12 + 4
-1
-2 -4 -8
an  20  2n1
5
n1
20

2

n1
3 9 27

 ...
Rewrite the following using sigma notation: 
5 10 15
Numerator is geometric, r = 3
Denominator is arithmetic d= 5
NUMERATOR: 3  9  27  ...  an  3 3 
n1
DENOMINATOR: 5  10  15  ...  an  5  n  1 5  an  5n

SIGMA NOTATION:

n1
3 3 
n1
5n
RATIONAL EXPRESSIONS
Simplifying Rational Expressions
• A “rational expression” is the quotient of two
polynomials. (division)
2n
n2
x  3x  10
3x  2
2
Simplifying Rational Expressions
• A “rational expression” is the quotient of two
polynomials. (division)
• A rational expression is in simplest form when the
numerator and denominator have no common
factors (other than 1)
9
Simplify
15
3 3

3 5
3

5
Simplifying Rational Expressions
• A “rational expression” is the quotient of two
polynomials. (division)
• A rational expression is in simplest Form when the
numerator and denominator have no common
factors (other than 1)
3 3
Simplify
3 5
How to get a rational expression in simplest form…
• Factor the numerator completely (factor out a common factor,
difference of 2 squares, bottoms up)
• Factor the denominator completely (factor out a common factor,
difference of 2 squares, bottoms up)
• Cancel out any common factors (not addends)
Difference between a factor and an addend
• A factor is in between a multiplication sign
• An addend is in between an addition or subtraction sign
Example:
x+3
3x + 9
x–9
6x + 3
Factor
5 x  10 5 x  2 
Simplify :

5x
5x

x  2

x
x2

x
y  3y  2
Simplify :
2
y 1
2
 y  1 y  2
 y  1 y  1
y2

y 1
12 y  24
Simplify :
48 y
12  y  2 
12 4 y 
y2

4y
2x  x
Simplify : 2
3x  2 x
2
 x  2 x  1
 x  3x  2 
2x 1

3x  2
a 9
Simplify: 2
a  2a  15
2
 a  3 a  3
 a  5 a  3
a 1

2a  1
x4
Simplify :
4 x
x  4
 1x  4
1

 1
1
Multiplying and Dividing
Rational Expressions
Remember that a rational number
can be expressed as a quotient of
two integers. A rational expression
can be expressed as a quotient of
two polynomials.
Examples of rational expressions
4
,
3x
x 8
,
x3
4y  7
2
y  5y  9
Remember, denominators
can not = 0.
Now,lets go through the steps to
simplify a rational expression.
Simplify:
7x  7
2
x 1
Step 1: Factor the numerator and
the denominator completely looking
for common factors.
7x  7  7(x  1)
x  1  (x  1)(x  1)
2
Next
7x  7
7(x  1)

2
x  1 (x  1)(x  1)
What is the common factor?
x 1
Step 2: Divide the numerator and
denominator by the common factor.
1
7(x  1)
7(x  1)

(x  1)(x  1) (x  1)(x  1)
1
Step 3: Multiply to get your answer.
7
Answer:
x 1
Looking at the answer from the
previous example, what value of x
would make the denominator 0?
x= -1
The expression is undefined when the
values make the denominator equal to 0
How do I find the values
that make an expression
undefined?
Completely factor the original
denominator.
2ab(a  2)(b  3)
Ex:
2
3ab(a  4)
Factor the denominator
2
3ab(a  4)  3ab(a  2)(a  2)
The expression is undefined when:
a= 0, 2, and -2 and b= 0.
Lets go through another example.
3a  a
3
2
2a  6a
3
4
3a  a
a (3  a)
3
2 
2
2a  6a
2a (a  3)
3
4
3
Next
Factor
out the
GCF
a (3  a )

2
2a ( a  3)
3
3  a  factored is
1(a  3)
a3  1(a  3)

2
2a (a  3)
1
1a3 (a  3)

2
2a (a  3)
1
cancel like factors
a3
1a
2
2a
cancel out the like factor a 2
1
1a
2
answer
What values is the original expression undefined?
Now try to do some on your own.
x  5x  6
1)
2
x 9
3
2
5 x  10 x
2) 3
2
x  6 x  16 x
2
Also find the values that make
each expression undefined?
The same method can be used
to multiply rational expressions.
1
2
1 1 1
1
4a
3bc
4  a  a 3  b c
Ex:
3 
3 
5 a  b  b b 12  a  a a
5ab 12a
1
1
c
 2 2
5b a
1
1
Let’s do another one.
x  3x
x  10x  9
Ex: 2
 2
x  5x  6 x  6x  27
Step #1: Factor the numerator
and the denominator.
3
2
2
x (x  3) (x 1)(x  9)

(x  6)(x 1) (x  9)(x 3)
2
Next
Step #2: Divide the numerator and
denominator by the common factors.
1
1
1
1
1
x (x  3) (x 1)(x  9)

(x  6)(x 1) (x  9)(x 3)
2
1
Step #3: Multiply the numerator
and the denominator.
2
x
x6
Remember how to divide fractions?
Multiply by the reciprocal of the
divisor.
4 16 4 25

 
5 25 5 16
1
5
1
4
5
4  25


4
516
Dividing rational expressions uses
the same procedure.
Ex: Simplify
y2
y  2y
 2
2
y 10 y  24 y  2y  8
2
y2
y 2  2y
 2

2
y 10 y  24 y  2y  8
y2
y  2y  8
 2

2
y 10 y  24 y  2y
2
1 2)
12
y
( y  4)( y 


( y  12 )( y  2)
y( y  2 )
1
1
y4
y ( y  12)
Next
Now you try to simplify the
expression:
x3
2x  6x

2
x  4x  12
x2
2
1
Answer:
2x(x  6)
Now try these on your own.
x +3
x  7x  6
1)
 2
3
2
2x  2x x  10x  21
2
3x  6 5x  10
2)

7x  7 14x  14
Here are the answers:
x6
1)
2
2x (x  7)
6(x  1)
2)
5(x  1)
RATIONAL EXPRESSIONS
Adding and Subtracting Rational Expressions with the Same Denominator and Least
Common Denominators
Rational Expressions
If P, Q and R are polynomials and Q  0,
P Q PQ
 
R R
R
P Q P Q
 
R R
R
Adding Rational Expressions
Example
Add the following rational expressions.
4 p 3 3p 8 7 p  5
4 p 3 3p 8



2p 7
2p 7
2p 7 2p 7
Subtracting Rational Expressions
Example
Subtract the following rational expressions.
8 y  16 8( y  2)
8y
16



 8
y2
y2 y2
y2
Subtracting Rational Expressions
Example
Subtract the following rational expressions.
3y  6
3y
6

 2
 2
2
y  3 y  10
y  3 y  10 y  3 y  10
3( y  2)

( y  5)( y  2)
3
y5
Least Common Denominators
To add or subtract rational expressions with unlike
denominators, you have to change them to
equivalent forms that have the same denominator
(a common denominator).
This involves finding the least common
denominator of the two original rational
expressions.
Least Common Denominators
To find a Least Common Denominator:
1) Factor the given denominators.
2) Take the product of all the unique factors.
Each factor should be raised to a power equal
to the greatest number of times that factor
appears in any one of the factored
denominators.
Least Common Denominators
Example
Find the LCD of the following rational expressions.
1
3x
,
6 y 4 y  12
6 y  2 3y
4 y  12  4( y  3)  2 ( y  3)
2
So the LCD is 2  3 y( y  3)  12 y( y  3)
2
Least Common Denominators
Example
Find the LCD of the following rational expressions.
4
4x  2
, 2
2
x  4 x  3 x  10 x  21
x  4 x  3  ( x  3)( x  1)
2
x  10 x  21  ( x  3)( x  7)
2
So the LCD is (x  3)(x  1)(x  7)
Least Common Denominators
Example
Find the LCD of the following rational expressions.
2
3x
4x
, 2
2
5x  5 x  2 x  1
5x  5  5( x  1)  5( x  1)( x  1)
2
2
x  2 x  1  ( x  1)
2
2
So the LCD is 5(x  1)(x -1)
2
Least Common Denominators
Example
Find the LCD of the following rational expressions.
1
2
,
x 3 3 x
Both of the denominators are already factored.
Since each is the opposite of the other, you can
use either x – 3 or 3 – x as the LCD.
Multiplying by 1
To change rational expressions into equivalent
forms, we use the principal that multiplying by 1 (or
any form of 1), will give you an equivalent
expression.
P P
P R PR
 1   
Q Q
Q R QR
Equivalent Expressions
Example
Rewrite the rational expression as an equivalent rational
expression with the given denominator.
3

5
9y
72 y 9
4
3 8y
3
24 y 4
 4 

5
5
9
9y 8y
9y
72 y
RATIONAL EXPRESSIONS
Adding and Subtracting Rational Expressions with Different Denominators
Unlike Denominators
As stated in the previous section, to add or subtract rational
expressions with different denominators, we have to change them to
equivalent forms first.
Unlike Denominators
Adding or Subtracting Rational Expressions with
Unlike Denominators
1) Find the LCD of all the rational expressions.
2) Rewrite each rational expression as an equivalent
one with the LCD as the denominator.
3) Add or subtract numerators and write result over
the LCD.
4) Simplify rational expression, if possible.
Adding with Unlike Denominators
Example
Add the following rational expressions.
15 8
,
7 a 6a
15 8
6 15 7  8




7 a 6 a 6  7 a 7  6a
73
90
56
146



21a
42a 42a 42a
Subtracting with Unlike Denominators
Example
Subtract the following rational expressions.
5
3
,
2x  6 6  2x
5
3
5
3




2x  6 6  2x 2x  6 2x  6
4
222
8


2 x  6 2( x  3) x  3
Subtracting with Unlike Denominators
Example
Subtract the following rational expressions.
7
and 3
2x  3
7
3(2 x  3)
7


3 
2x  3
2x  3
2x  3
7
6 x  9 7  6 x  9 16  6 x



2x  3
2x  3 2x  3
2x  3
Adding with Unlike Denominators
Example
Add the following rational expressions.
4
x
, 2
2
x  x  6 x  5x  6
4
x
4
x
 2



2
x  x  6 x  5 x  6 ( x  3)( x  2) ( x  3)( x  2)
4( x  3)
x( x  3)


( x  3)( x  2)( x  3) ( x  3)( x  2)( x  3)
2
2
x
 x  12
4 x  12  x  3x

( x  2)( x  3)( x  3) ( x  2)( x  3)( x  3)
RATIONAL EXPRESSIONS
Solving Equations Containing Rational Expressions
Solving Equations
First note that an equation contains an equal sign and
an expression does not.
To solve EQUATIONS containing rational expressions,
clear the fractions by multiplying both sides of the
equation by the LCD of all the fractions.
Then solve as in previous sections.
Note: this works for equations only, not simplifying
expressions.
Solving Equations
Example
Solve the following rational equation.
5
7
1 
3x
6
 5
 7
6 x  1   6 x
 3x   6 
10  6x  7 x
10  x
Check in the original
equation.
5
7
1 
3 10
6
5
7
1 
30
6
1
7
1 
6
6
true
Solving Equations
Example
Solve the following rational equation.
1
1
1

 2
2 x x  1 3x  3x

1  
1
 1
6 xx  1
6 xx  1 
  
 2 x x  1   3x( x  1) 
3x  1  6x  2
3x  3  6x  2
3  3x  2
 3x  1
x 1
3
Continued.
Solving Equations
Example Continued
Substitute the value for x into the original
equation, to check the solution.
1
1
1


2
1
1
2
1 3 1
3 1
3
3
3
3
   
   
3 3
1
 
2 4 1 1
3
6 3 3
true
 
4 4 4
So the solution is x  1
3
Solving Equations
Example
Solve the following rational equation.
x2
1
1


2
x  7 x  10 3x  6 x  5
x2
1 

  1
3x  2x  5 2


3x  2x  5
 x  7 x  10   3x  6 x  5 
3x  2  x  5  3x  2
3x  6  x  5  3x  6
3x  x  3x  5  6  6
5x  7
x  7
5
Continued.
Solving Equations
Example Continued
Substitute the value for x into the original
equation, to check the solution.

7 2
1
1
5


2
7
7
7
 7  7  10 3  5  6  5  5
5
5
3
1
1
5


49  49  10  21  6 18
5
25
5
5




5
5 5
 
18 9 18


true
So the solution is x   7 5

Solving Equations
Example
Solve the following rational equation.
1
2

x 1 x  1
 1   2 
x  1x  1

 x  1 x  1
 x 1   x 1 
x  1  2x 1
x  1  2x  2
3 x
Continued.
Solving Equations
Example Continued
Substitute the value for x into the original
equation, to check the solution.
1
2

3 1 3 1
1 2

2 4
true
So the solution is x = 3.
Solving Equations
Example
Solve the following rational equation.
12
3
2


2
9a
3 a 3 a
3   2 
 12
3  a 3  a 


3  a 3  a 
2
3 a   3 a 
9a
12  33  a   23  a
12  9  3a  6  2a
21  3a  6  2a
15  5a
3a
Continued.
Solving Equations
Example Continued
Substitute the value for x into the original
equation, to check the solution.
12
3
2
2  33  33
93
12 3 2
 
0 5 0
Since substituting the suggested value of a into the
equation produced undefined expressions, the
solution is .
Solving Equations with Multiple Variables
Solving an Equation With Multiple Variables for
One of the Variables
1) Multiply to clear fractions.
2) Use distributive property to remove grouping
symbols.
3) Combine like terms to simplify each side.
4) Get all terms containing the specified variable on
the same side of the equation, other terms on the
opposite side.
5) Isolate the specified variable.
Solving Equations with Multiple Variables
Example
Solve the following equation for R1
1 1
1
 
R R1 R2
1 
1  1
RR1R2       RR1R2
 R   R1 R2 
R1R2  RR2  RR1
R1R2  RR1  RR2
R1 R2  R  RR2
RR2
R1 
R2  R
RATIONAL EXPRESSIONS
Problem Solving with Rational Equations
Ratios and Rates
Ratio is the quotient of two numbers or two
quantities.
The ratio of the numbers a and b can also be
a
written as a:b, or .
b
The units associated with the ratio are
important.
The units should match.
If the units do not match, it is called a rate,
rather than a ratio.
Proportions
Proportion is two ratios (or rates) that are equal to
each other.
a c

b d
We can rewrite the proportion by multiplying
by the LCD, bd.
This simplifies the proportion to ad = bc.
This is commonly referred to as the cross product.
Solving Proportions
Example
Solve the proportion for x.
x 1 5

x2 3
3x  1  5x  2
3x  3  5x 10
 2x  7
x  7
2
Continued.
Solving Proportions
Example Continued
Substitute the value for x into the original
equation, to check the solution.
 7 1 5
2

7  2 3
2
5
25
3
3
2
true
So the solution is x   7
2
Solving Proportions
Example
If a 170-pound person weighs approximately 65 pounds
on Mars, how much does a 9000-pound satellite
weigh?
170 - pound person on Earth
65 - pound person on Mars

9000 - pound satellite on Earth x - pound satellite on Mars
170 x  9000  65  585,000
x  585000 / 170  3441 pounds
Solving Proportions
Example
Given the following prices charged for various sizes
of picante sauce, find the best buy.
• 10 ounces for $0.99
• 16 ounces for $1.69
• 30 ounces for $3.29
Continued.
Solving Proportions
Example Continued
Size
Price
Unit Price
10 ounces
$0.99
$0.99/10 = $0.099
16 ounces
$1.69
$1.69/16 = $0.105625
30 ounces
$3.29
$3.29/30  $0.10967
The 10 ounce size has the lower unit price, so it is the
best buy.
Similar Triangles
In similar triangles, the measures of corresponding
angles are equal, and corresponding sides are in
proportion.
Given information about two similar triangles, you
can often set up a proportion that will allow you to
solve for the missing lengths of sides.
Similar Triangles
Example
Given the following triangles, find the unknown
length y.
12 m
10 m
5m
y
Continued
Similar Triangles
Example
1.) Understand
Read and reread the problem. We look for the corresponding sides in the 2
triangles. Then set up a proportion that relates the unknown side, as well.
2.) Translate
By setting up a proportion relating lengths of corresponding sides of the two
triangles, we get
12 10

5
y
Continued
Similar Triangles
Example continued
3.) Solve
12 10

5
y
12 y  5 10  50
y  50
12
 25
6
meters
Continued
Similar Triangles
Example continued
4.) Interpret
Check: We substitute the value we found from
the proportion calculation back into the
problem.
12 10
60


5 25
25
6
true
State: The missing length of the triangle is 25 6 meters
Finding an Unknown Number
Example
The quotient of a number and 9 times its reciprocal
is 1. Find the number.
1.) Understand
Read and reread the problem. If we let
n = the number, then
= the reciprocal of the number
1
n
Continued
Finding an Unknown Number
Example continued
2.) Translate
The quotient of
is
a number
and 9 times its reciprocal
n
1
9 
n

1
=
1
Continued
Finding an Unknown Number
Example continued
3.) Solve
 1
n  9    1
 n
9
n   1
n
n
n 1
9
n2  9
n  3,3
Continued
Finding an Unknown Number
Example continued
4.) Interpret
Check: We substitute the values we found from the
equation back into the problem. Note that nothing in
the problem indicates that we are restricted to
positive values.
 1
 1 
3  9   1
 3  9 
 1
 3
 3
3  3  1 true
 3  3  1 true
State: The missing number is 3 or –3.
Solving a Work Problem
Example
An experienced roofer can roof a house in 26 hours. A
beginner needs 39 hours to do the same job. How long will it
take if the two roofers work together?
1.) Understand
Read and reread the problem. By using the times for each roofer to complete the
job alone, we can figure out their corresponding work rates in portion of the job
done per hour.
Time in hrs
Experienced roofer
Beginner roofer
39
Together
26
/39
t
Portion job/hr
1/26
1/t
Continued
Solving a Work Problem
Example continued
2.) Translate
Since the rate of the two roofers working together
would be equal to the sum of the rates of the two
roofers working independently,
1
1 1


26 39 t
Continued
Solving a Work Problem
Example continued
3.) Solve
1
1 1


26 39 t
1  1
 1
78t      78t
 26 39   t 
3t  2t  78
5t  78
t  78 / 5 or 15.6 hours
Continued
Solving a Work Problem
Example continued
4.) Interpret
Check: We substitute the value we found from the
proportion calculation back into the problem.
1
1
1


26 39 78
5
3
2
5


78 78 78
true
State: The roofers would take 15.6 hours working
together to finish the job.
Solving a Rate Problem
Example
The speed of Lazy River’s current is 5 mph. A boat travels 20
miles downstream in the same time as traveling 10 miles
upstream. Find the speed of the boat in still water.
1.) Understand
Read and reread the problem. By using the formula d=rt, we can rewrite the formula
to find that t = d/r.
We note that the rate of the boat downstream would be the rate in still water + the
water current and the rate of the boat upstream would be the rate in still water –
the water current.
Distance
Down
Up
20
10
rate
time = d/r
r + 5 20/(r + 5)
r – 5 10/(r – 5)
Continued
Solving a Rate Problem
Example continued
2.) Translate
Since the problem states that the time to travel
downstairs was the same as the time to travel
upstairs, we get the equation
20
10

r 5 r 5
Continued
Solving a Rate Problem
Example continued
3.) Solve
20
10

r 5 r 5
 20   10 
r  5r  5

r  5r  5
 r 5  r 5
20r  5  10r  5
20r 100  10r  50
10r  150
r  15 mph
Continued
Solving a Rate Problem
Example continued
4.) Interpret
Check: We substitute the value we found from the
proportion calculation back into the problem.
20
10

15  5 15  5
20 10

true
20 10
State: The speed of the boat in still water is 15 mph.
RATIONAL EXPRESSIONS
Simplifying Complex Fractions
Complex Rational Fractions
Complex rational expressions (complex fraction) are
rational expressions whose numerator, denominator,
or both contain one or more rational expressions.
There are two methods that can be used when
simplifying complex fractions.
Simplifying Complex Fractions
Simplifying a Complex Fraction (Method 1)
1) Simplify the numerator and denominator of the
complex fraction so that each is a single fraction.
2) Multiply the numerator of the complex fraction by
the reciprocal of the denominator of the complex
fraction.
3) Simplify, if possible.
Simplifying Complex Fractions
Example
x
2
2

x
2
2
x 4
x4

2 2
2  x4 2  x4
2 x4 x4
x 4
x4

2 2
2
Simplifying Complex Fractions
Method 2 for simplifying a complex fraction
1) Find the LCD of all the fractions in both the
numerator and the denominator.
2) Multiply both the numerator and the denominator
by the LCD.
3) Simplify, if possible.
Simplifying Complex Fractions
Example
1 2

2
2
2
y
3 6y
6  4y
 2
2
1 5 6y
6y  5y

y 6
RADICALS
Introduction to Radicals
Square Roots
Opposite of squaring a number is taking the square
root of a number.
A number b is a square root of a number a if b2 = a.
In order to find a square root of a, you need a #
that, when squared, equals a.
Principal Square Roots
The principal (positive) square root is noted as
a
The negative square root is noted as
 a
Radicands
Radical expression is an expression containing a
radical sign.
Radicand is the expression under a radical sign.
Note that if the radicand of a square root is a
negative number, the radical is NOT a real number.
Radicands
Example
49 
7
5
25

16
4
 4  2
Perfect Squares
Square roots of perfect square radicands simplify to
rational numbers (numbers that can be written as a
quotient of integers).
Square roots of numbers that are not perfect
squares (like 7, 10, etc.) are irrational numbers.
IF REQUESTED, you can find a decimal
approximation for these irrational numbers.
Otherwise, leave them in radical form.
Perfect Square Roots
Radicands might also contain variables and powers
of variables.
To avoid negative radicands, assume for this chapter
that if a variable appears in the radicand, it
represents positive numbers only.
Example
64x10  8x 5
Cube Roots
The cube root of a real number a
3
a  b only if b 3  a
Note: a is not restricted to nonnegative numbers for cubes.
Cube Roots
Example
3
27  3
3
 8x 6   2x 2
nth Roots
Other roots can be found, as well.
The nth root of a is defined as
n
a  b only if b n  a
If the index, n, is even, the root is NOT a
real number when a is negative.
If the index is odd, the root will be a real
number.
nth Roots
Example
Simplify the following.
2 20
25a b
10
5ab

3
4
a
64
a
3
  3
9
b
b
RADICALS
Simplifying Radicals
Product Rule for Radicals
If
a and
b are real numbers,
ab  a  b
a
a

if
b
b
b 0
Simplifying Radicals
Example
Simplify the following radical expressions.
40 
4  10  2 10
5

16
5
5

4
16
15
No perfect square factor, so
the radical is already
simplified.
Simplifying Radicals
Example
Simplify the following radical expressions.
x 
6
x x 
x  x x
20

16
x
20
4 5
2 5

8
x
x8
7
x16

6
3
x
Quotient Rule for Radicals
If
n
a and n b are real numbers,
n
n
ab  n a  n b
a na
 n if
b
b
n
b 0
Simplifying Radicals
Example
Simplify the following radical expressions.
3
3
16  3 8  2 
3

64
3
3
3

64
3
3
8 3 2  2 3 2
3
4
RADICALS
Adding and Subtracting Radicals
Sums and Differences
Rules in the previous section allowed us to split
radicals that had a radicand which was a product or
a quotient.
We can NOT split sums or differences.
ab  a  b
a b  a  b
Like Radicals
In previous chapters, we’ve discussed the concept of “like”
terms.
These are terms with the same variables raised to the same
powers.
They can be combined through addition and subtraction.
Similarly, we can work with the concept of “like” radicals to
combine radicals with the same radicand.
Like radicals are radicals with the same index and the same
radicand.
Like radicals can also be combined with addition or
subtraction by using the distributive property.
Adding and Subtracting Radical Expressions
Example
37 3  8 3
10 2  4 2  6 2
3
2 4 2
Can not simplify
5 3
Can not simplify
Adding and Subtracting Radical Expressions
Example
Simplify the following radical expression.
 75  12  3 3 
 25  3  4  3  3 3 
 25  3  4  3  3 3 
5 3  2 3 3 3 
 5  2  3
3  6 3
Adding and Subtracting Radical Expressions
Example
Simplify the following radical expression.
3
64  3 14  9 
4  3 14  9   5  3 14
Adding and Subtracting Radical Expressions
Example
Simplify the following radical expression. Assume
that variables represent positive real numbers.
3 45x3  x 5x  3 9 x 2  5x  x 5x 
3 9 x  5x  x 5x 
2
3  3x 5 x  x 5 x 
9 x 5x  x 5x 
9 x  x 
5x 
10 x 5 x
RADICALS
Multiplying and Dividing Radicals
Multiplying and Dividing Radical Expressions
If
n
a and
n
bare real numbers,
n
a  n b  n ab
n
a n a

if b  0
b
b
n
Multiplying and Dividing Radical Expressions
Example
Simplify the following radical expressions.
3 y  5x 
7 6
ab
3 2
ab

15 xy
7 6
ab

3 2
ab
ab  ab
4 4
2 2
Rationalizing the Denominator
Many times it is helpful to rewrite a radical quotient
with the radical confined to ONLY the numerator.
If we rewrite the expression so that there is no
radical in the denominator, it is called rationalizing
the denominator.
This process involves multiplying the quotient by a
form of 1 that will eliminate the radical in the
denominator.
Rationalizing the Denominator
Example
Rationalize the denominator.
3
2


2
2
6
3 2

2
2 2
3
6 33
63 3
63 3
6 3
3




 2 3
3
3
3
3
3
3
27
3
9 3
9
Conjugates
Many rational quotients have a sum or difference of
terms in a denominator, rather than a single radical.
In that case, we need to multiply by the conjugate
of the numerator or denominator (which ever one
we are rationalizing).
The conjugate uses the same terms, but the
opposite operation (+ or ).
Rationalizing the Denominator
Example
Rationalize the denominator.
2 3
3  2 3 2 2  2 3
32



2  3 2 2  3  2  3 3
2 3
6 3 2 2  2 3

23
6 3 2 2  2 3

1
 6 3 2 2  2 3
RADICALS
Solving Equations Containing Radicals
Extraneous Solutions
Power Rule (text only talks about squaring, but
applies to other powers, as well).
If both sides of an equation are raised to the same power,
solutions of the new equation contain all the solutions of
the original equation, but might also contain additional
solutions.
A proposed solution of the new equation that is
NOT a solution of the original equation is an
extraneous solution.
Solving Radical Equations
Example
Solve the following radical equation.

x 1  5

2
x 1  5
2
x  1  25
x  24
Substitute into the
original equation.
24  1  5
25  5
So the solution is x = 24.
true
Solving Radical Equations
Example
Solve the following radical equation.
Substitute into the
5x  5
original equation.
 5x 
2
  5
5x  25
2
5  5  5
25  5
Does NOT check, since the left side
of the equation is asking for the
x5
principal square root.
So the solution is .
Solving Radical Equations
Steps for Solving Radical Equations
1) Isolate one radical on one side of equal sign.
2) Raise each side of the equation to a power equal to
the index of the isolated radical, and simplify. (With
square roots, the index is 2, so square both sides.)
3) If equation still contains a radical, repeat steps 1 and
2. If not, solve equation.
4) Check proposed solutions in the original equation.
Solving Radical Equations
Example
Solve the following radical equation.
x 1 1  0

x 1  1
Substitute into the
original equation.
x  1  12

2 1 1  0
x 1 1
1 1  0
2
x2
1 1  0 true
So the solution is x = 2.
Solving Radical Equations
Example
Solve the following radical equation.
2x  x 1  8

x 1  8  2x

x  1  8  2 x 
2
2
x  1  64  32 x  4 x 2
0  63  33x  4 x 2
0  (3  x)( 21  4 x)
21
x  3 or
4
Solving Radical Equations
Example continued
Substitute the value for x into the original equation, to
check the solution.
2(3)  3  1  8
6  4  8 true
So the solution is x = 3.
 
21
21
2

1  8
4
4
21
25

8
2
4
21 5
 8
2 2
26
8
2
false
Solving Radical Equations
Example
Solve the following radical equation.
y 5  2 y 4

 
2
y 5  2 y 4

2
y 5  44 y 4  y 4
5  4 y  4
5
  y4
4
2
 5
  
 4

y4

2
25
 y4
16
25 89
y  4

16 16
Solving Radical Equations
Example continued
Substitute the value for x into the original equation, to
check the solution.
89
89
5  2
4
16
16
169
25
 2
16
16
13
5
 2
4
4
13 3

4 4
false
So the solution is .
Solving Radical Equations
Example
Solve the following radical equation.
2 x  4  3x  4  2

2 x  4  2  3x  4
 
2
2 x  4   2  3x  4

2
2 x  4  4  4 3x  4  3x  4
2 x  4  8  3x  4 3x  4
x 2  24 x  80  0
 x  12  4 3x  4
x  20x  4  0
 x  12


2
  4 3x  4
x 2  24 x  144  16(3x  4)  48x  64
2
x  4 or 20
Solving Radical Equations
Example continued
Substitute the value for x into the original
equation, to check the solution.
2(4)  4  3(4)  4  2
2(20)  4  3(20)  4  2
4  16  2
36  64  2
2  4  2
6  8  2
true
true
So the solution is x = 4 or 20.