warm up evaluate each expression for the given values of the variables

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


Evaluate expressions containing zero and integer exponents.

Simplify expressions containing zero and integer exponents.

Objectives

Holt Algebra 1


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

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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


Base

x

Exponent

Remember!

4

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Holt Algebra 1


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


2–4 is read “2 to the negative fourth power.”

Reading Math

Holt Algebra 1


Example 1: Application

One cup is 2–4 gallons. Simplify this expression.

cup is equal to

Holt Algebra 1


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

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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

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In (–3)–4, the base is negative because the

negative sign is inside the parentheses. In –3–4

the base (3) is positive.

Caution

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Check It Out! Example 2 Simplify.

a. 10–4

b. (–2)–4

c. (–2)–5

d. –2–5

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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


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


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.


Holt Algebra 1


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.


Simplify.

Holt Algebra 1


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.

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Simplify.

Example 4: Simplifying Expressions with Zero and Negative Numbers

A. 7w–4 B.

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Simplify.

Example 4: Simplifying Expressions with Zero and Negative Numbers

C.

and

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Simplify.

a. 2r0m–3

b.

c.

rº = 1 and .

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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


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

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Evaluate and multiply by powers of 10.Convert between standard notation and scientific notation.

Objectives

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scientific notation

Vocabulary

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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


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


Holt Algebra 1


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


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

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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


If you do not see a decimal point in a number, it is understood to be at the end of the number.

Reading Math

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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



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.

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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

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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




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


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


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


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


Standard form refers to the usual way that numbers are written—not in scientific notation.

Reading Math

Holt Algebra 1



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



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


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


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

2 10-12, 4 10-3, 5.2 10-3, 3 1014, 4.5 1014, 4.5 1030

Holt Algebra 1


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




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


Use multiplication properties of exponents to evaluate and simplify expressions.

Objective

Holt Algebra 1


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


Holt Algebra 1


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


Holt Algebra 1


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


Simplify.

Example 1: Finding Products of Powers

C.

D.

1



Group the positive exponents and add since they have the same base

Add the like bases.

Holt Algebra 1


A number or variable written without an exponent actually has an exponent of 1.

Remember!

10 = 101

y = y1

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a. Simplify.

Since the powers have the same base, keep the base and add the exponents.

b.



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Simplify.

c.


Add.

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Simplify.

d.

Group the first two and second two terms.

Divide the first group and add the second group.

Multiply.

Holt Algebra 1


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



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


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


Holt Algebra 1


Simplify.

Example 3: Finding Powers of Powers

Use the Power of a Power Property.

Simplify.

1


Zero multiplied by any number is zero

Any number raised to the zero power is 1.

Holt Algebra 1


Simplify.

Example 3: Finding Powers of Powers


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



Simplify.


Simplify.

1


Zero multiplied by any number is zero.

Any number raised to the zero power is 1.

Holt Algebra 1


Check It Out! Example 3c

Simplify.


Simplify the exponents of the two terms.

Since the powers have the same base, add the exponents.

c.

Holt Algebra 1


Powers of products can be found by using the meaning of an exponent.

Holt Algebra 1


Example 4: Finding Powers of Products

Simplify.

Use the Power of a Product Property.

Simplify.


Simplify.

A.

B.

Holt Algebra 1


Example 4: Finding Powers of Products

Simplify.



Simplify.

C.

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Simplify.


Simplify.



Simplify.

Holt Algebra 1



Simplify.



Combine like terms.

Write with a positive exponent.

c.

Holt Algebra 1


Lesson Quiz: Part I

Simplify.

1. 32• 34

3.

5.

7.

2.

4.

6.

(x3)2

Holt Algebra 1



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


Use division properties of exponents to evaluate and simplify expressions.

Objective

Holt Algebra 1


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


Holt Algebra 1


Simplify.

Example 1: Finding Quotients of Powers

A. B.

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C.

Simplify.

Example 1: Finding Quotients of Powers

D.

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Both and 729 are considered to be simplified.

Helpful Hint

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a.

Simplify.

b.

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Simplify.

c. d.

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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


You can “split up” a quotient of products into a product of quotients:

Example:

Writing Math

Holt Algebra 1



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.


Holt Algebra 1



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


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.


Write in standard form.

Holt Algebra 1



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


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.


Write in standard form.The average debt per person was $12,800.

Holt Algebra 1


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


Holt Algebra 1


Simplify.

Example 4A: Finding Positive Powers of Quotient

Use the Power of a Quotient Property.

Simplify.

Holt Algebra 1


Simplify.

Example 4B: Finding Positive Powers of Quotient


Use the Power of a Product Property:

Simplify and use the Power of a Power Property:

Holt Algebra 1


Simplify.

Example 4C: Finding Positive Powers of Quotient




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Simplify.

Example 4C Continued


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Simplify.


Simplify.

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Check It Out! Example 4b Simplify.

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Check It Out! Example 4c Simplify.

Holt Algebra 1


Therefore,

Write the fraction as division.


Multiply by the reciprocal.

Simplify.


Remember that What if x is a fraction?.

Holt Algebra 1


Holt Algebra 1


Simplify.

Example 5A: Finding Negative Powers of Quotients

Rewrite with a positive exponent.

and

Use the Powers of a Quotient Property .

Holt Algebra 1


Simplify.

Example 5B: Finding Negative Powers of Quotients

Holt Algebra 1


Simplify.

Example 5C: Finding Negative Powers of Quotients

Rewrite each fraction with a positive exponent.



(3)2 (2n)3 = 32 23n3

and (2)2 (6m)3 = 22 63m3

Holt Algebra 1


1

241

21

12

Divide out common factors.

Simplify.

Example 5C: Finding Negative Powers of Quotients

Simplify.

Square and cube terms.

Holt Algebra 1


Whenever all of the factors in the numerator or the denominator divide out, replace them with 1.

Helpful Hint

Holt Algebra 1



Simplify.

93=729 and 43 = 64.

Use the power of a Quotient Property.


Holt Algebra 1



Simplify.


Use the Power of a Power Property: (b2c3)4= b2•4c3•4 = b8c12 and (2a)4= 24a4= 16a4.


Holt Algebra 1


Check It Out! Example 5c Simplify.

Rewrite each fraction with a positive exponent.


Use the Power of a Product Property: (3)2= 9.

Add exponents and divide out common terms.

Holt Algebra 1


Lesson Quiz: Part I

1.

3. 4.

5.

2.

Simplify.

Holt Algebra 1



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


Find the degree of each monomial.

a. 1.5k2m

The degree is 3. Add the exponents of the variables: 2 + 1 = 3.

b. 4x


b. 2c3


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.



:degree 3 :degree 4

–5: degree 0

Holt Algebra 1

7-5 Polynomials



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



–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


Example 3B: Writing Polynomials in Standard Form


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



16 – 4x2 + x5 + 9x3


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




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



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?


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


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


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



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


Add and subtract polynomials.

Objective

Holt Algebra 1


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


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


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


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



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


Combine like terms.



Holt Algebra 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


Combine like terms.



Holt Algebra 1


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


Add.

Example 2: Adding Polynomials

A. (4m2 + 5) + (m2 – m + 6)

(4m2 + 5) + (m2 – m + 6)

(4m2 + m2) + (–m) +(5 + 6)

5m2 – m + 11


Group like terms together.

Combine like terms.

B. (10xy + x) + (–3xy + y)

(10xy + x) + (–3xy + y)

(10xy – 3xy) + x + y

7xy + x + y



Combine like terms.

Holt Algebra 1


Add.

Example 2C: Adding Polynomials

(6x2 – 4y) + (3x2 + 3y – 8x2 – 2y)


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


Add.

Example 2D: Adding Polynomials



Combine like terms.

Holt Algebra 1



Add (5a3 + 3a2 – 6a + 12a2) + (7a3 – 10a).

(5a3 + 3a2 – 6a + 12a2) + (7a3 – 10a)

(5a3 + 7a3) + (3a2 + 12a2) + (–10a – 6a)

12a3 + 15a2 – 16a



Combine like terms.

Holt Algebra 1


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


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.



Combine like terms.

Holt Algebra 1


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




Combine like terms.

Holt Algebra 1


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



Use the vertical method.Write 0x as a placeholder.Combine like terms.

Holt Algebra 1


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


Identify like terms.Use the vertical method.Write 0 and 0q as

placeholders.

Combine like terms.

Holt Algebra 1



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



Use the vertical method.Write 0x as a placeholder.

Combine like terms.

Holt Algebra 1


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

Plot A.Plot B.

Combine like terms.

+

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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


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



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


Multiply polynomials.

Objective

Holt Algebra 1


To multiply monomials and polynomials, you will use some of the properties of exponents that you learned earlier in this chapter.

Holt Algebra 1


Multiply.

Example 1: Multiplying Monomials

A. (6y3)(3y5)

(6y3)(3y5)

18y8

Group factors with like bases together.

B. (3mn2) (9m2n)

(3mn2)(9m2n)

27m3n3

Multiply.


Multiply.

(6 3)(y3 y5)

(3 9)(m m2)(n2 n)

Holt Algebra 1


Multiply.

Example 1C: Multiplying Monomials

2 2 2112

4s t st st

4 53s t


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


When multiplying powers with the same base, keep the base and add the exponents.

x2 x3 = x2+3 = x5

Remember!

Holt Algebra 1



Multiply.

a. (3x3)(6x2)

(3x3)(6x2)

(3 6)(x3 x2)

18x5


Multiply.


Multiply.

b. (2r2t)(5t3)

(2r2t)(5t3)

(2 5)(r2)(t3 t)

10r2t4

Holt Algebra 1



Multiply.


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


To multiply a polynomial by a monomial, use the Distributive Property.

Holt Algebra 1


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


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


Multiply.

Example 2C: Multiplying a Polynomial by a Monomial


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



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



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.


Multiply.

Holt Algebra 1



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.


Multiply.

Holt Algebra 1


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


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


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


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


Example 3C: Multiplying Binomials

Multiply.

(8m2 – n)(m2 – 3n)

8m2(m2) + 8m2(–3n) – n(m2) – n(–3n)

8m4 – 24m2n – m2n + 3n2

8m4 – 25m2n + 3n2


Multiply.

Combine like terms.

Holt Algebra 1


In the expression (x + 5)2, the base is (x + 5). (x + 5)2 = (x + 5)(x + 5)

Helpful Hint

Holt Algebra 1



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



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.


Multiply.

Combine like terms.

Holt Algebra 1



Multiply.

(2a – b2)(a + 4b2)

(2a – b2)(a + 4b2)

2a(a) + 2a(4b2) – b2(a) + (–b2)(4b2)

2a2 + 8ab2 – ab2 – 4b4

2a2 + 7ab2 – 4b4


Multiply.

Combine like terms.

Holt Algebra 1


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


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


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


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


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.


Holt Algebra 1


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


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


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


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



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



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


Holt Algebra 1


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


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.

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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



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


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. 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



Find special products of binomials.

Objective



Vocabularyperfect-square trinomialdifference of two squares



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.



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.



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


= 16s2 + 24st + 9t2

Identify a and b: a = 4s and b = 3t.

Simplify.



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


Identify a and b: a = 5 and b = m2.

Simplify.




Multiply.A. (x + 6)2

(a + b)2 = a2 + 2ab + b2 Identify a and b: a = x and b = 6.


(x + 6)2 = x2 + 2(x)(6) + 62

= x2 + 12x + 36 Simplify.

B. (5a + b)2

(a + b)2 = a2 + 2ab + b2


(5a + b)2 = (5a)2 + 2(5a)(b) + b2

Identify a and b: a = 5a and b = b.

= 25a2 + 10ab + b2 Simplify.



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.



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).



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.


Simplify.

B. (4m – 10)2

(a – b)2 = a2 – 2ab + b2

(4m – 10)2 = (4m)2 – 2(4m)(10) + (10)2

= 16m2 – 80m + 100


Identify a and b: a = 4m and b = 10.

Simplify.



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


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


Identify a and b: a = 7 and b = r3.

Simplify.




Multiply.

a. (x – 7)2

(a – b)2 = a2 – 2ab + b2

(x – 7)2 = x2 – 2(x)(7) + (7)2

= x2 – 14x + 49



Simplify.

b. (3b – 2c)2

(a – b)2 = a2 – 2ab + b2

(3b – 2c)2 = (3b)2 – 2(3b)(2c) + (2c)2

= 9b2 – 12bc + 4c2


Identify a and b: a = 3b and b = 2c.

Simplify.




Multiply.

(a2 – 4)2

(a – b)2 = a2 – 2ab + b2

(a2 – 4)2 = (a2)2 – 2(a2)(4) + (4)2

= a4 – 8a2 + 16


Identify a and b: a = a2 and b = 4.

Simplify.



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.



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).


Simplify.

B. (p2 + 8q)(p2 – 8q)

(a + b)(a – b) = a2 – b2

(p2 + 8q)(p2 – 8q) = (p2)2 – (8q)2

= p4 – 64q2


Identify a and b: a = p2 and b = 8q.

Simplify.



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


Identify a and b: a = 10 and b = b.

Simplify.




Multiply.a. (x + 8)(x – 8)

(a + b)(a – b) = a2 – b2

(x + 8)(x – 8) = x2 – 82

= x2 – 64



Simplify.

b. (3 + 2y2)(3 – 2y2)

(a + b)(a – b) = a2 – b2

(3 + 2y2)(3 – 2y2) = 32 – (2y2)2

= 9 – 4y4


Identify a and b: a = 3 and b = 2y2.

Simplify.




Multiply.

c. (9 + r)(9 – r)

(a + b)(a – b) = a2 – b2

(9 + r)(9 – r) = 92 – r2

= 81 – r2


Identify a and b: a = 9 and b = r.

Simplify.



Write a polynomial that represents the area of the yard around the pool shown below.

Example 4: Problem-Solving Application



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



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



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



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


Group like terms together

The area of the yard around the pool is 10x + 29.

Example 4 Continued

Solve3



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



To subtract a polynomial, add the opposite of each term.

Remember!




Write an expression that represents the area of the swimming pool.




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.



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.




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




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=


Group like terms together

The area of the pool is 25.

Solve3




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.




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




7. Write a polynomial that represents the shaded area of the figure below.

14x – 85

x + 6

x – 6x – 7

x – 7

warm up evaluate each expression for the given values of the variables

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