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7-1 Integer Exponents
Warm Up
Evaluate each expression for the given
values of the variables.
1. x3y2 for x = –1 and y = 10
2.
–100
for x = 4 and y = (–7)
Write each number as a power of the
given base.
3. 64; base 4
43
4. –27; base (–3)
Holt Algebra 1
(–3)3
7-1 Integer Exponents
Objectives
Evaluate expressions containing zero
and integer exponents.
Simplify expressions containing zero
and integer exponents.
Holt Algebra 1
7-1 Integer Exponents
You have seen positive exponents. Recall that
to simplify 32, use 3 as a factor 2 times: 32 =
3  3 = 9.
But what does it mean for an exponent to be
negative or 0? You can use a table and look for a
pattern to figure it out.
Power 55 54 53 52
Value 3125 625 125 25
5
Holt Algebra 1
5
5
5
51
5
50
5–1
5–2
7-1 Integer Exponents
When the exponent decreases by one, the value
of the power is divided by 5. Continue the
pattern of dividing by 5.
Holt Algebra 1
7-1 Integer Exponents
Remember!
Base
x
4
Exponent
Holt Algebra 1
7-1 Integer Exponents
Holt Algebra 1
7-1 Integer Exponents
Notice the phrase “nonzero number” in the
previous table. This is because 00 and 0 raised to
a negative power are both undefined. For
example, if you use the pattern given above the
table with a base of 0 instead of 5, you would
get 0º =
. Also 0–6 would be
=
. Since
division by 0 is undefined, neither value exists.
Holt Algebra 1
7-1 Integer Exponents
Reading Math
2–4 is read “2 to the negative fourth power.”
Holt Algebra 1
7-1 Integer Exponents
Example 1: Application
One cup is 2–4 gallons. Simplify this expression.
cup is equal to
Holt Algebra 1
7-1 Integer Exponents
Check It Out! Example 1
A sand fly may have a wingspan up to 5–3 m.
Simplify this expression.
5-3 m is equal to
Holt Algebra 1
7-1 Integer Exponents
Example 2: Zero and Negative Exponents
Simplify.
A. 4–3
B. 70
7º = 1
C. (–5)–4
D. –5–4
Holt Algebra 1
Any nonzero number raised to the zero
power is 1.
7-1 Integer Exponents
Caution
In (–3)–4, the base is negative because the
negative sign is inside the parentheses. In –3–4
the base (3) is positive.
Holt Algebra 1
7-1 Integer Exponents
Simplify.
a. 10–4
Check It Out! Example 2
b. (–2)–4
c. (–2)–5
d. –2–5
Holt Algebra 1
7-1 Integer Exponents
Example 3A: Evaluating Expressions with Zero and
Negative Exponents
Evaluate the expression for the given value of the
variables.
x–2 for x = 4
Substitute 4 for x.
Use the definition
Holt Algebra 1
7-1 Integer Exponents
Example 3B: Evaluating Expressions with Zero and
Negative Exponents
Evaluate the expression for the given values of
the variables.
–2a0b-4 for a = 5 and b = –3
Substitute 5 for a and –3 for b.
Evaluate expressions with
exponents.
Write the power in the
denominator as a product.
Evaluate the powers in
the product.
Simplify.
Holt Algebra 1
7-1 Integer Exponents
Check It Out! Example 3a
Evaluate the expression for the given value of
the variable.
p–3 for p = 4
Substitute 4 for p.
Evaluate exponent.
Write the power in the
denominator as a product.
Evaluate the powers in
the product.
Holt Algebra 1
7-1 Integer Exponents
Check It Out! Example 3b
Evaluate the expression for the given values
of the variables.
for a = –2 and b = 6
Substitute –2 for a and 6 for b.
Evaluate expressions with
exponents.
2
Holt Algebra 1
Write the power in the
denominator as a product.
Evaluate the powers in
the product.
Simplify.
7-1 Integer Exponents
What if you have an expression with a negative
exponent in a denominator, such as
?
or
Definition of a
negative exponent.
Substitute –8 for n.
Simplify the exponent
on the right side.
An ifexpression
that
contains exponent
negative or
zero
So
a base with
a negative
is in
a
exponents is it
not
to the
be simplified.
denominator,
is considered
equivalent to
same base with
Expressions
be exponent
rewritten with
positive
the
opposite should
(positive)
in theonly
numerator.
exponents.
Holt Algebra 1
7-1 Integer Exponents
Example 4: Simplifying Expressions with Zero and
Negative Numbers
Simplify.
A. 7w–4
Holt Algebra 1
B.
7-1 Integer Exponents
Example 4: Simplifying Expressions with Zero and
Negative Numbers
Simplify.
C.
and
Holt Algebra 1
7-1 Integer Exponents
Check It Out! Example 4
Simplify.
a. 2r0m–3
rº = 1 and
b.
Holt Algebra 1
c.
.
7-1 Integer Exponents
Lesson Quiz: Part I
1. A square foot is 3–2 square yards. Simplify this
expression.
Simplify.
2. 2–6
3. (–7)–3
4. 60
5. –112
Holt Algebra 1
1
–121
7-1 Integer Exponents
Lesson Quiz: Part II
Evaluate each expression for the given
value(s) of the variables(s).
6. x–4 for x =10
7.
Holt Algebra 1
for a = 6 and b = 3
7-2 Powers of 10 and Scientific Notation
Warm Up
Evaluate each expression.
1. 123  1,000
123,000
2. 123  1,000
0.123
3. 0.003  100
0.3
4. 0.003  100
0.00003
5. 104
10,000
6. 10–4 0.0001
7. 230 1
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Objectives
Evaluate and multiply by powers of 10.
Convert between standard notation and
scientific notation.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Vocabulary
scientific notation
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
The table shows relationships between
several powers of 10.
Each time you divide by 10, the exponent
decreases by 1 and the decimal point moves
one place to the left.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
The table shows relationships between
several powers of 10.
Each time you multiply by 10, the exponent
increases by 1 and the decimal point moves
one place to the right.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Example 1: Evaluating Powers of 10
Find the value of each power of 10.
A. 10–6
B. 104
C. 109
Start with 1
and move the
decimal point
six places to
the left.
0.000001
Start with 1
and move the
decimal point
four places to
the right.
10,000
Start with 1
and move the
decimal point
nine places to
the right.
1,000,000,000
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Writing Math
You may need to add zeros to the right or left of
a number in order to move the decimal point in
that direction.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Check It Out! Example 1
Find the value of each power of 10.
a. 10–2
b. 105
c. 1010
Start with 1
and move the
decimal point
two places to
the left.
0.01
Start with 1
and move the
decimal point
five places to
the right.
100,000
Start with 1
and move the
decimal point
ten places to
the right.
10,000,000,000
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Reading Math
If you do not see a decimal point in a number, it
is understood to be at the end of the number.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Example 2: Writing Powers of 10
Write each number as a power of 10.
A. 1,000,000
B. 0.0001
C. 1,000
The decimal
point is six
places to the
right of 1, so the
exponent is 6.
The decimal
point is four
places to the
left of 1, so the
exponent is –4.
The decimal
point is three
places to the
right of 1, so the
exponent is 3.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Check It Out! Example 2
Write each number as a power of 10.
a. 100,000,000
b. 0.0001
c. 0.1
The decimal
point is eight
places to the
right of 1, so the
exponent is 8.
The decimal
point is four
places to the
left of 1, so the
exponent is –4.
The decimal
point is one
place to the
left of 1, so the
exponent is –1.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
You can also move the decimal point to find
the value of any number multiplied by a
power of 10. You start with the number
rather than starting with 1.
Multiplying by Powers of 10
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Example 3: Multiplying by Powers of 10
Find the value of each expression.
A. 23.89  108
23.8 9 0 0 0 0 0 0
2,389,000,000
Move the decimal point 8
places to the right.
B. 467  10–3
467
0.467
Holt Algebra 1
Move the decimal point 3
places to the left.
7-2 Powers of 10 and Scientific Notation
Check It Out! Example 3
Find the value of each expression.
a. 853.4  105
853.4 0 0 0 0
85,340,000
Move the decimal point 5
places to the right.
b. 0.163  10–2
0.0 0163
0.00163
Holt Algebra 1
Move the decimal point 2
places to the left.
7-2 Powers of 10 and Scientific Notation
Scientific notation is a method of writing
numbers that are very large or very small. A
number written in scientific notation has two
parts that are multiplied.
The first part is a number that is greater than or equal
to 1 and less than 10.
The second part is a power of 10.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Example 4A: Astronomy Application
Saturn has a diameter of about
Its distance from the Sun is about
1,427,000,000 km.
km.
Write Saturn’s diameter in standard form.
120000
120,000 km
Holt Algebra 1
Move the decimal point 5 places to
the right.
7-2 Powers of 10 and Scientific Notation
Example 4B: Astronomy Application
Saturn has a diameter of about
Its distance from the Sun is about
1,427,000,000 km.
km.
Write Saturn’s distance from the Sun in
scientific notation.
Count the number of places
1,427,000,000
you need to move the
decimal point to get a
1,4 2 7,0 0 0,0 0 0
number between 1 and 10.
9 places
1.427  109 km
Holt Algebra 1
Use that number as the
exponent of 10.
7-2 Powers of 10 and Scientific Notation
Reading Math
Standard form refers to the usual way that
numbers are written—not in scientific notation.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Check It Out! Example 4a
Use the information above to write Jupiter’s
diameter in scientific notation.
143,000 km
143000
5 places
1.43  105 km
Holt Algebra 1
Count the number of places
you need to move the
decimal point to get a
number between 1 and 10.
Use that number as the
exponent of 10.
7-2 Powers of 10 and Scientific Notation
Check It Out! Example 4b
Use the information above to write Jupiter’s
orbital speed in standard form.
13000
13,000 m/s
Holt Algebra 1
Move the decimal point 4 places to
the right.
7-2 Powers of 10 and Scientific Notation
Example 5: Comparing and Ordering Numbers in
Scientific Notation
Order the list of numbers from least to greatest.
Step 1 List the numbers in order by powers of 10.
Step 2 Order the numbers that have the same
power of 10
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Check It Out! Example 5
Order the list of numbers from least to greatest.
Step 1 List the numbers in order by powers of 10.
2  10-12, 4  10-3, 5.2  10-3, 3  1014, 4.5  1014,
4.5  1030
Step 2 Order the numbers that have the same
power of 10
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Lesson Quiz: Part I
Find the value of each expression.
1.
3,745,000
2.
0.00293
3. The Pacific Ocean has an area of about 6.4 х 107
square miles. Its volume is about 170,000,000
cubic miles.
a. Write the area of the Pacific Ocean in standard
form.
b. Write the volume of the Pacific Ocean in scientific
notation. 1.7  108 mi3
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Lesson Quiz: Part II
Find the value of each expression.
4. Order the list of numbers from least to
greatest
Holt Algebra 1
Multiplication
Properties
7-3
7-3 Multiplication Properties of Exponents
of Exponents
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
11
7-3 Multiplication Properties of Exponents
Warm Up
Write each expression using an exponent.
1. 2 • 2 • 2 23
2. x • x • x • x
3.
Write each expression without using an
exponent.
4. 43 4 • 4 • 4
5. y2 y • y
6. m–4
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Objective
Use multiplication properties of
exponents to evaluate and simplify
expressions.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
You have seen that exponential expressions are
useful when writing very small or very large
numbers. To perform operations on these numbers,
you can use properties of exponents. You can also
use these properties to simplify your answer.
In this lesson, you will learn some properties that
will help you simplify exponential expressions
containing multiplication.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Products of powers with the same base can be
found by writing each power as a repeated
multiplication.
Notice the relationship between the exponents in
the factors and the exponents in the product
5 + 2 = 7.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Example 1: Finding Products of Powers
Simplify.
A.
Since the powers have the same
base, keep the base and add the
exponents.
B.
Group powers with the same base
together.
Add the exponents of powers with
the same base.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Example 1: Finding Products of Powers
Simplify.
C.
Group powers with the same base
together.
Add the exponents of powers with
the same base.
D.
Group the positive exponents and add
since they have the same base
1
Holt Algebra 1
Add the like bases.
7-3 Multiplication Properties of Exponents
Remember!
A number or variable written without an exponent
actually has an exponent of 1.
10 = 101
y = y1
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Check It Out! Example 1
Simplify.
a.
Since the powers have the same
base, keep the base and add the
exponents.
b.
Group powers with the same base
together.
Add the exponents of powers with
the same base.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Check It Out! Example 1
Simplify.
c.
Group powers with the same base
together.
Add.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Check It Out! Example 1
Simplify.
d.
Group the first two and second
two terms.
Divide the first group and add the
second group.
Multiply.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Example 2: Astronomy Application
Light from the Sun travels at about
miles per second. It takes about 15,000 seconds
for the light to reach Neptune. Find the
approximate distance from the Sun to Neptune.
Write your answer in scientific notation.
distance = rate  time
mi
Holt Algebra 1
Write 15,000 in
scientific notation.
Use the Commutative
and Associative
Properties to group.
Multiply within each
group.
7-3 Multiplication Properties of Exponents
Check It Out! Example 2
Light travels at about
miles per
second. Find the approximate distance that
light travels in one hour. Write your answer in
scientific notation.
distance = rate  time
Write 3,600 in
scientific notation.
Use the Commutative
and Associative
Properties to group.
Multiply within each
group.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
To find a power of a power, you can use the
meaning of exponents.
Notice the relationship between the exponents in
the original power and the exponent in the final
power:
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Example 3: Finding Powers of Powers
Simplify.
Use the Power of a Power Property.
Simplify.
Use the Power of a Power Property.
Zero multiplied by any number is
zero
1
Holt Algebra 1
Any number raised to the zero
power is 1.
7-3 Multiplication Properties of Exponents
Example 3: Finding Powers of Powers
Simplify.
C.
Use the Power of a Power Property.
Simplify the exponent of the first
term.
Since the powers have the same
base, add the exponents.
Write with a positive exponent.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Check It Out! Example 3
Simplify.
Use the Power of a Power Property.
Simplify.
Use the Power of a Power Property.
Zero multiplied by any number is
zero.
1
Holt Algebra 1
Any number raised to the zero
power is 1.
7-3 Multiplication Properties of Exponents
Check It Out! Example 3c
Simplify.
c.
Use the Power of a Power Property.
Simplify the exponents of the two
terms.
Since the powers have the same
base, add the exponents.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Powers of products can be found by using the
meaning of an exponent.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Example 4: Finding Powers of Products
Simplify.
A.
Use the Power of a Product Property.
Simplify.
B.
Use the Power of a Product Property.
Simplify.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Example 4: Finding Powers of Products
Simplify.
C.
Use the Power of a Product Property.
Use the Power of a Product Property.
Simplify.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Check It Out! Example 4
Simplify.
Use the Power of a Product Property.
Simplify.
Use the Power of a Product Property.
Use the Power of a Product Property.
Simplify.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Check It Out! Example 4
Simplify.
c.
Use the Power of a Product
Property.
Use the Power of a Product
Property.
Combine like terms.
Write with a positive
exponent.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Lesson Quiz: Part I
Simplify.
1. 32• 34
2.
3. (x3)2
4.
5.
6.
7.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Lesson Quiz: Part II
7. The islands of Samoa have an approximate
area of 2.9  103 square kilometers. The area
of Texas is about 2.3  102 times as great as
that of the islands. What is the approximate
area of Texas? Write your answer in scientific
notation.
Holt Algebra 1
7-4
7-4 Division
DivisionProperties
PropertiesofofExponents
Exponents
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
11
7-4 Division Properties of Exponents
Warm Up
Simplify.
1. (x2)3
2.
3.
4.
5.
6.
Write in Scientific Notation.
7.
8.
Holt Algebra 1
7-4 Division Properties of Exponents
Objective
Use division properties of exponents to
evaluate and simplify expressions.
Holt Algebra 1
7-4 Division Properties of Exponents
A quotient of powers with the same base
can be found by writing the powers in a
factored form and dividing out common
factors.
Notice the relationship between the
exponents in the original quotient and the
exponent in the final answer: 5 – 3 = 2.
Holt Algebra 1
7-4 Division Properties of Exponents
Holt Algebra 1
7-4 Division Properties of Exponents
Example 1: Finding Quotients of Powers
Simplify.
A.
Holt Algebra 1
B.
7-4 Division Properties of Exponents
Example 1: Finding Quotients of Powers
Simplify.
C.
Holt Algebra 1
D.
7-4 Division Properties of Exponents
Helpful Hint
Both
Holt Algebra 1
and 729 are considered to be simplified.
7-4 Division Properties of Exponents
Check It Out! Example 1
Simplify.
a.
Holt Algebra 1
b.
7-4 Division Properties of Exponents
Check It Out! Example 1
Simplify.
c.
Holt Algebra 1
d.
7-4 Division Properties of Exponents
Example 2: Dividing Numbers in Scientific Notation
Simplify
and write the
answer in scientific notation
Write as a product of quotients.
Simplify each quotient.
Simplify the exponent.
Write 0.5 in scientific notation
as 5 x 10 .
The second two terms have
the same base, so add the
exponents.
Simplify the exponent.
Holt Algebra 1
7-4 Division Properties of Exponents
Writing Math
You can “split up” a quotient of products into a
product of quotients:
Example:
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 2
Simplify
and write the
answer in scientific notation.
Write as a product of quotients.
Simplify each quotient.
Simplify the exponent.
Write 1.1 in scientific notation
as 11 x 10 .
The second two terms have
the same base, so add the
exponents.
Simplify the exponent.
Holt Algebra 1
7-4 Division Properties of Exponents
Example 3: Application
The Colorado Department of Education spent
about
dollars in fiscal year 2004-05
on public schools. There were about
students enrolled in public school. What was
the average spending per student? Write your
answer in standard form.
To find the average spending per student, divide
the total debt by the number of students.
Write as a product of
quotients.
Holt Algebra 1
7-4 Division Properties of Exponents
Example 3 Continued
The Colorado Department of Education spent
about
dollars in fiscal year 2004-05
on public schools. There were about
students enrolled in public school. What was
the average spending per student? Write your
answer in standard form.
To find the average spending per student, divide
the total debt by the number of students.
Simplify each quotient.
Simplify the exponent.
Write in standard form.
The average spending per student is $5,800.
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 3
In 1990, the United States public debt was
about
dollars. The population of the
United States was about
people. What
was the average debt per person? Write your
answer in standard form.
To find the average debt per person, divide the
total debt by the number of people.
Write as a product of
quotients.
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 3 Continued
In 1990, the United States public debt was
about
dollars. The population of the
United States was about
people. What
was the average debt per person? Write your
answer in standard form.
To find the average debt per person, divide the
total debt by the number of people.
Simplify each quotient.
Simplify the exponent.
Write in standard form.
The average debt per person was $12,800.
Holt Algebra 1
7-4 Division Properties of Exponents
A power of a quotient can be found by first
writing the numerator and denominator as
powers.
Notice that the exponents in the final answer
are the same as the exponent in the original
expression.
Holt Algebra 1
7-4 Division Properties of Exponents
Holt Algebra 1
7-4 Division Properties of Exponents
Example 4A: Finding Positive Powers of Quotient
Simplify.
Use the Power of a Quotient
Property.
Simplify.
Holt Algebra 1
7-4 Division Properties of Exponents
Example 4B: Finding Positive Powers of Quotient
Simplify.
Use the Power of a Product
Property.
Use the Power of a Product
Property:
Simplify and use the Power
of a Power Property:
Holt Algebra 1
7-4 Division Properties of Exponents
Example 4C: Finding Positive Powers of Quotient
Simplify.
Use the Power of a Product
Property.
Use the Power of a Product
Property:
Use the Power of a Product
Property:
Holt Algebra 1
7-4 Division Properties of Exponents
Example 4C Continued
Simplify.
Use the Power of a Product
Property:
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 4a
Simplify.
Use the Power of a Quotient
Property.
Simplify.
Holt Algebra 1
7-4 Division Properties of Exponents
Simplify.
Holt Algebra 1
Check It Out! Example 4b
7-4 Division Properties of Exponents
Simplify.
Holt Algebra 1
Check It Out! Example 4c
7-4 Division Properties of Exponents
Remember that
. What if x is a fraction?
Write the fraction as division.
Use the Power of a Quotient
Property.
Multiply by the reciprocal.
Simplify.
Use the Power of a Quotient
Property.
Therefore,
Holt Algebra 1
7-4 Division Properties of Exponents
Holt Algebra 1
7-4 Division Properties of Exponents
Example 5A: Finding Negative Powers of Quotients
Simplify.
Rewrite with a positive exponent.
Use the Powers of a Quotient
Property .
and
Holt Algebra 1
7-4 Division Properties of Exponents
Example 5B: Finding Negative Powers of Quotients
Simplify.
Holt Algebra 1
7-4 Division Properties of Exponents
Example 5C: Finding Negative Powers of Quotients
Simplify.
Rewrite each fraction with a
positive exponent.
Use the Power of a Quotient
Property.
Use the Power of a Product
Property:
(3)2 (2n)3 = 32  23n3
and (2)2  (6m)3 = 22  63m3
Holt Algebra 1
7-4 Division Properties of Exponents
Example 5C: Finding Negative Powers of Quotients
Simplify.
Square and cube terms.
1
1
1
2
24
Divide out common
factors.
12
Simplify.
Holt Algebra 1
7-4 Division Properties of Exponents
Helpful Hint
Whenever all of the factors in the numerator or
the denominator divide out, replace them with 1.
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 5a
Simplify.
Rewrite with a positive
exponent.
Use the power of a Quotient
Property.
93=729 and 43 = 64.
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 5b
Simplify.
Rewrite with a positive
exponent.
Use the Power of a Quotient
Property.
Use the Power of a Power
Property: (b2c3)4= b2•4c3•4 =
b8c12 and (2a)4= 24a4= 16a4.
Holt Algebra 1
7-4 Division Properties of Exponents
Simplify.
Check It Out! Example 5c
Rewrite each fraction with a
positive exponent.
Use the Power of a
Quotient Property.
Use the Power of a Product
Property: (3)2= 9.
Add exponents and divide
out common terms.
Holt Algebra 1
7-4 Division Properties of Exponents
Lesson Quiz: Part I
Simplify.
1.
2.
3.
4.
5.
Holt Algebra 1
7-4 Division Properties of Exponents
Lesson Quiz: Part II
Simplify.
6. Simplify (3  1012) ÷ (5  105) and write the
answer in scientific notation. 6  106
7. The Republic of Botswana has an area of 6 
105 square kilometers. Its population is about
1.62  106. What is the population density of
Botswana? Write your answer in standard
form. 2.7 people/km2
Holt Algebra 1
7-5 Polynomials
Warm Up
Evaluate each expression for the given value
of x.
1. 2x + 3; x = 2 7
3. –4x – 2; x = –1 2
2. x2 + 4; x = –3 13
4. 7x2 + 2x = 3 69
Identify the coefficient in each term.
5. 4x3 4
6. y3 1
7. 2n7 2
Holt Algebra 1
8. –54 –1
7-5 Polynomials
Objectives
Classify polynomials and write
polynomials in standard form.
Evaluate polynomial expressions.
Holt Algebra 1
7-5 Polynomials
Vocabulary
monomial
degree of a monomial
polynomial
degree of a polynomial
standard form of a
polynomial
leading coefficient
Holt Algebra 1
quadratic
cubic
binomial
trinomial
7-5 Polynomials
A monomial is a number, a variable, or a product
of numbers and variables with whole-number
exponents.
The degree of a monomial is the sum of the
exponents of the variables. A constant has
degree 0.
Holt Algebra 1
7-5 Polynomials
Example 1: Finding the Degree of a Monomial
Find the degree of each monomial.
A. 4p4q3
The degree is 7.
B. 7ed
The degree is 2.
C. 3
The degree is 0.
Holt Algebra 1
Add the exponents of the
variables: 4 + 3 = 7.
Add the exponents of the
variables: 1+ 1 = 2.
Add the exponents of the
variables: 0 = 0.
7-5 Polynomials
Remember!
The terms of an expression are the parts being
added or subtracted. See Lesson 1-7.
Holt Algebra 1
7-5 Polynomials
Check It Out! Example 1
Find the degree of each monomial.
a. 1.5k2m
The degree is 3.
b. 4x
The degree is 1.
b. 2c3
The degree is 3.
Holt Algebra 1
Add the exponents of the
variables: 2 + 1 = 3.
Add the exponents of the
variables: 1 = 1.
Add the exponents of the
variables: 3 = 3.
7-5 Polynomials
A polynomial is a monomial or a sum or
difference of monomials.
The degree of a polynomial is the
degree of the term with the greatest
degree.
Holt Algebra 1
7-5 Polynomials
Example 2: Finding the Degree of a Polynomial
Find the degree of each polynomial.
A. 11x7 + 3x3
11x7: degree 7
3x3: degree 3
The degree of the polynomial is
the greatest degree, 7.
Find the degree of
each term.
B.
:degree 3
–5: degree 0
:degree 4
Find the degree of
each term.
The degree of the polynomial is the greatest degree, 4.
Holt Algebra 1
7-5 Polynomials
Check It Out! Example 2
Find the degree of each polynomial.
a. 5x – 6
5x: degree 1
–6: degree 0
The degree of the polynomial
is the greatest degree, 1.
Find the degree of
each term.
b. x3y2 + x2y3 – x4 + 2
x3y2: degree 5
–x4: degree 4
x2y3: degree 5
2: degree 0
The degree of the polynomial is
the greatest degree, 5.
Holt Algebra 1
Find the degree of
each term.
7-5 Polynomials
The terms of a polynomial may be written in
any order. However, polynomials that
contain only one variable are usually written
in standard form.
The standard form of a polynomial that
contains one variable is written with the
terms in order from greatest degree to
least degree. When written in standard
form, the coefficient of the first term is
called the leading coefficient.
Holt Algebra 1
7-5 Polynomials
Example 3A: Writing Polynomials in Standard Form
Write the polynomial in standard form. Then
give the leading coefficient.
6x – 7x5 + 4x2 + 9
Find the degree of each term. Then arrange them in
descending order:
6x – 7x5 + 4x2 + 9
Degree
1
5
2
0
–7x5 + 4x2 + 6x + 9
5
2
1
0
The standard form is –7x5 + 4x2 + 6x + 9. The leading
coefficient is –7.
Holt Algebra 1
7-5 Polynomials
Example 3B: Writing Polynomials in Standard Form
Write the polynomial in standard form. Then
give the leading coefficient.
y2 + y6 − 3y
Find the degree of each term. Then arrange them in
descending order:
y2 + y6 – 3y
Degree
2
6
1
y6 + y2 – 3y
6
2
1
The standard form is y6 + y2 – 3y. The leading
coefficient is 1.
Holt Algebra 1
7-5 Polynomials
Remember!
A variable written without a coefficient has a
coefficient of 1.
y5 = 1y5
Holt Algebra 1
7-5 Polynomials
Check It Out! Example 3a
Write the polynomial in standard form. Then
give the leading coefficient.
16 – 4x2 + x5 + 9x3
Find the degree of each term. Then arrange them in
descending order:
16 – 4x2 + x5 + 9x3
Degree 0
2
5
3
x5 + 9x3 – 4x2 + 16
5
3
2
0
The standard form is x5 + 9x3 – 4x2 + 16. The leading
coefficient is 1.
Holt Algebra 1
7-5 Polynomials
Check It Out! Example 3b
Write the polynomial in standard form. Then
give the leading coefficient.
18y5 – 3y8 + 14y
Find the degree of each term. Then arrange them in
descending order:
18y5 – 3y8 + 14y
Degree
5
8
1
–3y8 + 18y5 + 14y
8
5
1
The standard form is –3y8 + 18y5 + 14y. The leading
coefficient is –3.
Holt Algebra 1
7-5 Polynomials
Some polynomials have special names based on
their degree and the number of terms they have.
Degree
Name
Terms
Name
0
Constant
1
Monomial
1
Linear
2
Binomial
2
Quadratic
Trinomial
3
4
Cubic
Quartic
3
4 or
more
5
Quintic
6 or more
Holt Algebra 1
6th,7th,degree
and so on
Polynomial
7-5 Polynomials
Example 4: Classifying Polynomials
Classify each polynomial according to its
degree and number of terms.
A. 5n3 + 4n
Degree 3 Terms 2
5n3 + 4n is a cubic
binomial.
B. 4y6 – 5y3 + 2y – 9
Degree 6 Terms 4
4y6 – 5y3 + 2y – 9 is a
C. –2x
Degree 1 Terms 1
–2x is a linear monomial.
Holt Algebra 1
6th-degree polynomial.
7-5 Polynomials
Check It Out! Example 4
Classify each polynomial according to its
degree and number of terms.
a. x3 + x2 – x + 2
Degree 3 Terms 4
x3 + x2 – x + 2 is a
cubic polymial.
b. 6
Degree 0 Terms 1
6 is a constant monomial.
c. –3y8 + 18y5 + 14y
Degree 8 Terms 3
–3y8 + 18y5 + 14y is an
8th-degree trinomial.
Holt Algebra 1
7-5 Polynomials
Example 5: Application
A tourist accidentally drops her lip balm off the
Golden Gate Bridge. The bridge is 220 feet from
the water of the bay. The height of the lip balm
is given by the polynomial –16t2 + 220, where t
is time in seconds. How far above the water will
the lip balm be after 3 seconds?
Substitute the time for t to find the lip balm’s
height.
–16t2 + 220
–16(3)2 + 200
The time is 3 seconds.
–16(9) + 200
–144 + 200
76
Holt Algebra 1
Evaluate the polynomial by using
the order of operations.
7-5 Polynomials
Example 5: Application Continued
A tourist accidentally drops her lip balm off the
Golden Gate Bridge. The bridge is 220 feet from
the water of the bay. The height of the lip balm
is given by the polynomial –16t2 + 220, where t
is time in seconds. How far above the water will
the lip balm be after 3 seconds?
After 3 seconds the lip balm will be 76 feet
from the water.
Holt Algebra 1
7-5 Polynomials
Check It Out! Example 5
What if…? Another firework with a 5-second
fuse is launched from the same platform at a
speed of 400 feet per second. Its height is
given by –16t2 +400t + 6. How high will this
firework be when it explodes?
Substitute the time t to find the firework’s height.
–16t2 + 400t + 6
–16(5)2 + 400(5) + 6
The time is 5 seconds.
–16(25) + 400(5) + 6
–400 + 2000 + 6
–400 + 2006
1606
Holt Algebra 1
Evaluate the polynomial by
using the order of
operations.
7-5 Polynomials
Check It Out! Example 5 Continued
What if…? Another firework with a 5-second
fuse is launched from the same platform at a
speed of 400 feet per second. Its height is
given by –16t2 +400t + 6. How high will this
firework be when it explodes?
When the firework explodes, it will be 1606
feet above the ground.
Holt Algebra 1
7-5 Polynomials
Lesson Quiz: Part I
Find the degree of each polynomial.
1. 7a3b2 – 2a4 + 4b – 15
2. 25x2 – 3x4
5
4
Write each polynomial in standard form. Then
give the leading coefficient.
3. 24g3 + 10 + 7g5 – g2 7g5 + 24g3 – g2 + 10; 7
4. 14 – x4 + 3x2
Holt Algebra 1
–x4 + 3x2 + 14; –1
7-5 Polynomials
Lesson Quiz: Part II
Classify each polynomial according to its
degree and number of terms.
5. 18x2 – 12x + 5
6. 2x4 – 1
quadratic trinomial
quartic binomial
7. The polynomial 3.675v + 0.096v2 is used to
estimate the stopping distance in feet for a car
whose speed is y miles per hour on flat dry
pavement. What is the stopping distance for a
car traveling at 70 miles per hour?
727.65 ft
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Warm Up
Simplify each expression by combining like
terms.
1. 4x + 2x 6x
2. 3y + 7y 10y
3. 8p – 5p 3p
4. 5n + 6n2 not like terms
Simplify each expression.
5. 3(x + 4) 3x + 12
6. –2(t + 3) –2t – 6
7. –1(x2 – 4x – 6) –x2 + 4x + 6
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Objective
Add and subtract polynomials.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Just as you can perform operations on
numbers, you can perform operations on
polynomials. To add or subtract
polynomials, combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Example 1: Adding and Subtracting Monomials
Add or Subtract..
A. 12p3 + 11p2 + 8p3
12p3 + 11p2 + 8p3
12p3 + 8p3 + 11p2
20p3 + 11p2
B. 5x2 – 6 – 3x + 8
5x2 – 6 – 3x + 8
5x2 – 3x + 8 – 6
5x2 – 3x + 2
Holt Algebra 1
Identify like terms.
Rearrange terms so that like
terms are together.
Combine like terms.
Identify like terms.
Rearrange terms so that like
terms are together.
Combine like terms.
7-6 Adding and Subtracting Polynomials
Example 1: Adding and Subtracting Monomials
Add or Subtract..
C. t2 + 2s2 – 4t2 – s2
t2 + 2s2 – 4t2 – s2
t2 – 4t2 + 2s2 – s2
–3t2 + s2
Identify like terms.
Rearrange terms so that
like terms are together.
Combine like terms.
D. 10m2n + 4m2n – 8m2n
10m2n + 4m2n – 8m2n
Identify like terms.
6m2n
Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Remember!
Like terms are constants or terms with the same
variable(s) raised to the same power(s). To
review combining like terms, see lesson 1-7.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Check It Out! Example 1
Add or subtract.
a. 2s2 + 3s2 + s
2s2 + 3s2 + s
5s2 + s
Identify like terms.
Combine like terms.
b. 4z4 – 8 + 16z4 + 2
4z4 – 8 + 16z4 + 2
4z4 + 16z4 – 8 + 2
20z4 – 6
Holt Algebra 1
Identify like terms.
Rearrange terms so that
like terms are together.
Combine like terms.
7-6 Adding and Subtracting Polynomials
Check It Out! Example 1
Add or subtract.
c. 2x8 + 7y8 – x8 – y8
2x8
7y8
x8
y8
+
–
–
2x8 – x8 + 7y8 – y8
x8 + 6y8
Identify like terms.
Rearrange terms so that
like terms are together.
Combine like terms.
d. 9b3c2 + 5b3c2 – 13b3c2
9b3c2 + 5b3c2 – 13b3c2
b3c2
Holt Algebra 1
Identify like terms.
Combine like terms.
7-6 Adding and Subtracting Polynomials
Polynomials can be added in either vertical or
horizontal form.
In vertical form, align
the like terms and add:
5x2 + 4x + 1
+ 2x2 + 5x + 2
7x2 + 9x + 3
In horizontal form, use the
Associative and
Commutative Properties to
regroup and combine like
terms.
(5x2 + 4x + 1) + (2x2 + 5x + 2)
= (5x2 + 2x2 + 1) + (4x + 5x) + (1 + 2)
= 7x2 + 9x + 3
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Example 2: Adding Polynomials
Add.
A. (4m2 + 5) + (m2 – m + 6)
(4m2 + 5) + (m2 – m + 6)
Identify like terms.
(4m2 + m2) + (–m) +(5 + 6)
Group like terms
together.
Combine like terms.
5m2 – m + 11
B. (10xy + x) + (–3xy + y)
(10xy + x) + (–3xy + y)
Identify like terms.
(10xy – 3xy) + x + y
Group like terms
together.
Combine like terms.
7xy + x + y
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Example 2C: Adding Polynomials
Add.
(6x2 – 4y) + (3x2 + 3y – 8x2 – 2y)
(6x2 – 4y) + (3x2 + 3y – 8x2 – 2y) Identify like terms.
(6x2 + 3x2 – 8x2) + (3y – 4y – 2y)
6x2 – 4y
+ –5x2 + y
x2 – 3y
Holt Algebra 1
Group like terms together
within each polynomial.
Use the vertical method.
Combine like terms.
Simplify.
7-6 Adding and Subtracting Polynomials
Example 2D: Adding Polynomials
Add.
Identify like terms.
Group like terms
together.
Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Check It Out! Example 2
Add (5a3 + 3a2 – 6a + 12a2) + (7a3 – 10a).
(5a3 + 3a2 – 6a + 12a2) + (7a3 – 10a)
Identify like terms.
(5a3 + 7a3) + (3a2 + 12a2) + (–10a – 6a) Group like terms
together.
12a3 + 15a2 – 16a
Holt Algebra 1
Combine like terms
7-6 Adding and Subtracting Polynomials
To subtract polynomials, remember that
subtracting is the same as adding the
opposite. To find the opposite of a
polynomial, you must write the opposite
of each term in the polynomial:
–(2x3 – 3x + 7)= –2x3 + 3x – 7
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Example 3A: Subtracting Polynomials
Subtract.
(x3 + 4y) – (2x3)
(x3 + 4y) + (–2x3)
Rewrite subtraction as addition
of the opposite.
Identify like terms.
(x3 – 2x3) + 4y
Group like terms together.
–x3 + 4y
Combine like terms.
(x3 + 4y) + (–2x3)
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Example 3B: Subtracting Polynomials
Subtract.
(7m4 – 2m2) – (5m4 – 5m2 + 8)
(7m4 – 2m2) + (–5m4 + 5m2 – 8) Rewrite subtraction as
addition of the opposite.
(7m4 – 2m2) + (–5m4 + 5m2 – 8) Identify like terms.
(7m4 – 5m4) + (–2m2 + 5m2) – 8 Group like terms together.
2m4 + 3m2 – 8
Holt Algebra 1
Combine like terms.
7-6 Adding and Subtracting Polynomials
Example 3C: Subtracting Polynomials
Subtract.
(–10x2 – 3x + 7) – (x2 – 9)
(–10x2 – 3x + 7) + (–x2 + 9)
(–10x2 – 3x + 7) + (–x2 + 9)
–10x2 – 3x + 7
–x2 + 0x + 9
–11x2 – 3x + 16
Holt Algebra 1
Rewrite subtraction as
addition of the opposite.
Identify like terms.
Use the vertical method.
Write 0x as a placeholder.
Combine like terms.
7-6 Adding and Subtracting Polynomials
Example 3D: Subtracting Polynomials
Subtract.
(9q2 – 3q) – (q2 – 5)
(9q2 – 3q) + (–q2 + 5)
(9q2 – 3q) + (–q2 + 5)
9q2 – 3q + 0
+ − q2 – 0q + 5
8q2 – 3q + 5
Holt Algebra 1
Rewrite subtraction as
addition of the opposite.
Identify like terms.
Use the vertical method.
Write 0 and 0q as
placeholders.
Combine like terms.
7-6 Adding and Subtracting Polynomials
Check It Out! Example 3
Subtract.
(2x2 – 3x2 + 1) – (x2 + x + 1)
(2x2 – 3x2 + 1) + (–x2 – x – 1)
Rewrite subtraction as
addition of the opposite.
(2x2 – 3x2 + 1) + (–x2 – x – 1)
Identify like terms.
–x2 + 0x + 1
+ –x2 – x – 1
–2x2 – x
Holt Algebra 1
Use the vertical method.
Write 0x as a placeholder.
Combine like terms.
7-6 Adding and Subtracting Polynomials
Example 4: Application
A farmer must add the areas of two plots of
land to determine the amount of seed to
plant. The area of plot A can be represented
by 3x2 + 7x – 5 and the area of plot B can
be represented by 5x2 – 4x + 11. Write a
polynomial that represents the total area of
both plots of land.
(3x2 + 7x – 5)
+ (5x2 – 4x + 11)
8x2 + 3x + 6
Holt Algebra 1
Plot A.
Plot B.
Combine like terms.
7-6 Adding and Subtracting Polynomials
Check It Out! Example 4
The profits of two different
manufacturing plants can
be modeled as shown,
where x is the number of
units produced at each
plant.
Use the information above to write a polynomial
that represents the total profits from both plants.
–0.03x2 + 25x – 1500
+ –0.02x2 + 21x – 1700
–0.05x2 + 46x – 3200
Holt Algebra 1
Eastern plant profit.
Southern plant profit.
Combine like terms.
7-6 Adding and Subtracting Polynomials
Lesson Quiz: Part I
Add or subtract.
1. 7m2 + 3m + 4m2
11m2 + 3m
2. (r2 + s2) – (5r2 + 4s2)
(–4r2 – 3s2)
3. (10pq + 3p) + (2pq – 5p + 6pq) 18pq – 2p
4. (14d2 – 8) + (6d2 – 2d +1) 20d2 – 2d – 7
5. (2.5ab + 14b) – (–1.5ab + 4b)
Holt Algebra 1
4ab + 10b
7-6 Adding and Subtracting Polynomials
Lesson Quiz: Part II
6. A painter must add the areas of two walls to
determine the amount of paint needed. The area
of the first wall is modeled by 4x2 + 12x + 9, and
the area of the second wall is modeled by
36x2 – 12x + 1. Write a polynomial that
represents the total area of the two walls.
40x2 + 10
Holt Algebra 1
7-7 Multiplying Polynomials
Warm Up
Evaluate.
1. 32
2. 24 16
9
3. 102 100
Simplify.
4. 23  24 27
5. y5  y4 y9
6. (53)2
7. (x2)4
56
8. –4(x – 7)
Holt Algebra 1
–4x + 28
x8
7-7 Multiplying Polynomials
Objective
Multiply polynomials.
Holt Algebra 1
7-7 Multiplying Polynomials
To multiply monomials and polynomials,
you will use some of the properties of
exponents that you learned earlier in this
chapter.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 1: Multiplying Monomials
Multiply.
A. (6y3)(3y5)
(6y3)(3y5)
(6  3)(y3  y5)
18y8
Group factors with like bases
together.
Multiply.
B. (3mn2) (9m2n)
(3mn2)(9m2n)
(3  9)(m  m2)(n2  n)
27m3n3
Holt Algebra 1
Group factors with like bases
together.
Multiply.
7-7 Multiplying Polynomials
Example 1C: Multiplying Monomials
Multiply.
 1 2 2
2
s
t
t

12
t
s


s

÷
4



1
 2g g

12
÷s s s
4 g



Holt Algebra 1

t
2
Group factors with like
bases together.
gt gt 2

Multiply.
7-7 Multiplying Polynomials
Remember!
When multiplying powers with the same base,
keep the base and add the exponents.
x2  x3 = x2+3 = x5
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 1
Multiply.
a. (3x3)(6x2)
(3x3)(6x2)
(3  6)(x3  x2)
18x5
Group factors with like bases
together.
Multiply.
b. (2r2t)(5t3)
(2r2t)(5t3)
(2  5)(r2)(t3  t)
10r2t4
Holt Algebra 1
Group factors with like bases
together.
Multiply.
7-7 Multiplying Polynomials
Check It Out! Example 1
Multiply.
1 2 
3 2
4 5
x
y
12
x
z
yz


c. 
÷
3


1 2 
3 2
x
y
12
x
z
÷
3


y z 
1 g  2 g 3
 3 12÷ x x


y gy z


4x 5y 5 z 7
Holt Algebra 1
4
5
4
2
g z5

Group factors with
like bases
together.
Multiply.
7-7 Multiplying Polynomials
To multiply a polynomial by a monomial, use
the Distributive Property.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 2A: Multiplying a Polynomial by a Monomial
Multiply.
4(3x2 + 4x – 8)
4(3x2 + 4x – 8)
Distribute 4.
(4)3x2 +(4)4x – (4)8
Multiply.
12x2 + 16x – 32
Holt Algebra 1
7-7 Multiplying Polynomials
Example 2B: Multiplying a Polynomial by a Monomial
Multiply.
6pq(2p – q)
(6pq)(2p – q)
Distribute 6pq.
(6pq)2p + (6pq)(–q)
(6

2)(p

Group like bases
together.
p)(q) + (–1)(6)(p)(q  q)
12p2q – 6pq2
Holt Algebra 1
Multiply.
7-7 Multiplying Polynomials
Example 2C: Multiplying a Polynomial by a Monomial
Multiply.
1 2
2
2
x y6xy + 8 x y 
2
1 2
2 2
xy
x y 6
+ 8x y
2

1 2
Distribute x y .
2

1 2 
1 2 
2 2
Group like bases
x
y
6
xy
+
x
y
8
x
y



÷
2
÷

2


together.
 2
1  2
1
 • 6 ÷x • x  y • y +  • 8÷ x • x2 y • y2

2 
2



3x3y2 + 4x4y3
Holt Algebra 1



Multiply.

7-7 Multiplying Polynomials
Check It Out! Example 2
Multiply.
a. 2(4x2 + x + 3)
2(4x2 + x + 3)
Distribute 2.
2(4x2) + 2(x) + 2(3)
Multiply.
8x2 + 2x + 6
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 2
Multiply.
b. 3ab(5a2 + b)
3ab(5a2 + b)
Distribute 3ab.
(3ab)(5a2) + (3ab)(b)
(3  5)(a  a2)(b) + (3)(a)(b  b)
15a3b + 3ab2
Holt Algebra 1
Group like bases
together.
Multiply.
7-7 Multiplying Polynomials
Check It Out! Example 2
Multiply.
c. 5r2s2(r – 3s)
5r2s2(r – 3s)
Distribute 5r2s2.
(5r2s2)(r) – (5r2s2)(3s)
(5)(r2  r)(s2) – (5  3)(r2)(s2  s)
5r3s2 – 15r2s3
Holt Algebra 1
Group like bases
together.
Multiply.
7-7 Multiplying Polynomials
To multiply a binomial by a binomial, you can
apply the Distributive Property more than once:
(x + 3)(x + 2) = x(x + 2) + 3(x + 2)
Distribute x and 3.
= x(x + 2) + 3(x + 2)
Distribute x and 3
again.
= x(x) + x(2) + 3(x) + 3(2)
Multiply.
= x2 + 2x + 3x + 6
Combine like terms.
= x2 + 5x + 6
Holt Algebra 1
7-7 Multiplying Polynomials
Another method for multiplying binomials is
called the FOIL method.
F
1. Multiply the First terms. (x + 3)(x + 2)
x  x = x2
O
2. Multiply the Outer terms. (x + 3)(x + 2)
I
3. Multiply the Inner terms. (x + 3)(x + 2)
L
4. Multiply the Last terms. (x + 3)(x + 2)
x  2 = 2x
3  x = 3x
3 2 = 6
(x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6
F
Holt Algebra 1
O
I
L
7-7 Multiplying Polynomials
Example 3A: Multiplying Binomials
Multiply.
(s + 4)(s – 2)
(s + 4)(s – 2)
s(s – 2) + 4(s – 2)
Distribute s and 4.
s(s) + s(–2) + 4(s) + 4(–2)
s2 – 2s + 4s – 8
Distribute s and 4
again.
Multiply.
s2 + 2s – 8
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 3B: Multiplying Binomials
Multiply.
(x –
4)2
(x – 4)(x – 4)
Write as a product of
two binomials.
Use the FOIL method.
(x  x) + (x  (–4)) + (–4  x) + (–4  (–4))
x2 – 4x – 4x + 8
Multiply.
x2 – 8x + 8
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 3C: Multiplying Binomials
Multiply.
(8m2 – n)(m2 – 3n)
Use the FOIL method.
8m2(m2) + 8m2(–3n) – n(m2) – n(–3n)
8m4 – 24m2n – m2n + 3n2
Multiply.
8m4 – 25m2n + 3n2
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Helpful Hint
In the expression (x + 5)2, the base is (x + 5).
(x + 5)2 = (x + 5)(x + 5)
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 3a
Multiply.
(a + 3)(a – 4)
(a + 3)(a – 4)
Distribute a and 3.
a(a – 4)+3(a – 4)
Distribute a and 3
again.
a(a) + a(–4) + 3(a) + 3(–4)
a2 – 4a + 3a – 12
Multiply.
a2 – a – 12
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 3b
Multiply.
(x – 3)2
(x – 3)(x – 3)
Write as a product of
two binomials.
Use the FOIL method.
(x ●x) + (x(–3)) + (–3  x)+ (–3)(–3)
Holt Algebra 1
x2 – 3x – 3x + 9
Multiply.
x2 – 6x + 9
Combine like terms.
7-7 Multiplying Polynomials
Check It Out! Example 3c
Multiply.
(2a – b2)(a + 4b2)
(2a – b2)(a + 4b2)
Use the FOIL method.
2a(a) + 2a(4b2) – b2(a) + (–b2)(4b2)
2a2 + 8ab2 – ab2 – 4b4
Multiply.
2a2 + 7ab2 – 4b4
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
To multiply polynomials with more than two terms,
you can use the Distributive Property several times.
Multiply (5x + 3) by (2x2 + 10x – 6):
(5x + 3)(2x2 + 10x – 6) = 5x(2x2 + 10x – 6) + 3(2x2 + 10x – 6)
= 5x(2x2 + 10x – 6) + 3(2x2 + 10x – 6)
= 5x(2x2) + 5x(10x) + 5x(–6) + 3(2x2) + 3(10x) + 3(–6)
= 10x3 + 50x2 – 30x + 6x2 + 30x – 18
= 10x3 + 56x2 – 18
Holt Algebra 1
7-7 Multiplying Polynomials
You can also use a rectangle model to multiply
polynomials with more than two terms. This is
similar to finding the area of a rectangle with
length (2x2 + 10x – 6) and width (5x + 3):
2x2
5x
+3
10x3
6x2
+10x
–6
50x2 –30x
30x
–18
Write the product of the
monomials in each row and
column:
To find the product, add all of the terms inside the
rectangle by combining like terms and simplifying
if necessary.
10x3 + 6x2 + 50x2 + 30x – 30x – 18
10x3 + 56x2 – 18
Holt Algebra 1
7-7 Multiplying Polynomials
Another method that can be used to multiply
polynomials with more than two terms is the
vertical method. This is similar to methods used to
multiply whole numbers.
2x2 + 10x – 6
Multiply each term in the top
polynomial by 3.
Multiply each term in the top

5x + 3
polynomial by 5x, and align
6x2 + 30x – 18
like terms.
+ 10x3 + 50x2 – 30x
10x3 + 56x2 + 0x – 18 Combine like terms by adding
vertically.
10x3 + 56x2 +
– 18 Simplify.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 4A: Multiplying Polynomials
Multiply.
(x – 5)(x2 + 4x – 6)
(x – 5 )(x2 + 4x – 6)
Distribute x and –5.
x(x2 + 4x – 6) – 5(x2 + 4x – 6) Distribute x and −5
again.
x(x2) + x(4x) + x(–6) – 5(x2) – 5(4x) – 5(–6)
x3 + 4x2 – 5x2 – 6x – 20x + 30
Simplify.
x3 – x2 – 26x + 30
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 4B: Multiplying Polynomials
Multiply.
(2x – 5)(–4x2 – 10x + 3)
(2x – 5)(–4x2 – 10x + 3)
–4x2 – 10x + 3
x
2x – 5
20x2 + 50x – 15
+ –8x3 – 20x2 + 6x
–8x3
+ 56x – 15
Holt Algebra 1
Multiply each term in the
top polynomial by –5.
Multiply each term in the
top polynomial by 2x,
and align like terms.
Combine like terms by
adding vertically.
7-7 Multiplying Polynomials
Example 4C: Multiplying Polynomials
Multiply.
(x + 3)3
[(x + 3)(x + 3)](x + 3)
Write as the product of
three binomials.
[x(x+3) + 3(x+3)](x + 3)
Use the FOIL method on
the first two factors.
(x2 + 3x + 3x + 9)(x + 3) Multiply.
(x2 + 6x + 9)(x + 3)
Holt Algebra 1
Combine like terms.
7-7 Multiplying Polynomials
Example 4C: Multiplying Polynomials
Multiply.
(x + 3)3
(x + 3)(x2 + 6x + 9)
Use the Commutative
Property of
Multiplication.
x(x2 + 6x + 9) + 3(x2 + 6x + 9)
Distribute the x and 3.
x(x2) + x(6x) + x(9) + 3(x2) +
3(6x) + 3(9)
Distribute the x and 3
again.
x3 + 6x2 + 9x + 3x2 + 18x + 27
Combine like terms.
x3 + 9x2 + 27x + 27
Holt Algebra 1
7-7 Multiplying Polynomials
Example 4D: Multiplying Polynomials
Multiply.
(3x + 1)(x3 – 4x2 – 7)
x3
4x2
3x
3x4
12x3
+1
x3
4x2
–7
–21x
–7
Write the product of the
monomials in each
row and column.
Add all terms inside the
rectangle.
3x4 + 12x3 + x3 + 4x2 – 21x – 7
3x4 + 13x3 + 4x2 – 21x – 7
Holt Algebra 1
Combine like terms.
7-7 Multiplying Polynomials
Helpful Hint
A polynomial with m terms multiplied by a
polynomial with n terms has a product that,
before simplifying has mn terms. In Example 4A,
there are 2  3, or 6 terms before simplifying.
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 4a
Multiply.
(x + 3)(x2 – 4x + 6)
(x + 3 )(x2 – 4x + 6)
Distribute x and 3.
x(x2 – 4x + 6) + 3(x2 – 4x + 6)
Distribute x and 3
again.
x(x2) + x(–4x) + x(6) +3(x2) +3(–4x) +3(6)
x3 – 4x2 + 3x2 +6x – 12x + 18
Simplify.
x3 – x2 – 6x + 18
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 4b
Multiply.
(3x + 2)(x2 – 2x + 5)
(3x + 2)(x2 – 2x + 5)
x2 – 2x + 5

3x + 2
2x2 – 4x + 10
+ 3x3 – 6x2 + 15x
3x3 – 4x2 + 11x + 10
Holt Algebra 1
Multiply each term in the
top polynomial by 2.
Multiply each term in the
top polynomial by 3x,
and align like terms.
Combine like terms by
adding vertically.
7-7 Multiplying Polynomials
Example 5: Application
The width of a rectangular prism is 3 feet less
than the height, and the length of the prism is
4 feet more than the height.
a. Write a polynomial that represents the area of the
base of the prism.
A = lw
Write the formula for the
area of a rectangle.
A = lw
Substitute h – 3 for w
A = (h + 4)(h – 3)
and h + 4 for l.
A = h2 + 4h – 3h – 12 Multiply.
A = h2 + h – 12
Combine like terms.
The area is represented by h2 + h – 12.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 5: Application
The width of a rectangular prism is 3 feet less
than the height, and the length of the prism is
4 feet more than the height.
b. Find the area of the base when the height is 5 ft.
A = h2 + h – 12
A = h2 + h – 12
Write the formula for the area
the base of the prism.
A = 52 + 5 – 12
Substitute 5 for h.
A = 25 + 5 – 12
Simplify.
A = 18
Combine terms.
The area is 18 square feet.
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 5
The length of a rectangle is 4 meters shorter
than its width.
a. Write a polynomial that represents the area of the
rectangle.
Write the formula for the
A = lw
area of a rectangle.
A = lw
A = x(x – 4)
A = x2 – 4x
Substitute x – 4 for l and
x for w.
Multiply.
The area is represented by x2 – 4x.
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 5
The length of a rectangle is 4 meters shorter
than its width.
b. Find the area of a rectangle when the width is 6
meters.
A = x2 – 4x
Write the formula for the area of a
rectangle whose length is 4
A = x2 – 4x
meters shorter than width .
Substitute 6 for x.
A = 62 – 4  6
A = 36 – 24
Simplify.
A = 12
Combine terms.
The area is 12 square meters.
Holt Algebra 1
7-7 Multiplying Polynomials
Lesson Quiz: Part I
Multiply.
1. (6s2t2)(3st) 18s3t3
2. 4xy2(x + y) 4x2y2 + 4xy3
3. (x + 2)(x – 8) x2 – 6x – 16
4. (2x – 7)(x2 + 3x – 4) 2x3 – x2 – 29x + 28
5. 6mn(m2 + 10mn – 2)
6m3n + 60m2n2 – 12mn
6. (2x – 5y)(3x + y) 6x2 – 13xy – 5y2
Holt Algebra 1
7-7 Multiplying Polynomials
Lesson Quiz: Part II
7. A triangle has a base that is 4cm longer than its
height.
a. Write a polynomial that represents the area
of the triangle.
1 2
h + 2h
2
b. Find the area when the height is 8 cm.
48 cm2
Holt Algebra 1
7-9 Special Products of Binomials
Warm Up
Simplify.
1. 42 16
3. (–2)2 4
5. –(5y2) –25y2
4. (x)2 x2
7. 2(6xy) 12xy
8. 2(8x2) 16x2
Holt McDougal Algebra 1
2. 72 49
6. (m2)2 m4
7-9 Special Products of Binomials
Objective
Find special products of binomials.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Vocabulary
perfect-square trinomial
difference of two squares
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Imagine a square with sides of length (a + b):
The area of this square is (a + b)(a + b) or (a + b)2.
The area of this square can also be found by adding the
areas of the smaller squares and the rectangles inside.
The sum of the areas inside is a2 + ab + ab + b2.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
This means that (a + b)2 = a2+ 2ab + b2.
You can use the FOIL method to verify this:
F
L
(a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2
I
= a2 + 2ab + b2
O
A trinomial of the form a2 + 2ab + b2 is called a
perfect-square trinomial. A perfect-square
trinomial is a trinomial that is the result of
squaring a binomial.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Example 1: Finding Products in the Form (a + b)2
Multiply.
A. (x +3)2
(a + b)2 = a2 + 2ab + b2
(x + 3)2 = x2 + 2(x)(3) + 32
= x2 + 6x + 9
Use the rule for (a + b)2.
Identify a and b: a = x and
b = 3.
Simplify.
B. (4s + 3t)2
(a + b)2 = a2 + 2ab + b2
Use the rule for (a + b)2.
Identify a and b: a = 4s
and b = 3t.
2
2
2
(4s + 3t) = (4s) + 2(4s)(3t) + (3t)
= 16s2 + 24st + 9t2
Holt McDougal Algebra 1
Simplify.
7-9 Special Products of Binomials
Example 1C: Finding Products in the Form (a + b)2
Multiply.
C. (5 + m2)2
(a + b)2 = a2 + 2ab + b2
Use the rule for (a + b)2.
Identify a and b: a = 5 and
b = m2.
(5 + m2)2 = 52 + 2(5)(m2) + (m2)2
= 25 + 10m2 + m4
Holt McDougal Algebra 1
Simplify.
7-9 Special Products of Binomials
Check It Out! Example 1
Multiply.
A. (x + 6)2
(a + b)2 = a2 + 2ab + b2
(x + 6)2 = x2 + 2(x)(6) + 62
= x2 + 12x + 36
B. (5a + b)2
Use the rule for (a + b)2.
Identify a and b: a = x and
b = 6.
Simplify.
Use the rule for (a + b)2.
(a + b)2 = a2 + 2ab + b2
Identify a and b: a = 5a
and b = b.
(5a + b)2 = (5a)2 + 2(5a)(b) + b2
= 25a2 + 10ab + b2
Holt McDougal Algebra 1
Simplify.
7-9 Special Products of Binomials
Check It Out! Example 1C
Multiply.
(1 + c3)2
(a + b)2 = a2 + 2ab + b2
Use the rule for (a + b)2.
Identify a and b: a = 1 and
b = c3.
(1 + c3)2 = 12 + 2(1)(c3) + (c3)2
= 1 + 2c3 + c6
Holt McDougal Algebra 1
Simplify.
7-9 Special Products of Binomials
You can use the FOIL method to find products in
the form of (a – b)2.
F
L
(a – b)2 = (a – b)(a – b) = a2 – ab – ab + b2
I
O
= a2 – 2ab + b2
A trinomial of the form a2 – ab + b2 is also a
perfect-square trinomial because it is the result
of squaring the binomial (a – b).
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Example 2: Finding Products in the Form (a – b)2
Multiply.
A. (x – 6)2
(a – b)2 = a2 – 2ab + b2
(x – 6)2 = x2 – 2x(6) + (6)2
= x2 – 12x + 36
Use the rule for (a – b)2.
Identify a and b: a = x and
b = 6.
Simplify.
B. (4m – 10)2
Use the rule for (a – b)2.
Identify a and b: a = 4m
(a – b)2 = a2 – 2ab + b2
and b = 10.
(4m – 10)2 = (4m)2 – 2(4m)(10) + (10)2
= 16m2 – 80m + 100
Holt McDougal Algebra 1
Simplify.
7-9 Special Products of Binomials
Example 2: Finding Products in the Form (a – b)2
Multiply.
C. (2x – 5y)2
Use the rule for (a – b)2.
(a – b)2 = a2 – 2ab + b2
Identify a and b: a = 2x
and b = 5y.
2
2
2
(2x – 5y) = (2x) – 2(2x)(5y) + (5y)
= 4x2 – 20xy +25y2
D. (7 – r3)2
(a – b)2 = a2 – 2ab + b2
Simplify.
Use the rule for (a – b)2.
Identify a and b: a = 7 and
3.
b
=
r
2
(7 – r3)2 = 72 – 2(7)(r3) + (r3)
Simplify.
= 49 – 14r3 + r6
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Check It Out! Example 2
Multiply.
a. (x – 7)2
(a – b)2 = a2 – 2ab + b2
(x – 7)2 = x2 – 2(x)(7) + (7)2
= x2 – 14x + 49
b. (3b – 2c)2
Use the rule for (a – b)2.
Identify a and b: a = x
and b = 7.
Simplify.
Use the rule for (a – b)2.
(a – b)2 = a2 – 2ab + b2
Identify a and b: a = 3b
and b = 2c.
2
2
2
(3b – 2c) = (3b) – 2(3b)(2c) + (2c)
= 9b2 – 12bc + 4c2
Holt McDougal Algebra 1
Simplify.
7-9 Special Products of Binomials
Check It Out! Example 2c
Multiply.
(a2 – 4)2
Use the rule for (a – b)2.
(a – b)2 = a2 – 2ab + b2
Identify a and b: a = a2
(a2 – 4)2 = (a2)2 – 2(a2)(4) + (4)2 and b = 4.
= a4 – 8a2 + 16
Holt McDougal Algebra 1
Simplify.
7-9 Special Products of Binomials
You can use an area model to see that
(a + b)(a–b)= a2 – b2.
Begin with a square
with area a2. Remove
a square with area b2.
The area of the new
figure is a2 – b2.
So (a + b)(a – b)
a2 – b2 is called a
Holt McDougal Algebra 1
Remove the
rectangle on the
bottom. Turn it and
slide it up next to
the top rectangle.
The new arrangement is a rectangle
with length a + b and
width a – b. Its area
is (a + b)(a – b).
= a2 – b2. A binomial of the form
difference of two squares.
7-9 Special Products of Binomials
Example 3: Finding Products in the Form (a + b)(a – b)
Multiply.
A. (x + 4)(x – 4)
(a + b)(a – b) = a2 – b2
(x + 4)(x – 4) = x2 – 42
= x2 – 16
B. (p2 + 8q)(p2 – 8q)
Use the rule for (a + b)(a – b).
Identify a and b: a = x
and b = 4.
Simplify.
Use the rule for (a + b)(a – b).
(a + b)(a – b) = a2 – b2
(p2 + 8q)(p2 – 8q) = (p2)2 – (8q)2 Identify a and b: a = p2
and b = 8q.
= p4 – 64q2 Simplify.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Example 3: Finding Products in the Form (a + b)(a – b)
Multiply.
C. (10 + b)(10 – b)
Use the rule for (a + b)(a – b).
(a + b)(a – b) = a2 – b2
(10 + b)(10 – b) = 102 – b2
= 100 – b2
Holt McDougal Algebra 1
Identify a and b: a = 10
and b = b.
Simplify.
7-9 Special Products of Binomials
Check It Out! Example 3
Multiply.
a. (x + 8)(x – 8)
Use the rule for (a + b)(a – b).
(a + b)(a – b) = a2 – b2
(x + 8)(x – 8) =
x2
–
82
= x2 – 64
b. (3 + 2y2)(3 – 2y2)
(a + b)(a – b) = a2 – b2
(3 + 2y2)(3 – 2y2) = 32 – (2y2)2
= 9 – 4y4
Holt McDougal Algebra 1
Identify a and b: a = x and
b = 8.
Simplify.
Use the rule for (a + b)(a – b).
Identify a and b: a = 3 and
b = 2y2.
Simplify.
7-9 Special Products of Binomials
Check It Out! Example 3
Multiply.
c. (9 + r)(9 – r)
(a + b)(a – b) = a2 – b2
(9 + r)(9 – r) = 92 – r2
= 81 – r2
Holt McDougal Algebra 1
Use the rule for (a + b)(a – b).
Identify a and b: a = 9 and
b = r.
Simplify.
7-9 Special Products of Binomials
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Example 4: Problem-Solving Application
Write a polynomial that represents the
area of the yard around the pool
shown below.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Example 4 Continued
1
Understand the Problem
The answer will be an expression that shows
the area of the yard less the area of the pool.
List important information:
• The yard is a square with a side length of x + 5.
• The pool has side lengths of x + 2 and x – 2.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Example 4 Continued
2
Make a Plan
The area of the yard is (x + 5)2. The area of the
pool is (x + 2) (x – 2). You can subtract the
area of the pool from the yard to find the area
of the yard surrounding the pool.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Example 4 Continued
3
Solve
Step 1 Find the total area.
(x +5)2 = x2 + 2(x)(5) + 52
= x2 + 10x + 25
Use the rule for (a +
b)2: a = x and b =
5.
Step 2 Find the area of the pool.
(x + 2)(x – 2) = x2 – 2x + 2x – 4 Use the rule for (a
+ b)(a – b): a = x
= x2 – 4
and b = 2.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Example 4 Continued
3
Solve
Step 3 Find the area of the yard around the pool.
Area of yard =
a
total area
– area of pool
= x2 + 10x + 25 –
(x2 – 4)
Identify like
=
+ 10x + 25 –
+4
terms.
= (x2 – x2) + 10x + ( 25 + 4) Group like
terms
= 10x + 29
together
x2
x2
The area of the yard around the pool is 10x + 29.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Example 4 Continued
4
Look Back
Suppose that x = 20. Then the total area in the
back yard would be 252 or 625. The area of the
pool would be 22  18 or 396. The area of the yard
around the pool would be 625 – 396 = 229.
According to the solution, the area of the yard
around the pool is 10x + 29. If x = 20, then
10x +29 = 10(20) + 29 = 229.

Holt McDougal Algebra 1
7-9 Special Products of Binomials
Remember!
To subtract a polynomial, add the opposite of
each term.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Check It Out! Example 4
Write an expression that represents
the area of the swimming pool.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Check It Out! Example 4 Continued
1
Understand the Problem
The answer will be an expression that shows
the area of the two rectangles combined.
List important information:
• The upper rectangle has side lengths of 5 + x
and 5 – x .
• The lower rectangle is a square with side
length of x.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Check It Out! Example 4 Continued
2
Make a Plan
The area of the upper rectangle is (5 + x)(5 – x).
The area of the lower square is x2. Added
together they give the total area of the pool.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Check It Out! Example 4 Continued
3
Solve
Step 1 Find the area of the upper rectangle.
(5 + x)(5 – x) = 25 – 5x + 5x –
x2
Use the rule for (a + b)
(a – b): a = 5 and b = x.
= –x2 + 25
Step 2 Find the area of the lower square.
= xx
= x2
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Check It Out! Example 4 Continued
3
Solve
Step 3 Find the area of the pool.
Area of pool =
rectangle area + square area
–x2 + 25
a =
=
–x2 + 25 + x2
= (x2 – x2) + 25
= 25
The area of the pool is 25.
Holt McDougal Algebra 1
+
x2
Identify like
terms.
Group like
terms
together
7-9 Special Products of Binomials
Check It Out! Example 4 Continued
4
Look Back
Suppose that x = 2. Then the area of the
upper rectangle would be 21. The area of the
lower square would be 4. The area of the
pool would be 21 + 4 = 25.
According to the solution, the area of the
pool is 25. 
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Lesson Quiz: Part I
Multiply.
1. (x + 7)2 x2 + 14x + 49
2. (x – 2)2
x2 – 4x + 4
3. (5x + 2y)2
25x2 + 20xy + 4y2
4. (2x – 9y)2
4x2 – 36xy + 81y2
5. (4x + 5y)(4x – 5y) 16x2 – 25y2
6. (m2 + 2n)(m2 – 2n)
Holt McDougal Algebra 1
m4 – 4n2
7-9 Special Products of Binomials
Lesson Quiz: Part II
7. Write a polynomial that represents the shaded
area of the figure below.
x+6
x–7
x–6
x–7
14x – 85
Holt McDougal Algebra 1