warm up evaluate each expression for the given values of the variables
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Warm Up Evaluate each expression for the given values of the variables. 1. x 3 y 2 for x = –1 and y = 10 2. for x = 4 and y = (–7) Write each number as a power of the given base. –100. 3. 64; base 4. 4 3. 4. – 27; base ( – 3). (–3) 3. Objectives. - PowerPoint PPT PresentationTRANSCRIPT
Holt Algebra 1
7-1 Integer Exponents
Warm UpEvaluate each expression for the given values of the variables.
1. x3y2 for x = –1 and y = 10
2. for x = 4 and y = (–7)
Write each number as a power of the given base.
–100
433. 64; base 4
(–3)34. –27; base (–3)
Holt Algebra 1
7-1 Integer Exponents
Evaluate expressions containing zero and integer exponents.
Simplify expressions containing zero and integer exponents.
Objectives
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.
3125 625 125 25 5
5
Power
Value
55 54 53 52 51 5–150 5–2
5 5 5
Holt Algebra 1
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
Base
x
Exponent
Remember!
4
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
2–4 is read “2 to the negative fourth power.”
Reading Math
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 = 1Any nonzero number raised to the zero
power is 1.
C. (–5)–4
D. –5–4
Holt Algebra 1
7-1 Integer Exponents
In (–3)–4, the base is negative because the
negative sign is inside the parentheses. In –3–4
the base (3) is positive.
Caution
Holt Algebra 1
7-1 Integer Exponents
Check It Out! Example 2 Simplify.
a. 10–4
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
2
Substitute –2 for a and 6 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
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.
So if a base with a negative exponent is in a denominator, it is equivalent to the same base with the opposite (positive) exponent in the numerator.
An expression that contains negative or zero exponents is not considered to be simplified. Expressions should be rewritten with only positive exponents.
Holt Algebra 1
7-1 Integer Exponents
Simplify.
Example 4: Simplifying Expressions with Zero and Negative Numbers
A. 7w–4 B.
Holt Algebra 1
7-1 Integer Exponents
Simplify.
Example 4: Simplifying Expressions with Zero and Negative Numbers
C.
and
Holt Algebra 1
7-1 Integer Exponents
Check It Out! Example 4
Simplify.
a. 2r0m–3
b.
c.
rº = 1 and .
Holt Algebra 1
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
1
–121
Holt Algebra 1
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. for a = 6 and b = 3
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Warm UpEvaluate each expression.
1. 123 1,000
2. 123 1,000
3. 0.003 100
4. 0.003 100
5. 104
6. 10–4
7. 230
123,000
0.123
0.3
0.00003
10,000
0.0001
1
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Evaluate and multiply by powers of 10.Convert between standard notation and scientific notation.
Objectives
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
scientific notation
Vocabulary
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
Find the value of each power of 10.
Example 1: Evaluating Powers of 10
Start with 1 and move the decimal point six places to the left.
A. 10–6 C. 109 B. 104
1,000,000,000
Start with 1 and move the decimal point four places to the right.
Start with 1 and move the decimal point nine places to the right.
10,0000.000001
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
You may need to add zeros to the right or left of a number in order to move the decimal point in that direction.
Writing Math
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 c. 1010 b. 105
10,000,000,000100,0000.01
Start with 1 and move the decimal point two places to the left.
Start with 1 and move the decimal point five places to the right.
Start with 1 and move the decimal point ten places to the right.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
If you do not see a decimal point in a number, it is understood to be at the end of the number.
Reading Math
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Write each number as a power of 10.
Example 2: Writing Powers of 10
A. 1,000,000
The decimal point is six places to the right of 1, so the exponent is 6.
B. 0.0001 C. 1,000
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
Find the value of each expression.
Example 3: Multiplying by Powers of 10
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
4 6 7
0.467
Move the decimal point 3 places to the left.
Holt Algebra 1
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 Move the decimal point 5 places to the right.
85,340,000
b. 0.163 10–2
0.0 0163
0.00163
Move the decimal point 2 places to the left.
Holt Algebra 1
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 km. Its distance from the Sun is about 1,427,000,000 km.
Write Saturn’s diameter in standard form.
1 2 0 0 0 0
120,000 km
Move the decimal point 5 places to the right.
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Write Saturn’s distance from the Sun in scientific notation.
1,427,000,000
1,4 2 7,0 0 0,0 0 0
9 places
Example 4B: Astronomy Application
Saturn has a diameter of about km. Its distance from the Sun is about 1,427,000,000 km.
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. 1.427 109 km
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Standard form refers to the usual way that numbers are written—not in scientific notation.
Reading Math
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
1 4 3 0 0 0
5 places
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. 1.43 105 km
Holt Algebra 1
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.
1 3 0 0 0 Move the decimal point 4 places to the right.
13,000 m/s
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Order the list of numbers from least to greatest.
Example 5: Comparing and Ordering Numbers in Scientific Notation
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
Order the list of numbers from least to greatest.
Check It Out! Example 5
Step 1 List the numbers in order by powers of 10.
Step 2 Order the numbers that have the same power of 10
2 10-12, 4 10-3, 5.2 10-3, 3 1014, 4.5 1014, 4.5 1030
Holt Algebra 1
7-2 Powers of 10 and Scientific Notation
Lesson Quiz: Part IFind the value of each expression.
1.
2.
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
0.00293
3,745,000
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
7-3 Multiplication Properties of Exponents7-3Multiplication Properties of Exponents
Holt Algebra 1
Warm Up
Lesson Presentation
Lesson Quiz
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Warm Up
Write each expression using an exponent.1. 2 • 2 • 2
2. x • x • x • x
3.
Write each expression without using an exponent.
4. 43
5. y2
6. m–4
23
4 • 4 • 4
y • y
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Use multiplication properties of exponents to evaluate and simplify expressions.
Objective
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
Simplify.
Example 1: Finding Products of Powers
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
Simplify.
Example 1: Finding Products of Powers
C.
D.
1
Group powers with the same base together.
Add the exponents of powers with the same base.
Group the positive exponents and add since they have the same base
Add the like bases.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
A number or variable written without an exponent actually has an exponent of 1.
Remember!
10 = 101
y = y1
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Check It Out! Example 1
a. Simplify.
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 Write 15,000 in scientific notation.
Use the Commutative and Associative Properties to group.
Multiply within each group.mi
Holt Algebra 1
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.
Multiply within each group.
Use the Commutative and Associative Properties to 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
Simplify.
Example 3: Finding Powers of Powers
Use the Power of a Power Property.
Simplify.
1
Use the Power of a Power Property.
Zero multiplied by any number is zero
Any number raised to the zero power is 1.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Simplify.
Example 3: Finding Powers of Powers
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.
C.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Check It Out! Example 3
Simplify.
Use the Power of a Power Property.
Simplify.
1
Use the Power of a Power Property.
Zero multiplied by any number is zero.
Any number raised to the zero power is 1.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Check It Out! Example 3c
Simplify.
Use the Power of a Power Property.
Simplify the exponents of the two terms.
Since the powers have the same base, add the exponents.
c.
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.
Use the Power of a Product Property.
Simplify.
Use the Power of a Product Property.
Simplify.
A.
B.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Example 4: Finding Powers of Products
Simplify.
Use the Power of a Product Property.
Use the Power of a Product Property.
Simplify.
C.
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.
Use the Power of a Product Property.
Use the Power of a Product Property.
Combine like terms.
Write with a positive exponent.
c.
Holt Algebra 1
7-3 Multiplication Properties of Exponents
Lesson Quiz: Part I
Simplify.
1. 32• 34
3.
5.
7.
2.
4.
6.
(x3)2
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 Division Properties of Exponents7-4 Division Properties of Exponents
Holt Algebra 1
Warm Up
Lesson Quiz
Lesson Presentation
Holt Algebra 1
7-4 Division Properties of Exponents
Warm UpSimplify.
1. (x2)3
3.
5.
2.
4.
6.
7.
Write in Scientific Notation.
8.
Holt Algebra 1
7-4 Division Properties of Exponents
Use division properties of exponents to evaluate and simplify expressions.
Objective
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
Simplify.
Example 1: Finding Quotients of Powers
A. B.
Holt Algebra 1
7-4 Division Properties of Exponents
C.
Simplify.
Example 1: Finding Quotients of Powers
D.
Holt Algebra 1
7-4 Division Properties of Exponents
Both and 729 are considered to be simplified.
Helpful Hint
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 1
a.
Simplify.
b.
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 1
Simplify.
c. d.
Holt Algebra 1
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
You can “split up” a quotient of products into a product of quotients:
Example:
Writing Math
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.
The average spending per student is $5,800.
Simplify each quotient.
Simplify the exponent.
Write in standard form.
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
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.
Check It Out! Example 3 Continued
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
Simplify.
Example 4A: Finding Positive Powers of Quotient
Use the Power of a Quotient Property.
Simplify.
Holt Algebra 1
7-4 Division Properties of Exponents
Simplify.
Example 4B: Finding Positive Powers of Quotient
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
Simplify.
Example 4C: Finding Positive Powers of Quotient
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
Simplify.
Example 4C Continued
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
Check It Out! Example 4b Simplify.
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 4c Simplify.
Holt Algebra 1
7-4 Division Properties of Exponents
Therefore,
Write the fraction as division.
Use the Power of a Quotient Property.
Multiply by the reciprocal.
Simplify.
Use the Power of a Quotient Property.
Remember that What if x is a fraction?.
Holt Algebra 1
7-4 Division Properties of Exponents
Holt Algebra 1
7-4 Division Properties of Exponents
Simplify.
Example 5A: Finding Negative Powers of Quotients
Rewrite with a positive exponent.
and
Use the Powers of a Quotient Property .
Holt Algebra 1
7-4 Division Properties of Exponents
Simplify.
Example 5B: Finding Negative Powers of Quotients
Holt Algebra 1
7-4 Division Properties of Exponents
Simplify.
Example 5C: Finding Negative Powers of Quotients
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
1
241
21
12
Divide out common factors.
Simplify.
Example 5C: Finding Negative Powers of Quotients
Simplify.
Square and cube terms.
Holt Algebra 1
7-4 Division Properties of Exponents
Whenever all of the factors in the numerator or the denominator divide out, replace them with 1.
Helpful Hint
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 5a
Simplify.
93=729 and 43 = 64.
Use the power of a Quotient Property.
Rewrite with a positive exponent.
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 Power Property: (b2c3)4= b2•4c3•4 = b8c12 and (2a)4= 24a4= 16a4.
Use the Power of a Quotient Property.
Holt Algebra 1
7-4 Division Properties of Exponents
Check It Out! Example 5c Simplify.
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
1.
3. 4.
5.
2.
Simplify.
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 UpEvaluate each expression for the given value of x.
1. 2x + 3; x = 2 2. x2 + 4; x = –3
3. –4x – 2; x = –1 4. 7x2 + 2x = 3
Identify the coefficient in each term.
5. 4x3 6. y3
7. 2n7 8. –54
7 13
2 69
4 1
2 –1
Holt Algebra 1
7-5 Polynomials
Classify polynomials and write polynomials in standard form. Evaluate polynomial expressions.
Objectives
Holt Algebra 1
7-5 Polynomials
monomialdegree of a monomialpolynomialdegree of a polynomialstandard form of a polynomialleading coefficient
Vocabulary
binomialtrinomial
quadraticcubic
Holt Algebra 1
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. Add the exponents of the variables: 4 + 3 = 7.
B. 7ed
The degree is 2. Add the exponents of the variables: 1+ 1 = 2.C. 3
The degree is 0. Add the exponents of the variables: 0 = 0.
Holt Algebra 1
7-5 Polynomials
The terms of an expression are the parts being added or subtracted. See Lesson 1-7.
Remember!
Holt Algebra 1
7-5 Polynomials
Check It Out! Example 1
Find the degree of each monomial.
a. 1.5k2m
The degree is 3. Add the exponents of the variables: 2 + 1 = 3.
b. 4x
The degree is 1. Add the exponents of the variables: 1 = 1.
b. 2c3
The degree is 3. Add the exponents of the variables: 3 = 3.
Holt Algebra 1
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
Find the degree of each polynomial.
Example 2: Finding the Degree of a 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.
Find the degree of each term.
The degree of the polynomial is the greatest degree, 4.
:degree 3 :degree 4
–5: degree 0
Holt Algebra 1
7-5 Polynomials
Check It Out! Example 2
Find the degree of each polynomial.
a. 5x – 6
5x: degree 1Find the degree of
each term.The degree of the polynomial is the greatest degree, 1.
b. x3y2 + x2y3 – x4 + 2
x3y2: degree 5
The degree of the polynomial is the greatest degree, 5.
Find the degree of each term.
–6: degree 0
x2y3: degree 5–x4: degree 4 2: degree 0
Holt Algebra 1
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
Write the polynomial in standard form. Then give the leading coefficient.
Example 3A: Writing Polynomials in Standard Form
6x – 7x5 + 4x2 + 9
Find the degree of each term. Then arrange them in descending order:
6x – 7x5 + 4x2 + 9 –7x5 + 4x2 + 6x + 9
Degree 1 5 2 0 5 2 1 0
–7x5 + 4x2 + 6x + 9.The standard form is The leading coefficient is –7.
Holt Algebra 1
7-5 Polynomials
Write the polynomial in standard form. Then give the leading coefficient.
Example 3B: Writing Polynomials in Standard Form
Find the degree of each term. Then arrange them in descending order:
y2 + y6 − 3y
y2 + y6 – 3y y6 + y2 – 3y
Degree 2 6 1 2 16
The standard form is The leading coefficient is 1.
y6 + y2 – 3y.
Holt Algebra 1
7-5 Polynomials
A variable written without a coefficient has a coefficient of 1.
Remember!
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 x5 + 9x3 – 4x2 + 16
Degree 0 2 5 3 0235
The standard form is The leading coefficient is 1.
x5 + 9x3 – 4x2 + 16.
Holt Algebra 1
7-5 Polynomials
Check It Out! Example 3b
Write the polynomial in standard form. Then give the leading coefficient.
Find the degree of each term. Then arrange them in descending order:
18y5 – 3y8 + 14y
18y5 – 3y8 + 14y –3y8 + 18y5 + 14y
Degree 5 8 1 8 5 1
The standard form is The leading coefficient is –3.
–3y8 + 18y5 + 14y.
Holt Algebra 1
7-5 Polynomials
Some polynomials have special names based on their degree and the number of terms they have.
Degree Name
0
1
2
Constant
Linear
Quadratic
3
4
5
6 or more 6th,7th,degree and so on
Cubic
Quartic
Quintic
NameTerms
Monomial
Binomial
Trinomial
Polynomial4 or more
1
2
3
Holt Algebra 1
7-5 Polynomials
Classify each polynomial according to its degree and number of terms.
Example 4: Classifying Polynomials
A. 5n3 + 4nDegree 3 Terms 2
5n3 + 4n is a cubic binomial.
B. 4y6 – 5y3 + 2y – 9
Degree 6 Terms 4
4y6 – 5y3 + 2y – 9 is a
6th-degree polynomial.
C. –2xDegree 1 Terms 1
–2x is a linear monomial.
Holt Algebra 1
7-5 Polynomials
Classify each polynomial according to its degree and number of terms.
Check It Out! Example 4
a. x3 + x2 – x + 2Degree 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 + 14yDegree 8 Terms 3
–3y8 + 18y5 + 14y is an 8th-degree trinomial.
Holt Algebra 1
7-5 Polynomials
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?
Example 5: Application
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) + 200Evaluate the polynomial by using
the order of operations.–144 + 20076
Holt Algebra 1
7-5 Polynomials
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?
Example 5: Application Continued
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 Evaluate the polynomial by using the order of operations.
–400 + 20061606
Holt Algebra 1
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
Write each polynomial in standard form. Then
give the leading coefficient.
3. 24g3 + 10 + 7g5 – g2
4. 14 – x4 + 3x2
4
5
–x4 + 3x2 + 14; –1
7g5 + 24g3 – g2 + 10; 7
Holt Algebra 1
7-5 Polynomials
Lesson Quiz: Part II
Classify each polynomial according to its degree and number of terms.
5. 18x2 – 12x + 5 quadratic trinomial
6. 2x4 – 1 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 UpSimplify each expression by combining like terms.
1. 4x + 2x
2. 3y + 7y
3. 8p – 5p
4. 5n + 6n2
Simplify each expression.
5. 3(x + 4)
6. –2(t + 3)
7. –1(x2 – 4x – 6)
6x
10y
3p
not like terms
3x + 12
–2t – 6
–x2 + 4x + 6
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Add and subtract polynomials.
Objective
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
Add or Subtract..
Example 1: Adding and Subtracting Monomials
A. 12p3 + 11p2 + 8p3
12p3 + 11p2 + 8p3
12p3 + 8p3 + 11p2
20p3 + 11p2
Identify like terms.Rearrange terms so that like
terms are together.Combine like terms.
B. 5x2 – 6 – 3x + 8
5x2 – 6 – 3x + 8
5x2 – 3x + 8 – 6
5x2 – 3x + 2
Identify like terms.Rearrange terms so that like
terms are together.Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Add or Subtract..
Example 1: Adding and Subtracting Monomials
C. t2 + 2s2 – 4t2 – s2
t2 – 4t2 + 2s2 – s2
t2 + 2s2 – 4t2 – s2
–3t2 + s2
Identify like terms.Rearrange terms so that
like terms are together.Combine like terms.
D. 10m2n + 4m2n – 8m2n
10m2n + 4m2n – 8m2n
6m2n
Identify like terms.
Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
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.
Remember!
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Check It Out! Example 1
a. 2s2 + 3s2 + s
Add or subtract.
2s2 + 3s2 + s
5s2 + s
b. 4z4 – 8 + 16z4 + 2
4z4 – 8 + 16z4 + 2
4z4 + 16z4 – 8 + 2
20z4 – 6
Identify like terms.
Combine like terms.
Identify like terms.Rearrange terms so that
like terms are together.Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Check It Out! Example 1
c. 2x8 + 7y8 – x8 – y8
Add or subtract.
2x8 + 7y8 – x8 – y8
2x8 – x8 + 7y8 – y8
x8 + 6y8
d. 9b3c2 + 5b3c2 – 13b3c2
9b3c2 + 5b3c2 – 13b3c2
b3c2
Identify like terms.
Combine like terms.
Identify like terms.Rearrange terms so that
like terms are together.Combine like terms.
Holt Algebra 1
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:
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
5x2 + 4x + 1+ 2x2 + 5x + 2
7x2 + 9x + 3
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Add.
Example 2: Adding Polynomials
A. (4m2 + 5) + (m2 – m + 6)
(4m2 + 5) + (m2 – m + 6)
(4m2 + m2) + (–m) +(5 + 6)
5m2 – m + 11
Identify like terms.
Group like terms together.
Combine like terms.
B. (10xy + x) + (–3xy + y)
(10xy + x) + (–3xy + y)
(10xy – 3xy) + x + y
7xy + x + y
Identify like terms.
Group like terms together.
Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Add.
Example 2C: Adding Polynomials
(6x2 – 4y) + (3x2 + 3y – 8x2 – 2y)
Identify like terms.
Group like terms together within each polynomial.
Combine like terms.
(6x2 – 4y) + (3x2 + 3y – 8x2 – 2y)
(6x2 + 3x2 – 8x2) + (3y – 4y – 2y)
Use the vertical method. 6x2 – 4y+ –5x2 + y
x2 – 3y Simplify.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Add.
Example 2D: Adding Polynomials
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)
(5a3 + 7a3) + (3a2 + 12a2) + (–10a – 6a)
12a3 + 15a2 – 16a
Identify like terms.
Group like terms together.
Combine like terms.
Holt Algebra 1
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
Subtract.
Example 3A: Subtracting Polynomials
(x3 + 4y) – (2x3)
(x3 + 4y) + (–2x3)
(x3 + 4y) + (–2x3)
(x3 – 2x3) + 4y
–x3 + 4y
Rewrite subtraction as addition of the opposite.
Identify like terms.
Group like terms together.
Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Subtract.
Example 3B: Subtracting Polynomials
(7m4 – 2m2) – (5m4 – 5m2 + 8)
(7m4 – 2m2) + (–5m4 + 5m2 – 8)
(7m4 – 5m4) + (–2m2 + 5m2) – 8
(7m4 – 2m2) + (–5m4 + 5m2 – 8)
2m4 + 3m2 – 8
Rewrite subtraction as addition of the opposite.
Identify like terms.
Group like terms together.
Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Subtract.
Example 3C: Subtracting Polynomials
(–10x2 – 3x + 7) – (x2 – 9)
(–10x2 – 3x + 7) + (–x2 + 9)
(–10x2 – 3x + 7) + (–x2 + 9)
–10x2 – 3x + 7–x2 + 0x + 9
–11x2 – 3x + 16
Rewrite subtraction as addition of the opposite.
Identify like terms.
Use the vertical method.Write 0x as a placeholder.Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Subtract.
Example 3D: Subtracting Polynomials
(9q2 – 3q) – (q2 – 5)
(9q2 – 3q) + (–q2 + 5)
(9q2 – 3q) + (–q2 + 5)
9q2 – 3q + 0+ − q2 – 0q + 5
8q2 – 3q + 5
Rewrite subtraction as addition of the opposite.
Identify like terms.Use the vertical method.Write 0 and 0q as
placeholders.
Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Check It Out! Example 3
Subtract.
(2x2 – 3x2 + 1) – (x2 + x + 1)
(2x2 – 3x2 + 1) + (–x2 – x – 1)
(2x2 – 3x2 + 1) + (–x2 – x – 1)
–x2 + 0x + 1 + –x2 – x – 1–2x2 – x
Rewrite subtraction as addition of the opposite.
Identify like terms.
Use the vertical method.Write 0x as a placeholder.
Combine like terms.
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
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.
Example 4: Application
(3x2 + 7x – 5)(5x2 – 4x + 11)
8x2 + 3x + 6
Plot A.Plot B.
Combine like terms.
+
Holt Algebra 1
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 Eastern plant profit.
–0.02x2 + 21x – 1700 Southern plant profit.Combine like terms.
+–0.05x2 + 46x – 3200
Holt Algebra 1
7-6 Adding and Subtracting Polynomials
Lesson Quiz: Part I
Add or subtract.
1. 7m2 + 3m + 4m2
2. (r2 + s2) – (5r2 + 4s2)
3. (10pq + 3p) + (2pq – 5p + 6pq)
4. (14d2 – 8) + (6d2 – 2d +1)
(–4r2 – 3s2)
11m2 + 3m
18pq – 2p
20d2 – 2d – 7
5. (2.5ab + 14b) – (–1.5ab + 4b) 4ab + 10b
Holt Algebra 1
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 UpEvaluate.
1. 32
3. 102
Simplify.
4. 23 24
6. (53)2
9 16
100
27
2. 24
5. y5 y4
56 7. (x2)4
8. –4(x – 7) –4x + 28
y9
x8
Holt Algebra 1
7-7 Multiplying Polynomials
Multiply polynomials.
Objective
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
Multiply.
Example 1: Multiplying Monomials
A. (6y3)(3y5)
(6y3)(3y5)
18y8
Group factors with like bases together.
B. (3mn2) (9m2n)
(3mn2)(9m2n)
27m3n3
Multiply.
Group factors with like bases together.
Multiply.
(6 3)(y3 y5)
(3 9)(m m2)(n2 n)
Holt Algebra 1
7-7 Multiplying Polynomials
Multiply.
Example 1C: Multiplying Monomials
2 2 2112
4s t st st
4 53s t
Group factors with like bases together.
Multiply.
( )( )æçè
- 22 2112
4ts tt s s
ö÷ø
( )( )g gg g gæ - öçè
2 2112
4t s ts ts÷
ø2
Holt Algebra 1
7-7 Multiplying Polynomials
When multiplying powers with the same base, keep the base and add the exponents.
x2 x3 = x2+3 = x5
Remember!
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.
Group factors with like bases together.
Multiply.
b. (2r2t)(5t3)
(2r2t)(5t3)
(2 5)(r2)(t3 t)
10r2t4
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 1
Multiply.
Group factors with like bases together.
Multiply.
c.
( )( )æçè
4 52 2112
3x zy zx y
ö÷ø
3
( )( )æçè
2112
3x y x z y z
2 4 53ö÷ø
( )( )( )g gg gæçè
3 22 4 5112 z
3zx x y y
ö÷ø
7554x y z
Holt Algebra 1
7-7 Multiplying Polynomials
To multiply a polynomial by a monomial, use the Distributive Property.
Holt Algebra 1
7-7 Multiplying Polynomials
Multiply.
Example 2A: Multiplying a Polynomial by a Monomial
4(3x2 + 4x – 8)
4(3x2 + 4x – 8)
(4)3x2 +(4)4x – (4)8
12x2 + 16x – 32
Distribute 4.
Multiply.
Holt Algebra 1
7-7 Multiplying Polynomials
6pq(2p – q)
(6pq)(2p – q)
Multiply.
Example 2B: Multiplying a Polynomial by a Monomial
(6pq)2p + (6pq)(–q)
(6 2)(p p)(q) + (–1)(6)(p)(q q)
12p2q – 6pq2
Distribute 6pq.
Group like bases together.
Multiply.
Holt Algebra 1
7-7 Multiplying Polynomials
Multiply.
Example 2C: Multiplying a Polynomial by a Monomial
Group like bases together.
Multiply.
( )+216
2x y xy x y8 2 2
x y( )+ 22 612
xyyx 8 2
x y x y( ) ( )æçè
+2 216
28xy x y 22ö
÷ø
12
æçè
ö÷ø
x2 • x( )( ) ( )( )æ+ç
è
1• 6
2y • y x2 • x2 y • y2• 8
ö÷ø
æçè
ö÷ø
12
3x3y2 + 4x4y3
Distribute .21x y
2
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 2
Multiply.
a. 2(4x2 + x + 3)
2(4x2 + x + 3)
2(4x2) + 2(x) + 2(3)
8x2 + 2x + 6
Distribute 2.
Multiply.
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 2
Multiply.
b. 3ab(5a2 + b)
3ab(5a2 + b)
(3ab)(5a2) + (3ab)(b)
(3 5)(a a2)(b) + (3)(a)(b b)
15a3b + 3ab2
Distribute 3ab.
Group like bases together.
Multiply.
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 2
Multiply.
c. 5r2s2(r – 3s)
5r2s2(r – 3s)
(5r2s2)(r) – (5r2s2)(3s)
(5)(r2 r)(s2) – (5 3)(r2)(s2 s)
5r3s2 – 15r2s3
Distribute 5r2s2.
Group like bases together.
Multiply.
Holt Algebra 1
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.
Distribute x and 3 again.
Multiply.
Combine like terms.
= x(x + 2) + 3(x + 2)
= x(x) + x(2) + 3(x) + 3(2)
= x2 + 2x + 3x + 6
= x2 + 5x + 6
Holt Algebra 1
7-7 Multiplying Polynomials
Another method for multiplying binomials is called the FOIL method.
4. Multiply the Last terms. (x + 3)(x + 2) 3 2 = 6
3. Multiply the Inner terms. (x + 3)(x + 2) 3 x = 3x
2. Multiply the Outer terms. (x + 3)(x + 2) x 2 = 2x
F
O
I
L
(x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6
F O I L
1. Multiply the First terms. (x + 3)(x + 2) x x = x2
Holt Algebra 1
7-7 Multiplying Polynomials
Multiply.
Example 3A: Multiplying Binomials
(s + 4)(s – 2)
(s + 4)(s – 2)
s(s – 2) + 4(s – 2)
s(s) + s(–2) + 4(s) + 4(–2)
s2 – 2s + 4s – 8
s2 + 2s – 8
Distribute s and 4.
Distribute s and 4 again.
Multiply.
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Multiply.
Example 3B: Multiplying Binomials
(x – 4)2
(x – 4)(x – 4)
(x x) + (x (–4)) + (–4 x) + (–4 (–4))
x2 – 4x – 4x + 8
x2 – 8x + 8
Write as a product of two binomials.
Use the FOIL method.
Multiply.
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 3C: Multiplying Binomials
Multiply.
(8m2 – n)(m2 – 3n)
8m2(m2) + 8m2(–3n) – n(m2) – n(–3n)
8m4 – 24m2n – m2n + 3n2
8m4 – 25m2n + 3n2
Use the FOIL method.
Multiply.
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
In the expression (x + 5)2, the base is (x + 5). (x + 5)2 = (x + 5)(x + 5)
Helpful Hint
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 3a
Multiply.
(a + 3)(a – 4)
(a + 3)(a – 4)
a(a – 4)+3(a – 4)
a(a) + a(–4) + 3(a) + 3(–4)
a2 – a – 12
a2 – 4a + 3a – 12
Distribute a and 3.
Distribute a and 3 again.
Multiply.
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 3b
Multiply.
(x – 3)2
(x – 3)(x – 3)
(x x) + (x(–3)) + (–3 x)+ (–3)(–3) ●
x2 – 3x – 3x + 9
x2 – 6x + 9
Write as a product of two binomials.
Use the FOIL method.
Multiply.
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 3c
Multiply.
(2a – b2)(a + 4b2)
(2a – b2)(a + 4b2)
2a(a) + 2a(4b2) – b2(a) + (–b2)(4b2)
2a2 + 8ab2 – ab2 – 4b4
2a2 + 7ab2 – 4b4
Use the FOIL method.
Multiply.
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 +10x –6
10x3 50x2 –30x
30x6x2 –18
5x
+3
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
5x + 36x2 + 30x – 18
+ 10x3 + 50x2 – 30x 10x3 + 56x2 + 0x – 18
10x3 + 56x2 + – 18
Multiply each term in the top polynomial by 3.
Multiply each term in the top polynomial by 5x, and align like terms.
Combine like terms by adding vertically.
Simplify.
Holt Algebra 1
7-7 Multiplying Polynomials
Multiply.
Example 4A: Multiplying Polynomials
(x – 5)(x2 + 4x – 6)
(x – 5 )(x2 + 4x – 6)
x(x2 + 4x – 6) – 5(x2 + 4x – 6)
x(x2) + x(4x) + x(–6) – 5(x2) – 5(4x) – 5(–6)
x3 + 4x2 – 5x2 – 6x – 20x + 30
x3 – x2 – 26x + 30
Distribute x and –5.
Distribute x and −5 again.
Simplify.
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Multiply.
Example 4B: Multiplying Polynomials
(2x – 5)(–4x2 – 10x + 3)
(2x – 5)(–4x2 – 10x + 3)
–4x2 – 10x + 32x – 5x
20x2 + 50x – 15+ –8x3 – 20x2 + 6x
–8x3 + 56x – 15
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.
Holt Algebra 1
7-7 Multiplying Polynomials
Multiply.
Example 4C: Multiplying Polynomials
(x + 3)3
[(x + 3)(x + 3)](x + 3)
[x(x+3) + 3(x+3)](x + 3)
(x2 + 3x + 3x + 9)(x + 3)
(x2 + 6x + 9)(x + 3)
Write as the product of three binomials.
Use the FOIL method on the first two factors.
Multiply.
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 4C: Multiplying Polynomials
Multiply.
(x + 3)3
x3 + 6x2 + 9x + 3x2 + 18x + 27
x3 + 9x2 + 27x + 27
x(x2) + x(6x) + x(9) + 3(x2) + 3(6x) + 3(9)
x(x2 + 6x + 9) + 3(x2 + 6x + 9)
Use the Commutative Property of Multiplication.
Distribute the x and 3.
Distribute the x and 3 again.
(x + 3)(x2 + 6x + 9)
Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 4D: Multiplying Polynomials
Multiply.(3x + 1)(x3 – 4x2 – 7)
x3 4x2 –7
3x4 12x3 –21x
4x2x3 –7
3x
+1
3x4 + 12x3 + x3 + 4x2 – 21x – 7
Write the product of the monomials in each row and column.
Add all terms inside the rectangle.
3x4 + 13x3 + 4x2 – 21x – 7 Combine like terms.
Holt Algebra 1
7-7 Multiplying Polynomials
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.
Helpful Hint
Holt Algebra 1
7-7 Multiplying Polynomials
Check It Out! Example 4a
Multiply.
(x + 3)(x2 – 4x + 6)
(x + 3 )(x2 – 4x + 6)
x(x2 – 4x + 6) + 3(x2 – 4x + 6)
Distribute x and 3.
Distribute x and 3 again.
x(x2) + x(–4x) + x(6) +3(x2) +3(–4x) +3(6)
x3 – 4x2 + 3x2 +6x – 12x + 18
x3 – x2 – 6x + 18
Simplify.
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
Multiply each term in the top polynomial by 2.
Multiply each term in the top polynomial by 3x, and align like terms.2x2 – 4x + 10
+ 3x3 – 6x2 + 15x3x3 – 4x2 + 11x + 10
Combine like terms by adding vertically.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 5: ApplicationThe 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.
Write the formula for the area of a rectangle.
Substitute h – 3 for w and h + 4 for l.
A = l w
A = l w
A = (h + 4)(h – 3)
Multiply.A = h2 + 4h – 3h – 12
Combine like terms.A = h2 + h – 12
The area is represented by h2 + h – 12.
Holt Algebra 1
7-7 Multiplying Polynomials
Example 5: ApplicationThe 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
A = 52 + 5 – 12
A = 25 + 5 – 12
A = 18
Write the formula for the area the base of the prism.
Substitute 5 for h.
Simplify.
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 area of a rectangle.
Substitute x – 4 for l and x for w.
A = l w
A = l w
A = x(x – 4)
Multiply.A = x2 – 4x
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
A = x2 – 4x
A = 36 – 24
A = 12
Write the formula for the area of a rectangle whose length is 4 meters shorter than width .
Substitute 6 for x.
Simplify.
Combine terms.
The area is 12 square meters.
A = 62 – 4 6
Holt Algebra 1
7-7 Multiplying Polynomials
Lesson Quiz: Part I
Multiply.
1. (6s2t2)(3st)
2. 4xy2(x + y)
3. (x + 2)(x – 8)
4. (2x – 7)(x2 + 3x – 4)
5. 6mn(m2 + 10mn – 2)
6. (2x – 5y)(3x + y)
4x2y2 + 4xy3
18s3t3
x2 – 6x – 16
2x3 – x2 – 29x + 28
6m3n + 60m2n2 – 12mn
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.
b. Find the area when the height is 8 cm.
48 cm2
12
h2 + 2h
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Warm UpSimplify.
1. 42
3. (–2)2 4. (x)2
5. –(5y2)
16 49
4 x2
2. 72
6. (m2)2 m4
7. 2(6xy) 2(8x2)8. 16x2
–25y2
12xy
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Find special products of binomials.
Objective
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Vocabularyperfect-square trinomialdifference 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:
(a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2
F L
I
O = a2 + 2ab + b2
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
Multiply.
Example 1: Finding Products in the Form (a + b)2
A. (x +3)2
(a + b)2 = a2 + 2ab + b2
Use the rule for (a + b)2.
(x + 3)2 = x2 + 2(x)(3) + 32
= x2 + 6x + 9
Identify a and b: a = x and b = 3.
Simplify.
B. (4s + 3t)2
(a + b)2 = a2 + 2ab + b2
(4s + 3t)2 = (4s)2 + 2(4s)(3t) + (3t)2
Use the rule for (a + b)2.
= 16s2 + 24st + 9t2
Identify a and b: a = 4s and b = 3t.
Simplify.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Multiply.
Example 1C: Finding Products in the Form (a + b)2
C. (5 + m2)2
(a + b)2 = a2 + 2ab + b2
(5 + m2)2 = 52 + 2(5)(m2) + (m2)2
= 25 + 10m2 + m4
Use the rule for (a + b)2.
Identify a and b: a = 5 and b = m2.
Simplify.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Check It Out! Example 1
Multiply.A. (x + 6)2
(a + b)2 = a2 + 2ab + b2 Identify a and b: a = x and b = 6.
Use the rule for (a + b)2.
(x + 6)2 = x2 + 2(x)(6) + 62
= x2 + 12x + 36 Simplify.
B. (5a + b)2
(a + b)2 = a2 + 2ab + b2
Use the rule for (a + b)2.
(5a + b)2 = (5a)2 + 2(5a)(b) + b2
Identify a and b: a = 5a and b = b.
= 25a2 + 10ab + b2 Simplify.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Check It Out! Example 1C
Multiply.
(1 + c3)2 Use the rule for (a + b)2.
(a + b)2 = a2 + 2ab + b2 Identify a and b: a = 1 and b = c3.
(1 + c3)2 = 12 + 2(1)(c3) + (c3)2
= 1 + 2c3 + c6 Simplify.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
You can use the FOIL method to find products in the form of (a – b)2.
(a – b)2 = (a – b)(a – b) = a2 – ab – ab + b2
F L
IO = 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
Multiply.
Example 2: Finding Products in the Form (a – b)2
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
(a – b)2 = a2 – 2ab + b2
(4m – 10)2 = (4m)2 – 2(4m)(10) + (10)2
= 16m2 – 80m + 100
Use the rule for (a – b)2.
Identify a and b: a = 4m and b = 10.
Simplify.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Multiply.
Example 2: Finding Products in the Form (a – b)2
C. (2x – 5y)2
(a – b)2 = a2 – 2ab + b2
(2x – 5y)2 = (2x)2 – 2(2x)(5y) + (5y)2
= 4x2 – 20xy +25y2
Use the rule for (a – b)2.
Identify a and b: a = 2x and b = 5y.
Simplify.
D. (7 – r3)2
(a – b)2 = a2 – 2ab + b2
(7 – r3)2 = 72 – 2(7)(r3) + (r3)2
= 49 – 14r3 + r6
Use the rule for (a – b)2.
Identify a and b: a = 7 and b = r3.
Simplify.
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
Use the rule for (a – b)2.
Identify a and b: a = x and b = 7.
Simplify.
b. (3b – 2c)2
(a – b)2 = a2 – 2ab + b2
(3b – 2c)2 = (3b)2 – 2(3b)(2c) + (2c)2
= 9b2 – 12bc + 4c2
Use the rule for (a – b)2.
Identify a and b: a = 3b and b = 2c.
Simplify.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Check It Out! Example 2c
Multiply.
(a2 – 4)2
(a – b)2 = a2 – 2ab + b2
(a2 – 4)2 = (a2)2 – 2(a2)(4) + (4)2
= a4 – 8a2 + 16
Use the rule for (a – b)2.
Identify a and b: a = a2 and b = 4.
Simplify.
Holt McDougal Algebra 1
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.
Remove the rectangle on the bottom. Turn it and slide it up next to the top rectangle.
The new arrange- ment is a rectangle with length a + b and width a – b. Its area is (a + b)(a – b).
So (a + b)(a – b) = a2 – b2. A binomial of the form a2 – b2 is called a difference of two squares.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Multiply.
Example 3: Finding Products in the Form (a + b)(a – b)
A. (x + 4)(x – 4)
(a + b)(a – b) = a2 – b2
(x + 4)(x – 4) = x2 – 42
= x2 – 16
Use the rule for (a + b)(a – b).
Identify a and b: a = x and b = 4.
Simplify.
B. (p2 + 8q)(p2 – 8q)
(a + b)(a – b) = a2 – b2
(p2 + 8q)(p2 – 8q) = (p2)2 – (8q)2
= p4 – 64q2
Use the rule for (a + b)(a – b).
Identify a and b: a = p2 and b = 8q.
Simplify.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Multiply.
Example 3: Finding Products in the Form (a + b)(a – b)
C. (10 + b)(10 – b)
(a + b)(a – b) = a2 – b2
(10 + b)(10 – b) = 102 – b2
= 100 – b2
Use the rule for (a + b)(a – b).
Identify a and b: a = 10 and b = b.
Simplify.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Check It Out! Example 3
Multiply.a. (x + 8)(x – 8)
(a + b)(a – b) = a2 – b2
(x + 8)(x – 8) = x2 – 82
= x2 – 64
Use the rule for (a + b)(a – b).
Identify a and b: a = x and b = 8.
Simplify.
b. (3 + 2y2)(3 – 2y2)
(a + b)(a – b) = a2 – b2
(3 + 2y2)(3 – 2y2) = 32 – (2y2)2
= 9 – 4y4
Use the rule for (a + b)(a – b).
Identify a and b: a = 3 and b = 2y2.
Simplify.
Holt McDougal Algebra 1
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
Use the rule for (a + b)(a – b).
Identify a and b: a = 9 and b = r.
Simplify.
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Write a polynomial that represents the area of the yard around the pool shown below.
Example 4: Problem-Solving Application
Holt McDougal Algebra 1
7-9 Special Products of Binomials
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.
1 Understand the Problem
The answer will be an expression that shows the area of the yard less the area of the pool.
Example 4 Continued
Holt McDougal Algebra 1
7-9 Special Products of Binomials
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.
Example 4 Continued
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Solve3
Step 1 Find the total area.
(x +5)2 = x2 + 2(x)(5) + 52 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 and b = 2.
x2 + 10x + 25=
x2 – 4 =
Example 4 Continued
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Step 3 Find the area of the yard around the pool.
Area of yard = total area – area of pool
a = x2 + 10x + 25 (x2 – 4) –
= x2 + 10x + 25 – x2 + 4
= (x2 – x2) + 10x + ( 25 + 4)
= 10x + 29
Identify like terms.
Group like terms together
The area of the yard around the pool is 10x + 29.
Example 4 Continued
Solve3
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Look Back4
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.
Example 4 Continued
Holt McDougal Algebra 1
7-9 Special Products of Binomials
To subtract a polynomial, add the opposite of each term.
Remember!
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
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.
Check It Out! Example 4 Continued
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Solve3
Step 1 Find the area of the upper rectangle.
Step 2 Find the area of the lower square.
(5 + x)(5 – x) = 25 – 5x + 5x – x2 Use the rule for (a + b)
(a – b): a = 5 and b = x.–x2 + 25=
x2 =
= x x
Check It Out! Example 4 Continued
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Step 3 Find the area of the pool.
Area of pool = rectangle area + square area
a x2 +
= –x2 + 25 + x2
= (x2 – x2) + 25
= 25
–x2 + 25=
Identify like terms.
Group like terms together
The area of the pool is 25.
Solve3
Check It Out! Example 4 Continued
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Look Back4
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.
Check It Out! Example 4 Continued
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Lesson Quiz: Part I
Multiply.
1. (x + 7)2
2. (x – 2)2
3. (5x + 2y)2
4. (2x – 9y)2
5. (4x + 5y)(4x – 5y)
6. (m2 + 2n)(m2 – 2n)
x2 – 4x + 4
x2 + 14x + 49
25x2 + 20xy + 4y2
4x2 – 36xy + 81y2
16x2 – 25y2
m4 – 4n2
Holt McDougal Algebra 1
7-9 Special Products of Binomials
Lesson Quiz: Part II
7. Write a polynomial that represents the shaded area of the figure below.
14x – 85
x + 6
x – 6x – 7
x – 7