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7-1 Integer Exponents Warm Up Evaluate each expression for the given values of the variables. 1. x3y2 for x = –1 and y = 10 2. –100 for x = 4 and y = (–7) Write each number as a power of the given base. 3. 64; base 4 43 4. –27; base (–3) Holt Algebra 1 (–3)3 7-1 Integer Exponents Objectives Evaluate expressions containing zero and integer exponents. Simplify expressions containing zero and integer exponents. Holt Algebra 1 7-1 Integer Exponents You have seen positive exponents. Recall that to simplify 32, use 3 as a factor 2 times: 32 = 3 3 = 9. But what does it mean for an exponent to be negative or 0? You can use a table and look for a pattern to figure it out. Power 55 54 53 52 Value 3125 625 125 25 5 Holt Algebra 1 5 5 5 51 5 50 5–1 5–2 7-1 Integer Exponents When the exponent decreases by one, the value of the power is divided by 5. Continue the pattern of dividing by 5. Holt Algebra 1 7-1 Integer Exponents Remember! Base x 4 Exponent Holt Algebra 1 7-1 Integer Exponents Holt Algebra 1 7-1 Integer Exponents Notice the phrase “nonzero number” in the previous table. This is because 00 and 0 raised to a negative power are both undefined. For example, if you use the pattern given above the table with a base of 0 instead of 5, you would get 0º = . Also 0–6 would be = . Since division by 0 is undefined, neither value exists. Holt Algebra 1 7-1 Integer Exponents Reading Math 2–4 is read “2 to the negative fourth power.” Holt Algebra 1 7-1 Integer Exponents Example 1: Application One cup is 2–4 gallons. Simplify this expression. cup is equal to Holt Algebra 1 7-1 Integer Exponents Check It Out! Example 1 A sand fly may have a wingspan up to 5–3 m. Simplify this expression. 5-3 m is equal to Holt Algebra 1 7-1 Integer Exponents Example 2: Zero and Negative Exponents Simplify. A. 4–3 B. 70 7º = 1 C. (–5)–4 D. –5–4 Holt Algebra 1 Any nonzero number raised to the zero power is 1. 7-1 Integer Exponents Caution In (–3)–4, the base is negative because the negative sign is inside the parentheses. In –3–4 the base (3) is positive. Holt Algebra 1 7-1 Integer Exponents Simplify. a. 10–4 Check It Out! Example 2 b. (–2)–4 c. (–2)–5 d. –2–5 Holt Algebra 1 7-1 Integer Exponents Example 3A: Evaluating Expressions with Zero and Negative Exponents Evaluate the expression for the given value of the variables. x–2 for x = 4 Substitute 4 for x. Use the definition Holt Algebra 1 7-1 Integer Exponents Example 3B: Evaluating Expressions with Zero and Negative Exponents Evaluate the expression for the given values of the variables. –2a0b-4 for a = 5 and b = –3 Substitute 5 for a and –3 for b. Evaluate expressions with exponents. Write the power in the denominator as a product. Evaluate the powers in the product. Simplify. Holt Algebra 1 7-1 Integer Exponents Check It Out! Example 3a Evaluate the expression for the given value of the variable. p–3 for p = 4 Substitute 4 for p. Evaluate exponent. Write the power in the denominator as a product. Evaluate the powers in the product. Holt Algebra 1 7-1 Integer Exponents Check It Out! Example 3b Evaluate the expression for the given values of the variables. for a = –2 and b = 6 Substitute –2 for a and 6 for b. Evaluate expressions with exponents. 2 Holt Algebra 1 Write the power in the denominator as a product. Evaluate the powers in the product. Simplify. 7-1 Integer Exponents What if you have an expression with a negative exponent in a denominator, such as ? or Definition of a negative exponent. Substitute –8 for n. Simplify the exponent on the right side. An ifexpression that contains exponent negative or zero So a base with a negative is in a exponents is it not to the be simplified. denominator, is considered equivalent to same base with Expressions be exponent rewritten with positive the opposite should (positive) in theonly numerator. exponents. Holt Algebra 1 7-1 Integer Exponents Example 4: Simplifying Expressions with Zero and Negative Numbers Simplify. A. 7w–4 Holt Algebra 1 B. 7-1 Integer Exponents Example 4: Simplifying Expressions with Zero and Negative Numbers Simplify. C. and Holt Algebra 1 7-1 Integer Exponents Check It Out! Example 4 Simplify. a. 2r0m–3 rº = 1 and b. Holt Algebra 1 c. . 7-1 Integer Exponents Lesson Quiz: Part I 1. A square foot is 3–2 square yards. Simplify this expression. Simplify. 2. 2–6 3. (–7)–3 4. 60 5. –112 Holt Algebra 1 1 –121 7-1 Integer Exponents Lesson Quiz: Part II Evaluate each expression for the given value(s) of the variables(s). 6. x–4 for x =10 7. Holt Algebra 1 for a = 6 and b = 3 7-2 Powers of 10 and Scientific Notation Warm Up Evaluate each expression. 1. 123 1,000 123,000 2. 123 1,000 0.123 3. 0.003 100 0.3 4. 0.003 100 0.00003 5. 104 10,000 6. 10–4 0.0001 7. 230 1 Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Objectives Evaluate and multiply by powers of 10. Convert between standard notation and scientific notation. Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Vocabulary scientific notation Holt Algebra 1 7-2 Powers of 10 and Scientific Notation The table shows relationships between several powers of 10. Each time you divide by 10, the exponent decreases by 1 and the decimal point moves one place to the left. Holt Algebra 1 7-2 Powers of 10 and Scientific Notation The table shows relationships between several powers of 10. Each time you multiply by 10, the exponent increases by 1 and the decimal point moves one place to the right. Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Example 1: Evaluating Powers of 10 Find the value of each power of 10. A. 10–6 B. 104 C. 109 Start with 1 and move the decimal point six places to the left. 0.000001 Start with 1 and move the decimal point four places to the right. 10,000 Start with 1 and move the decimal point nine places to the right. 1,000,000,000 Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Writing Math You may need to add zeros to the right or left of a number in order to move the decimal point in that direction. Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Check It Out! Example 1 Find the value of each power of 10. a. 10–2 b. 105 c. 1010 Start with 1 and move the decimal point two places to the left. 0.01 Start with 1 and move the decimal point five places to the right. 100,000 Start with 1 and move the decimal point ten places to the right. 10,000,000,000 Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Reading Math If you do not see a decimal point in a number, it is understood to be at the end of the number. Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Example 2: Writing Powers of 10 Write each number as a power of 10. A. 1,000,000 B. 0.0001 C. 1,000 The decimal point is six places to the right of 1, so the exponent is 6. The decimal point is four places to the left of 1, so the exponent is –4. The decimal point is three places to the right of 1, so the exponent is 3. Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Check It Out! Example 2 Write each number as a power of 10. a. 100,000,000 b. 0.0001 c. 0.1 The decimal point is eight places to the right of 1, so the exponent is 8. The decimal point is four places to the left of 1, so the exponent is –4. The decimal point is one place to the left of 1, so the exponent is –1. Holt Algebra 1 7-2 Powers of 10 and Scientific Notation You can also move the decimal point to find the value of any number multiplied by a power of 10. You start with the number rather than starting with 1. Multiplying by Powers of 10 Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Example 3: Multiplying by Powers of 10 Find the value of each expression. A. 23.89 108 23.8 9 0 0 0 0 0 0 2,389,000,000 Move the decimal point 8 places to the right. B. 467 10–3 467 0.467 Holt Algebra 1 Move the decimal point 3 places to the left. 7-2 Powers of 10 and Scientific Notation Check It Out! Example 3 Find the value of each expression. a. 853.4 105 853.4 0 0 0 0 85,340,000 Move the decimal point 5 places to the right. b. 0.163 10–2 0.0 0163 0.00163 Holt Algebra 1 Move the decimal point 2 places to the left. 7-2 Powers of 10 and Scientific Notation Scientific notation is a method of writing numbers that are very large or very small. A number written in scientific notation has two parts that are multiplied. The first part is a number that is greater than or equal to 1 and less than 10. The second part is a power of 10. Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Example 4A: Astronomy Application Saturn has a diameter of about Its distance from the Sun is about 1,427,000,000 km. km. Write Saturn’s diameter in standard form. 120000 120,000 km Holt Algebra 1 Move the decimal point 5 places to the right. 7-2 Powers of 10 and Scientific Notation Example 4B: Astronomy Application Saturn has a diameter of about Its distance from the Sun is about 1,427,000,000 km. km. Write Saturn’s distance from the Sun in scientific notation. Count the number of places 1,427,000,000 you need to move the decimal point to get a 1,4 2 7,0 0 0,0 0 0 number between 1 and 10. 9 places 1.427 109 km Holt Algebra 1 Use that number as the exponent of 10. 7-2 Powers of 10 and Scientific Notation Reading Math Standard form refers to the usual way that numbers are written—not in scientific notation. Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Check It Out! Example 4a Use the information above to write Jupiter’s diameter in scientific notation. 143,000 km 143000 5 places 1.43 105 km Holt Algebra 1 Count the number of places you need to move the decimal point to get a number between 1 and 10. Use that number as the exponent of 10. 7-2 Powers of 10 and Scientific Notation Check It Out! Example 4b Use the information above to write Jupiter’s orbital speed in standard form. 13000 13,000 m/s Holt Algebra 1 Move the decimal point 4 places to the right. 7-2 Powers of 10 and Scientific Notation Example 5: Comparing and Ordering Numbers in Scientific Notation Order the list of numbers from least to greatest. Step 1 List the numbers in order by powers of 10. Step 2 Order the numbers that have the same power of 10 Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Check It Out! Example 5 Order the list of numbers from least to greatest. Step 1 List the numbers in order by powers of 10. 2 10-12, 4 10-3, 5.2 10-3, 3 1014, 4.5 1014, 4.5 1030 Step 2 Order the numbers that have the same power of 10 Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Lesson Quiz: Part I Find the value of each expression. 1. 3,745,000 2. 0.00293 3. The Pacific Ocean has an area of about 6.4 х 107 square miles. Its volume is about 170,000,000 cubic miles. a. Write the area of the Pacific Ocean in standard form. b. Write the volume of the Pacific Ocean in scientific notation. 1.7 108 mi3 Holt Algebra 1 7-2 Powers of 10 and Scientific Notation Lesson Quiz: Part II Find the value of each expression. 4. Order the list of numbers from least to greatest Holt Algebra 1 Multiplication Properties 7-3 7-3 Multiplication Properties of Exponents of Exponents Warm Up Lesson Presentation Lesson Quiz Holt Algebra Holt Algebra 11 7-3 Multiplication Properties of Exponents Warm Up Write each expression using an exponent. 1. 2 • 2 • 2 23 2. x • x • x • x 3. Write each expression without using an exponent. 4. 43 4 • 4 • 4 5. y2 y • y 6. m–4 Holt Algebra 1 7-3 Multiplication Properties of Exponents Objective Use multiplication properties of exponents to evaluate and simplify expressions. Holt Algebra 1 7-3 Multiplication Properties of Exponents You have seen that exponential expressions are useful when writing very small or very large numbers. To perform operations on these numbers, you can use properties of exponents. You can also use these properties to simplify your answer. In this lesson, you will learn some properties that will help you simplify exponential expressions containing multiplication. Holt Algebra 1 7-3 Multiplication Properties of Exponents Holt Algebra 1 7-3 Multiplication Properties of Exponents Products of powers with the same base can be found by writing each power as a repeated multiplication. Notice the relationship between the exponents in the factors and the exponents in the product 5 + 2 = 7. Holt Algebra 1 7-3 Multiplication Properties of Exponents Holt Algebra 1 7-3 Multiplication Properties of Exponents Example 1: Finding Products of Powers Simplify. A. Since the powers have the same base, keep the base and add the exponents. B. Group powers with the same base together. Add the exponents of powers with the same base. Holt Algebra 1 7-3 Multiplication Properties of Exponents Example 1: Finding Products of Powers Simplify. C. Group powers with the same base together. Add the exponents of powers with the same base. D. Group the positive exponents and add since they have the same base 1 Holt Algebra 1 Add the like bases. 7-3 Multiplication Properties of Exponents Remember! A number or variable written without an exponent actually has an exponent of 1. 10 = 101 y = y1 Holt Algebra 1 7-3 Multiplication Properties of Exponents Check It Out! Example 1 Simplify. a. Since the powers have the same base, keep the base and add the exponents. b. Group powers with the same base together. Add the exponents of powers with the same base. Holt Algebra 1 7-3 Multiplication Properties of Exponents Check It Out! Example 1 Simplify. c. Group powers with the same base together. Add. Holt Algebra 1 7-3 Multiplication Properties of Exponents Check It Out! Example 1 Simplify. d. Group the first two and second two terms. Divide the first group and add the second group. Multiply. Holt Algebra 1 7-3 Multiplication Properties of Exponents Example 2: Astronomy Application Light from the Sun travels at about miles per second. It takes about 15,000 seconds for the light to reach Neptune. Find the approximate distance from the Sun to Neptune. Write your answer in scientific notation. distance = rate time mi Holt Algebra 1 Write 15,000 in scientific notation. Use the Commutative and Associative Properties to group. Multiply within each group. 7-3 Multiplication Properties of Exponents Check It Out! Example 2 Light travels at about miles per second. Find the approximate distance that light travels in one hour. Write your answer in scientific notation. distance = rate time Write 3,600 in scientific notation. Use the Commutative and Associative Properties to group. Multiply within each group. Holt Algebra 1 7-3 Multiplication Properties of Exponents To find a power of a power, you can use the meaning of exponents. Notice the relationship between the exponents in the original power and the exponent in the final power: Holt Algebra 1 7-3 Multiplication Properties of Exponents Holt Algebra 1 7-3 Multiplication Properties of Exponents Example 3: Finding Powers of Powers Simplify. Use the Power of a Power Property. Simplify. Use the Power of a Power Property. Zero multiplied by any number is zero 1 Holt Algebra 1 Any number raised to the zero power is 1. 7-3 Multiplication Properties of Exponents Example 3: Finding Powers of Powers Simplify. C. Use the Power of a Power Property. Simplify the exponent of the first term. Since the powers have the same base, add the exponents. Write with a positive exponent. Holt Algebra 1 7-3 Multiplication Properties of Exponents Check It Out! Example 3 Simplify. Use the Power of a Power Property. Simplify. Use the Power of a Power Property. Zero multiplied by any number is zero. 1 Holt Algebra 1 Any number raised to the zero power is 1. 7-3 Multiplication Properties of Exponents Check It Out! Example 3c Simplify. c. Use the Power of a Power Property. Simplify the exponents of the two terms. Since the powers have the same base, add the exponents. Holt Algebra 1 7-3 Multiplication Properties of Exponents Powers of products can be found by using the meaning of an exponent. Holt Algebra 1 7-3 Multiplication Properties of Exponents Example 4: Finding Powers of Products Simplify. A. Use the Power of a Product Property. Simplify. B. Use the Power of a Product Property. Simplify. Holt Algebra 1 7-3 Multiplication Properties of Exponents Example 4: Finding Powers of Products Simplify. C. Use the Power of a Product Property. Use the Power of a Product Property. Simplify. Holt Algebra 1 7-3 Multiplication Properties of Exponents Check It Out! Example 4 Simplify. Use the Power of a Product Property. Simplify. Use the Power of a Product Property. Use the Power of a Product Property. Simplify. Holt Algebra 1 7-3 Multiplication Properties of Exponents Check It Out! Example 4 Simplify. c. Use the Power of a Product Property. Use the Power of a Product Property. Combine like terms. Write with a positive exponent. Holt Algebra 1 7-3 Multiplication Properties of Exponents Lesson Quiz: Part I Simplify. 1. 32• 34 2. 3. (x3)2 4. 5. 6. 7. Holt Algebra 1 7-3 Multiplication Properties of Exponents Lesson Quiz: Part II 7. The islands of Samoa have an approximate area of 2.9 103 square kilometers. The area of Texas is about 2.3 102 times as great as that of the islands. What is the approximate area of Texas? Write your answer in scientific notation. Holt Algebra 1 7-4 7-4 Division DivisionProperties PropertiesofofExponents Exponents Warm Up Lesson Presentation Lesson Quiz Holt Algebra Holt Algebra 11 7-4 Division Properties of Exponents Warm Up Simplify. 1. (x2)3 2. 3. 4. 5. 6. Write in Scientific Notation. 7. 8. Holt Algebra 1 7-4 Division Properties of Exponents Objective Use division properties of exponents to evaluate and simplify expressions. Holt Algebra 1 7-4 Division Properties of Exponents A quotient of powers with the same base can be found by writing the powers in a factored form and dividing out common factors. Notice the relationship between the exponents in the original quotient and the exponent in the final answer: 5 – 3 = 2. Holt Algebra 1 7-4 Division Properties of Exponents Holt Algebra 1 7-4 Division Properties of Exponents Example 1: Finding Quotients of Powers Simplify. A. Holt Algebra 1 B. 7-4 Division Properties of Exponents Example 1: Finding Quotients of Powers Simplify. C. Holt Algebra 1 D. 7-4 Division Properties of Exponents Helpful Hint Both Holt Algebra 1 and 729 are considered to be simplified. 7-4 Division Properties of Exponents Check It Out! Example 1 Simplify. a. Holt Algebra 1 b. 7-4 Division Properties of Exponents Check It Out! Example 1 Simplify. c. Holt Algebra 1 d. 7-4 Division Properties of Exponents Example 2: Dividing Numbers in Scientific Notation Simplify and write the answer in scientific notation Write as a product of quotients. Simplify each quotient. Simplify the exponent. Write 0.5 in scientific notation as 5 x 10 . The second two terms have the same base, so add the exponents. Simplify the exponent. Holt Algebra 1 7-4 Division Properties of Exponents Writing Math You can “split up” a quotient of products into a product of quotients: Example: Holt Algebra 1 7-4 Division Properties of Exponents Check It Out! Example 2 Simplify and write the answer in scientific notation. Write as a product of quotients. Simplify each quotient. Simplify the exponent. Write 1.1 in scientific notation as 11 x 10 . The second two terms have the same base, so add the exponents. Simplify the exponent. Holt Algebra 1 7-4 Division Properties of Exponents Example 3: Application The Colorado Department of Education spent about dollars in fiscal year 2004-05 on public schools. There were about students enrolled in public school. What was the average spending per student? Write your answer in standard form. To find the average spending per student, divide the total debt by the number of students. Write as a product of quotients. Holt Algebra 1 7-4 Division Properties of Exponents Example 3 Continued The Colorado Department of Education spent about dollars in fiscal year 2004-05 on public schools. There were about students enrolled in public school. What was the average spending per student? Write your answer in standard form. To find the average spending per student, divide the total debt by the number of students. Simplify each quotient. Simplify the exponent. Write in standard form. The average spending per student is $5,800. Holt Algebra 1 7-4 Division Properties of Exponents Check It Out! Example 3 In 1990, the United States public debt was about dollars. The population of the United States was about people. What was the average debt per person? Write your answer in standard form. To find the average debt per person, divide the total debt by the number of people. Write as a product of quotients. Holt Algebra 1 7-4 Division Properties of Exponents Check It Out! Example 3 Continued In 1990, the United States public debt was about dollars. The population of the United States was about people. What was the average debt per person? Write your answer in standard form. To find the average debt per person, divide the total debt by the number of people. Simplify each quotient. Simplify the exponent. Write in standard form. The average debt per person was $12,800. Holt Algebra 1 7-4 Division Properties of Exponents A power of a quotient can be found by first writing the numerator and denominator as powers. Notice that the exponents in the final answer are the same as the exponent in the original expression. Holt Algebra 1 7-4 Division Properties of Exponents Holt Algebra 1 7-4 Division Properties of Exponents Example 4A: Finding Positive Powers of Quotient Simplify. Use the Power of a Quotient Property. Simplify. Holt Algebra 1 7-4 Division Properties of Exponents Example 4B: Finding Positive Powers of Quotient Simplify. Use the Power of a Product Property. Use the Power of a Product Property: Simplify and use the Power of a Power Property: Holt Algebra 1 7-4 Division Properties of Exponents Example 4C: Finding Positive Powers of Quotient Simplify. Use the Power of a Product Property. Use the Power of a Product Property: Use the Power of a Product Property: Holt Algebra 1 7-4 Division Properties of Exponents Example 4C Continued Simplify. Use the Power of a Product Property: Holt Algebra 1 7-4 Division Properties of Exponents Check It Out! Example 4a Simplify. Use the Power of a Quotient Property. Simplify. Holt Algebra 1 7-4 Division Properties of Exponents Simplify. Holt Algebra 1 Check It Out! Example 4b 7-4 Division Properties of Exponents Simplify. Holt Algebra 1 Check It Out! Example 4c 7-4 Division Properties of Exponents Remember that . What if x is a fraction? Write the fraction as division. Use the Power of a Quotient Property. Multiply by the reciprocal. Simplify. Use the Power of a Quotient Property. Therefore, Holt Algebra 1 7-4 Division Properties of Exponents Holt Algebra 1 7-4 Division Properties of Exponents Example 5A: Finding Negative Powers of Quotients Simplify. Rewrite with a positive exponent. Use the Powers of a Quotient Property . and Holt Algebra 1 7-4 Division Properties of Exponents Example 5B: Finding Negative Powers of Quotients Simplify. Holt Algebra 1 7-4 Division Properties of Exponents Example 5C: Finding Negative Powers of Quotients Simplify. Rewrite each fraction with a positive exponent. Use the Power of a Quotient Property. Use the Power of a Product Property: (3)2 (2n)3 = 32 23n3 and (2)2 (6m)3 = 22 63m3 Holt Algebra 1 7-4 Division Properties of Exponents Example 5C: Finding Negative Powers of Quotients Simplify. Square and cube terms. 1 1 1 2 24 Divide out common factors. 12 Simplify. Holt Algebra 1 7-4 Division Properties of Exponents Helpful Hint Whenever all of the factors in the numerator or the denominator divide out, replace them with 1. Holt Algebra 1 7-4 Division Properties of Exponents Check It Out! Example 5a Simplify. Rewrite with a positive exponent. Use the power of a Quotient Property. 93=729 and 43 = 64. Holt Algebra 1 7-4 Division Properties of Exponents Check It Out! Example 5b Simplify. Rewrite with a positive exponent. Use the Power of a Quotient Property. Use the Power of a Power Property: (b2c3)4= b2•4c3•4 = b8c12 and (2a)4= 24a4= 16a4. Holt Algebra 1 7-4 Division Properties of Exponents Simplify. Check It Out! Example 5c Rewrite each fraction with a positive exponent. Use the Power of a Quotient Property. Use the Power of a Product Property: (3)2= 9. Add exponents and divide out common terms. Holt Algebra 1 7-4 Division Properties of Exponents Lesson Quiz: Part I Simplify. 1. 2. 3. 4. 5. Holt Algebra 1 7-4 Division Properties of Exponents Lesson Quiz: Part II Simplify. 6. Simplify (3 1012) ÷ (5 105) and write the answer in scientific notation. 6 106 7. The Republic of Botswana has an area of 6 105 square kilometers. Its population is about 1.62 106. What is the population density of Botswana? Write your answer in standard form. 2.7 people/km2 Holt Algebra 1 7-5 Polynomials Warm Up Evaluate each expression for the given value of x. 1. 2x + 3; x = 2 7 3. –4x – 2; x = –1 2 2. x2 + 4; x = –3 13 4. 7x2 + 2x = 3 69 Identify the coefficient in each term. 5. 4x3 4 6. y3 1 7. 2n7 2 Holt Algebra 1 8. –54 –1 7-5 Polynomials Objectives Classify polynomials and write polynomials in standard form. Evaluate polynomial expressions. Holt Algebra 1 7-5 Polynomials Vocabulary monomial degree of a monomial polynomial degree of a polynomial standard form of a polynomial leading coefficient Holt Algebra 1 quadratic cubic binomial trinomial 7-5 Polynomials A monomial is a number, a variable, or a product of numbers and variables with whole-number exponents. The degree of a monomial is the sum of the exponents of the variables. A constant has degree 0. Holt Algebra 1 7-5 Polynomials Example 1: Finding the Degree of a Monomial Find the degree of each monomial. A. 4p4q3 The degree is 7. B. 7ed The degree is 2. C. 3 The degree is 0. Holt Algebra 1 Add the exponents of the variables: 4 + 3 = 7. Add the exponents of the variables: 1+ 1 = 2. Add the exponents of the variables: 0 = 0. 7-5 Polynomials Remember! The terms of an expression are the parts being added or subtracted. See Lesson 1-7. Holt Algebra 1 7-5 Polynomials Check It Out! Example 1 Find the degree of each monomial. a. 1.5k2m The degree is 3. b. 4x The degree is 1. b. 2c3 The degree is 3. Holt Algebra 1 Add the exponents of the variables: 2 + 1 = 3. Add the exponents of the variables: 1 = 1. Add the exponents of the variables: 3 = 3. 7-5 Polynomials A polynomial is a monomial or a sum or difference of monomials. The degree of a polynomial is the degree of the term with the greatest degree. Holt Algebra 1 7-5 Polynomials Example 2: Finding the Degree of a Polynomial Find the degree of each polynomial. A. 11x7 + 3x3 11x7: degree 7 3x3: degree 3 The degree of the polynomial is the greatest degree, 7. Find the degree of each term. B. :degree 3 –5: degree 0 :degree 4 Find the degree of each term. The degree of the polynomial is the greatest degree, 4. Holt Algebra 1 7-5 Polynomials Check It Out! Example 2 Find the degree of each polynomial. a. 5x – 6 5x: degree 1 –6: degree 0 The degree of the polynomial is the greatest degree, 1. Find the degree of each term. b. x3y2 + x2y3 – x4 + 2 x3y2: degree 5 –x4: degree 4 x2y3: degree 5 2: degree 0 The degree of the polynomial is the greatest degree, 5. Holt Algebra 1 Find the degree of each term. 7-5 Polynomials The terms of a polynomial may be written in any order. However, polynomials that contain only one variable are usually written in standard form. The standard form of a polynomial that contains one variable is written with the terms in order from greatest degree to least degree. When written in standard form, the coefficient of the first term is called the leading coefficient. Holt Algebra 1 7-5 Polynomials Example 3A: Writing Polynomials in Standard Form Write the polynomial in standard form. Then give the leading coefficient. 6x – 7x5 + 4x2 + 9 Find the degree of each term. Then arrange them in descending order: 6x – 7x5 + 4x2 + 9 Degree 1 5 2 0 –7x5 + 4x2 + 6x + 9 5 2 1 0 The standard form is –7x5 + 4x2 + 6x + 9. The leading coefficient is –7. Holt Algebra 1 7-5 Polynomials Example 3B: Writing Polynomials in Standard Form Write the polynomial in standard form. Then give the leading coefficient. y2 + y6 − 3y Find the degree of each term. Then arrange them in descending order: y2 + y6 – 3y Degree 2 6 1 y6 + y2 – 3y 6 2 1 The standard form is y6 + y2 – 3y. The leading coefficient is 1. Holt Algebra 1 7-5 Polynomials Remember! A variable written without a coefficient has a coefficient of 1. y5 = 1y5 Holt Algebra 1 7-5 Polynomials Check It Out! Example 3a Write the polynomial in standard form. Then give the leading coefficient. 16 – 4x2 + x5 + 9x3 Find the degree of each term. Then arrange them in descending order: 16 – 4x2 + x5 + 9x3 Degree 0 2 5 3 x5 + 9x3 – 4x2 + 16 5 3 2 0 The standard form is x5 + 9x3 – 4x2 + 16. The leading coefficient is 1. Holt Algebra 1 7-5 Polynomials Check It Out! Example 3b Write the polynomial in standard form. Then give the leading coefficient. 18y5 – 3y8 + 14y Find the degree of each term. Then arrange them in descending order: 18y5 – 3y8 + 14y Degree 5 8 1 –3y8 + 18y5 + 14y 8 5 1 The standard form is –3y8 + 18y5 + 14y. The leading coefficient is –3. Holt Algebra 1 7-5 Polynomials Some polynomials have special names based on their degree and the number of terms they have. Degree Name Terms Name 0 Constant 1 Monomial 1 Linear 2 Binomial 2 Quadratic Trinomial 3 4 Cubic Quartic 3 4 or more 5 Quintic 6 or more Holt Algebra 1 6th,7th,degree and so on Polynomial 7-5 Polynomials Example 4: Classifying Polynomials Classify each polynomial according to its degree and number of terms. A. 5n3 + 4n Degree 3 Terms 2 5n3 + 4n is a cubic binomial. B. 4y6 – 5y3 + 2y – 9 Degree 6 Terms 4 4y6 – 5y3 + 2y – 9 is a C. –2x Degree 1 Terms 1 –2x is a linear monomial. Holt Algebra 1 6th-degree polynomial. 7-5 Polynomials Check It Out! Example 4 Classify each polynomial according to its degree and number of terms. a. x3 + x2 – x + 2 Degree 3 Terms 4 x3 + x2 – x + 2 is a cubic polymial. b. 6 Degree 0 Terms 1 6 is a constant monomial. c. –3y8 + 18y5 + 14y Degree 8 Terms 3 –3y8 + 18y5 + 14y is an 8th-degree trinomial. Holt Algebra 1 7-5 Polynomials Example 5: Application A tourist accidentally drops her lip balm off the Golden Gate Bridge. The bridge is 220 feet from the water of the bay. The height of the lip balm is given by the polynomial –16t2 + 220, where t is time in seconds. How far above the water will the lip balm be after 3 seconds? Substitute the time for t to find the lip balm’s height. –16t2 + 220 –16(3)2 + 200 The time is 3 seconds. –16(9) + 200 –144 + 200 76 Holt Algebra 1 Evaluate the polynomial by using the order of operations. 7-5 Polynomials Example 5: Application Continued A tourist accidentally drops her lip balm off the Golden Gate Bridge. The bridge is 220 feet from the water of the bay. The height of the lip balm is given by the polynomial –16t2 + 220, where t is time in seconds. How far above the water will the lip balm be after 3 seconds? After 3 seconds the lip balm will be 76 feet from the water. Holt Algebra 1 7-5 Polynomials Check It Out! Example 5 What if…? Another firework with a 5-second fuse is launched from the same platform at a speed of 400 feet per second. Its height is given by –16t2 +400t + 6. How high will this firework be when it explodes? Substitute the time t to find the firework’s height. –16t2 + 400t + 6 –16(5)2 + 400(5) + 6 The time is 5 seconds. –16(25) + 400(5) + 6 –400 + 2000 + 6 –400 + 2006 1606 Holt Algebra 1 Evaluate the polynomial by using the order of operations. 7-5 Polynomials Check It Out! Example 5 Continued What if…? Another firework with a 5-second fuse is launched from the same platform at a speed of 400 feet per second. Its height is given by –16t2 +400t + 6. How high will this firework be when it explodes? When the firework explodes, it will be 1606 feet above the ground. Holt Algebra 1 7-5 Polynomials Lesson Quiz: Part I Find the degree of each polynomial. 1. 7a3b2 – 2a4 + 4b – 15 2. 25x2 – 3x4 5 4 Write each polynomial in standard form. Then give the leading coefficient. 3. 24g3 + 10 + 7g5 – g2 7g5 + 24g3 – g2 + 10; 7 4. 14 – x4 + 3x2 Holt Algebra 1 –x4 + 3x2 + 14; –1 7-5 Polynomials Lesson Quiz: Part II Classify each polynomial according to its degree and number of terms. 5. 18x2 – 12x + 5 6. 2x4 – 1 quadratic trinomial quartic binomial 7. The polynomial 3.675v + 0.096v2 is used to estimate the stopping distance in feet for a car whose speed is y miles per hour on flat dry pavement. What is the stopping distance for a car traveling at 70 miles per hour? 727.65 ft Holt Algebra 1 7-6 Adding and Subtracting Polynomials Warm Up Simplify each expression by combining like terms. 1. 4x + 2x 6x 2. 3y + 7y 10y 3. 8p – 5p 3p 4. 5n + 6n2 not like terms Simplify each expression. 5. 3(x + 4) 3x + 12 6. –2(t + 3) –2t – 6 7. –1(x2 – 4x – 6) –x2 + 4x + 6 Holt Algebra 1 7-6 Adding and Subtracting Polynomials Objective Add and subtract polynomials. Holt Algebra 1 7-6 Adding and Subtracting Polynomials Just as you can perform operations on numbers, you can perform operations on polynomials. To add or subtract polynomials, combine like terms. Holt Algebra 1 7-6 Adding and Subtracting Polynomials Example 1: Adding and Subtracting Monomials Add or Subtract.. A. 12p3 + 11p2 + 8p3 12p3 + 11p2 + 8p3 12p3 + 8p3 + 11p2 20p3 + 11p2 B. 5x2 – 6 – 3x + 8 5x2 – 6 – 3x + 8 5x2 – 3x + 8 – 6 5x2 – 3x + 2 Holt Algebra 1 Identify like terms. Rearrange terms so that like terms are together. Combine like terms. Identify like terms. Rearrange terms so that like terms are together. Combine like terms. 7-6 Adding and Subtracting Polynomials Example 1: Adding and Subtracting Monomials Add or Subtract.. C. t2 + 2s2 – 4t2 – s2 t2 + 2s2 – 4t2 – s2 t2 – 4t2 + 2s2 – s2 –3t2 + s2 Identify like terms. Rearrange terms so that like terms are together. Combine like terms. D. 10m2n + 4m2n – 8m2n 10m2n + 4m2n – 8m2n Identify like terms. 6m2n Combine like terms. Holt Algebra 1 7-6 Adding and Subtracting Polynomials Remember! Like terms are constants or terms with the same variable(s) raised to the same power(s). To review combining like terms, see lesson 1-7. Holt Algebra 1 7-6 Adding and Subtracting Polynomials Check It Out! Example 1 Add or subtract. a. 2s2 + 3s2 + s 2s2 + 3s2 + s 5s2 + s Identify like terms. Combine like terms. b. 4z4 – 8 + 16z4 + 2 4z4 – 8 + 16z4 + 2 4z4 + 16z4 – 8 + 2 20z4 – 6 Holt Algebra 1 Identify like terms. Rearrange terms so that like terms are together. Combine like terms. 7-6 Adding and Subtracting Polynomials Check It Out! Example 1 Add or subtract. c. 2x8 + 7y8 – x8 – y8 2x8 7y8 x8 y8 + – – 2x8 – x8 + 7y8 – y8 x8 + 6y8 Identify like terms. Rearrange terms so that like terms are together. Combine like terms. d. 9b3c2 + 5b3c2 – 13b3c2 9b3c2 + 5b3c2 – 13b3c2 b3c2 Holt Algebra 1 Identify like terms. Combine like terms. 7-6 Adding and Subtracting Polynomials Polynomials can be added in either vertical or horizontal form. In vertical form, align the like terms and add: 5x2 + 4x + 1 + 2x2 + 5x + 2 7x2 + 9x + 3 In horizontal form, use the Associative and Commutative Properties to regroup and combine like terms. (5x2 + 4x + 1) + (2x2 + 5x + 2) = (5x2 + 2x2 + 1) + (4x + 5x) + (1 + 2) = 7x2 + 9x + 3 Holt Algebra 1 7-6 Adding and Subtracting Polynomials Example 2: Adding Polynomials Add. A. (4m2 + 5) + (m2 – m + 6) (4m2 + 5) + (m2 – m + 6) Identify like terms. (4m2 + m2) + (–m) +(5 + 6) Group like terms together. Combine like terms. 5m2 – m + 11 B. (10xy + x) + (–3xy + y) (10xy + x) + (–3xy + y) Identify like terms. (10xy – 3xy) + x + y Group like terms together. Combine like terms. 7xy + x + y Holt Algebra 1 7-6 Adding and Subtracting Polynomials Example 2C: Adding Polynomials Add. (6x2 – 4y) + (3x2 + 3y – 8x2 – 2y) (6x2 – 4y) + (3x2 + 3y – 8x2 – 2y) Identify like terms. (6x2 + 3x2 – 8x2) + (3y – 4y – 2y) 6x2 – 4y + –5x2 + y x2 – 3y Holt Algebra 1 Group like terms together within each polynomial. Use the vertical method. Combine like terms. Simplify. 7-6 Adding and Subtracting Polynomials Example 2D: Adding Polynomials Add. Identify like terms. Group like terms together. Combine like terms. Holt Algebra 1 7-6 Adding and Subtracting Polynomials Check It Out! Example 2 Add (5a3 + 3a2 – 6a + 12a2) + (7a3 – 10a). (5a3 + 3a2 – 6a + 12a2) + (7a3 – 10a) Identify like terms. (5a3 + 7a3) + (3a2 + 12a2) + (–10a – 6a) Group like terms together. 12a3 + 15a2 – 16a Holt Algebra 1 Combine like terms 7-6 Adding and Subtracting Polynomials To subtract polynomials, remember that subtracting is the same as adding the opposite. To find the opposite of a polynomial, you must write the opposite of each term in the polynomial: –(2x3 – 3x + 7)= –2x3 + 3x – 7 Holt Algebra 1 7-6 Adding and Subtracting Polynomials Example 3A: Subtracting Polynomials Subtract. (x3 + 4y) – (2x3) (x3 + 4y) + (–2x3) Rewrite subtraction as addition of the opposite. Identify like terms. (x3 – 2x3) + 4y Group like terms together. –x3 + 4y Combine like terms. (x3 + 4y) + (–2x3) Holt Algebra 1 7-6 Adding and Subtracting Polynomials Example 3B: Subtracting Polynomials Subtract. (7m4 – 2m2) – (5m4 – 5m2 + 8) (7m4 – 2m2) + (–5m4 + 5m2 – 8) Rewrite subtraction as addition of the opposite. (7m4 – 2m2) + (–5m4 + 5m2 – 8) Identify like terms. (7m4 – 5m4) + (–2m2 + 5m2) – 8 Group like terms together. 2m4 + 3m2 – 8 Holt Algebra 1 Combine like terms. 7-6 Adding and Subtracting Polynomials Example 3C: Subtracting Polynomials Subtract. (–10x2 – 3x + 7) – (x2 – 9) (–10x2 – 3x + 7) + (–x2 + 9) (–10x2 – 3x + 7) + (–x2 + 9) –10x2 – 3x + 7 –x2 + 0x + 9 –11x2 – 3x + 16 Holt Algebra 1 Rewrite subtraction as addition of the opposite. Identify like terms. Use the vertical method. Write 0x as a placeholder. Combine like terms. 7-6 Adding and Subtracting Polynomials Example 3D: Subtracting Polynomials Subtract. (9q2 – 3q) – (q2 – 5) (9q2 – 3q) + (–q2 + 5) (9q2 – 3q) + (–q2 + 5) 9q2 – 3q + 0 + − q2 – 0q + 5 8q2 – 3q + 5 Holt Algebra 1 Rewrite subtraction as addition of the opposite. Identify like terms. Use the vertical method. Write 0 and 0q as placeholders. Combine like terms. 7-6 Adding and Subtracting Polynomials Check It Out! Example 3 Subtract. (2x2 – 3x2 + 1) – (x2 + x + 1) (2x2 – 3x2 + 1) + (–x2 – x – 1) Rewrite subtraction as addition of the opposite. (2x2 – 3x2 + 1) + (–x2 – x – 1) Identify like terms. –x2 + 0x + 1 + –x2 – x – 1 –2x2 – x Holt Algebra 1 Use the vertical method. Write 0x as a placeholder. Combine like terms. 7-6 Adding and Subtracting Polynomials Example 4: Application A farmer must add the areas of two plots of land to determine the amount of seed to plant. The area of plot A can be represented by 3x2 + 7x – 5 and the area of plot B can be represented by 5x2 – 4x + 11. Write a polynomial that represents the total area of both plots of land. (3x2 + 7x – 5) + (5x2 – 4x + 11) 8x2 + 3x + 6 Holt Algebra 1 Plot A. Plot B. Combine like terms. 7-6 Adding and Subtracting Polynomials Check It Out! Example 4 The profits of two different manufacturing plants can be modeled as shown, where x is the number of units produced at each plant. Use the information above to write a polynomial that represents the total profits from both plants. –0.03x2 + 25x – 1500 + –0.02x2 + 21x – 1700 –0.05x2 + 46x – 3200 Holt Algebra 1 Eastern plant profit. Southern plant profit. Combine like terms. 7-6 Adding and Subtracting Polynomials Lesson Quiz: Part I Add or subtract. 1. 7m2 + 3m + 4m2 11m2 + 3m 2. (r2 + s2) – (5r2 + 4s2) (–4r2 – 3s2) 3. (10pq + 3p) + (2pq – 5p + 6pq) 18pq – 2p 4. (14d2 – 8) + (6d2 – 2d +1) 20d2 – 2d – 7 5. (2.5ab + 14b) – (–1.5ab + 4b) Holt Algebra 1 4ab + 10b 7-6 Adding and Subtracting Polynomials Lesson Quiz: Part II 6. A painter must add the areas of two walls to determine the amount of paint needed. The area of the first wall is modeled by 4x2 + 12x + 9, and the area of the second wall is modeled by 36x2 – 12x + 1. Write a polynomial that represents the total area of the two walls. 40x2 + 10 Holt Algebra 1 7-7 Multiplying Polynomials Warm Up Evaluate. 1. 32 2. 24 16 9 3. 102 100 Simplify. 4. 23 24 27 5. y5 y4 y9 6. (53)2 7. (x2)4 56 8. –4(x – 7) Holt Algebra 1 –4x + 28 x8 7-7 Multiplying Polynomials Objective Multiply polynomials. Holt Algebra 1 7-7 Multiplying Polynomials To multiply monomials and polynomials, you will use some of the properties of exponents that you learned earlier in this chapter. Holt Algebra 1 7-7 Multiplying Polynomials Example 1: Multiplying Monomials Multiply. A. (6y3)(3y5) (6y3)(3y5) (6 3)(y3 y5) 18y8 Group factors with like bases together. Multiply. B. (3mn2) (9m2n) (3mn2)(9m2n) (3 9)(m m2)(n2 n) 27m3n3 Holt Algebra 1 Group factors with like bases together. Multiply. 7-7 Multiplying Polynomials Example 1C: Multiplying Monomials Multiply. 1 2 2 2 s t t 12 t s s ÷ 4 1 2g g 12 ÷s s s 4 g Holt Algebra 1 t 2 Group factors with like bases together. gt gt 2 Multiply. 7-7 Multiplying Polynomials Remember! When multiplying powers with the same base, keep the base and add the exponents. x2 x3 = x2+3 = x5 Holt Algebra 1 7-7 Multiplying Polynomials Check It Out! Example 1 Multiply. a. (3x3)(6x2) (3x3)(6x2) (3 6)(x3 x2) 18x5 Group factors with like bases together. Multiply. b. (2r2t)(5t3) (2r2t)(5t3) (2 5)(r2)(t3 t) 10r2t4 Holt Algebra 1 Group factors with like bases together. Multiply. 7-7 Multiplying Polynomials Check It Out! Example 1 Multiply. 1 2 3 2 4 5 x y 12 x z yz c. ÷ 3 1 2 3 2 x y 12 x z ÷ 3 y z 1 g 2 g 3 3 12÷ x x y gy z 4x 5y 5 z 7 Holt Algebra 1 4 5 4 2 g z5 Group factors with like bases together. Multiply. 7-7 Multiplying Polynomials To multiply a polynomial by a monomial, use the Distributive Property. Holt Algebra 1 7-7 Multiplying Polynomials Example 2A: Multiplying a Polynomial by a Monomial Multiply. 4(3x2 + 4x – 8) 4(3x2 + 4x – 8) Distribute 4. (4)3x2 +(4)4x – (4)8 Multiply. 12x2 + 16x – 32 Holt Algebra 1 7-7 Multiplying Polynomials Example 2B: Multiplying a Polynomial by a Monomial Multiply. 6pq(2p – q) (6pq)(2p – q) Distribute 6pq. (6pq)2p + (6pq)(–q) (6 2)(p Group like bases together. p)(q) + (–1)(6)(p)(q q) 12p2q – 6pq2 Holt Algebra 1 Multiply. 7-7 Multiplying Polynomials Example 2C: Multiplying a Polynomial by a Monomial Multiply. 1 2 2 2 x y6xy + 8 x y 2 1 2 2 2 xy x y 6 + 8x y 2 1 2 Distribute x y . 2 1 2 1 2 2 2 Group like bases x y 6 xy + x y 8 x y ÷ 2 ÷ 2 together. 2 1 2 1 • 6 ÷x • x y • y + • 8÷ x • x2 y • y2 2 2 3x3y2 + 4x4y3 Holt Algebra 1 Multiply. 7-7 Multiplying Polynomials Check It Out! Example 2 Multiply. a. 2(4x2 + x + 3) 2(4x2 + x + 3) Distribute 2. 2(4x2) + 2(x) + 2(3) Multiply. 8x2 + 2x + 6 Holt Algebra 1 7-7 Multiplying Polynomials Check It Out! Example 2 Multiply. b. 3ab(5a2 + b) 3ab(5a2 + b) Distribute 3ab. (3ab)(5a2) + (3ab)(b) (3 5)(a a2)(b) + (3)(a)(b b) 15a3b + 3ab2 Holt Algebra 1 Group like bases together. Multiply. 7-7 Multiplying Polynomials Check It Out! Example 2 Multiply. c. 5r2s2(r – 3s) 5r2s2(r – 3s) Distribute 5r2s2. (5r2s2)(r) – (5r2s2)(3s) (5)(r2 r)(s2) – (5 3)(r2)(s2 s) 5r3s2 – 15r2s3 Holt Algebra 1 Group like bases together. Multiply. 7-7 Multiplying Polynomials To multiply a binomial by a binomial, you can apply the Distributive Property more than once: (x + 3)(x + 2) = x(x + 2) + 3(x + 2) Distribute x and 3. = x(x + 2) + 3(x + 2) Distribute x and 3 again. = x(x) + x(2) + 3(x) + 3(2) Multiply. = x2 + 2x + 3x + 6 Combine like terms. = x2 + 5x + 6 Holt Algebra 1 7-7 Multiplying Polynomials Another method for multiplying binomials is called the FOIL method. F 1. Multiply the First terms. (x + 3)(x + 2) x x = x2 O 2. Multiply the Outer terms. (x + 3)(x + 2) I 3. Multiply the Inner terms. (x + 3)(x + 2) L 4. Multiply the Last terms. (x + 3)(x + 2) x 2 = 2x 3 x = 3x 3 2 = 6 (x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6 F Holt Algebra 1 O I L 7-7 Multiplying Polynomials Example 3A: Multiplying Binomials Multiply. (s + 4)(s – 2) (s + 4)(s – 2) s(s – 2) + 4(s – 2) Distribute s and 4. s(s) + s(–2) + 4(s) + 4(–2) s2 – 2s + 4s – 8 Distribute s and 4 again. Multiply. s2 + 2s – 8 Combine like terms. Holt Algebra 1 7-7 Multiplying Polynomials Example 3B: Multiplying Binomials Multiply. (x – 4)2 (x – 4)(x – 4) Write as a product of two binomials. Use the FOIL method. (x x) + (x (–4)) + (–4 x) + (–4 (–4)) x2 – 4x – 4x + 8 Multiply. x2 – 8x + 8 Combine like terms. Holt Algebra 1 7-7 Multiplying Polynomials Example 3C: Multiplying Binomials Multiply. (8m2 – n)(m2 – 3n) Use the FOIL method. 8m2(m2) + 8m2(–3n) – n(m2) – n(–3n) 8m4 – 24m2n – m2n + 3n2 Multiply. 8m4 – 25m2n + 3n2 Combine like terms. Holt Algebra 1 7-7 Multiplying Polynomials Helpful Hint In the expression (x + 5)2, the base is (x + 5). (x + 5)2 = (x + 5)(x + 5) Holt Algebra 1 7-7 Multiplying Polynomials Check It Out! Example 3a Multiply. (a + 3)(a – 4) (a + 3)(a – 4) Distribute a and 3. a(a – 4)+3(a – 4) Distribute a and 3 again. a(a) + a(–4) + 3(a) + 3(–4) a2 – 4a + 3a – 12 Multiply. a2 – a – 12 Combine like terms. Holt Algebra 1 7-7 Multiplying Polynomials Check It Out! Example 3b Multiply. (x – 3)2 (x – 3)(x – 3) Write as a product of two binomials. Use the FOIL method. (x ●x) + (x(–3)) + (–3 x)+ (–3)(–3) Holt Algebra 1 x2 – 3x – 3x + 9 Multiply. x2 – 6x + 9 Combine like terms. 7-7 Multiplying Polynomials Check It Out! Example 3c Multiply. (2a – b2)(a + 4b2) (2a – b2)(a + 4b2) Use the FOIL method. 2a(a) + 2a(4b2) – b2(a) + (–b2)(4b2) 2a2 + 8ab2 – ab2 – 4b4 Multiply. 2a2 + 7ab2 – 4b4 Combine like terms. Holt Algebra 1 7-7 Multiplying Polynomials To multiply polynomials with more than two terms, you can use the Distributive Property several times. Multiply (5x + 3) by (2x2 + 10x – 6): (5x + 3)(2x2 + 10x – 6) = 5x(2x2 + 10x – 6) + 3(2x2 + 10x – 6) = 5x(2x2 + 10x – 6) + 3(2x2 + 10x – 6) = 5x(2x2) + 5x(10x) + 5x(–6) + 3(2x2) + 3(10x) + 3(–6) = 10x3 + 50x2 – 30x + 6x2 + 30x – 18 = 10x3 + 56x2 – 18 Holt Algebra 1 7-7 Multiplying Polynomials You can also use a rectangle model to multiply polynomials with more than two terms. This is similar to finding the area of a rectangle with length (2x2 + 10x – 6) and width (5x + 3): 2x2 5x +3 10x3 6x2 +10x –6 50x2 –30x 30x –18 Write the product of the monomials in each row and column: To find the product, add all of the terms inside the rectangle by combining like terms and simplifying if necessary. 10x3 + 6x2 + 50x2 + 30x – 30x – 18 10x3 + 56x2 – 18 Holt Algebra 1 7-7 Multiplying Polynomials Another method that can be used to multiply polynomials with more than two terms is the vertical method. This is similar to methods used to multiply whole numbers. 2x2 + 10x – 6 Multiply each term in the top polynomial by 3. Multiply each term in the top 5x + 3 polynomial by 5x, and align 6x2 + 30x – 18 like terms. + 10x3 + 50x2 – 30x 10x3 + 56x2 + 0x – 18 Combine like terms by adding vertically. 10x3 + 56x2 + – 18 Simplify. Holt Algebra 1 7-7 Multiplying Polynomials Example 4A: Multiplying Polynomials Multiply. (x – 5)(x2 + 4x – 6) (x – 5 )(x2 + 4x – 6) Distribute x and –5. x(x2 + 4x – 6) – 5(x2 + 4x – 6) Distribute x and −5 again. x(x2) + x(4x) + x(–6) – 5(x2) – 5(4x) – 5(–6) x3 + 4x2 – 5x2 – 6x – 20x + 30 Simplify. x3 – x2 – 26x + 30 Combine like terms. Holt Algebra 1 7-7 Multiplying Polynomials Example 4B: Multiplying Polynomials Multiply. (2x – 5)(–4x2 – 10x + 3) (2x – 5)(–4x2 – 10x + 3) –4x2 – 10x + 3 x 2x – 5 20x2 + 50x – 15 + –8x3 – 20x2 + 6x –8x3 + 56x – 15 Holt Algebra 1 Multiply each term in the top polynomial by –5. Multiply each term in the top polynomial by 2x, and align like terms. Combine like terms by adding vertically. 7-7 Multiplying Polynomials Example 4C: Multiplying Polynomials Multiply. (x + 3)3 [(x + 3)(x + 3)](x + 3) Write as the product of three binomials. [x(x+3) + 3(x+3)](x + 3) Use the FOIL method on the first two factors. (x2 + 3x + 3x + 9)(x + 3) Multiply. (x2 + 6x + 9)(x + 3) Holt Algebra 1 Combine like terms. 7-7 Multiplying Polynomials Example 4C: Multiplying Polynomials Multiply. (x + 3)3 (x + 3)(x2 + 6x + 9) Use the Commutative Property of Multiplication. x(x2 + 6x + 9) + 3(x2 + 6x + 9) Distribute the x and 3. x(x2) + x(6x) + x(9) + 3(x2) + 3(6x) + 3(9) Distribute the x and 3 again. x3 + 6x2 + 9x + 3x2 + 18x + 27 Combine like terms. x3 + 9x2 + 27x + 27 Holt Algebra 1 7-7 Multiplying Polynomials Example 4D: Multiplying Polynomials Multiply. (3x + 1)(x3 – 4x2 – 7) x3 4x2 3x 3x4 12x3 +1 x3 4x2 –7 –21x –7 Write the product of the monomials in each row and column. Add all terms inside the rectangle. 3x4 + 12x3 + x3 + 4x2 – 21x – 7 3x4 + 13x3 + 4x2 – 21x – 7 Holt Algebra 1 Combine like terms. 7-7 Multiplying Polynomials Helpful Hint A polynomial with m terms multiplied by a polynomial with n terms has a product that, before simplifying has mn terms. In Example 4A, there are 2 3, or 6 terms before simplifying. Holt Algebra 1 7-7 Multiplying Polynomials Check It Out! Example 4a Multiply. (x + 3)(x2 – 4x + 6) (x + 3 )(x2 – 4x + 6) Distribute x and 3. x(x2 – 4x + 6) + 3(x2 – 4x + 6) Distribute x and 3 again. x(x2) + x(–4x) + x(6) +3(x2) +3(–4x) +3(6) x3 – 4x2 + 3x2 +6x – 12x + 18 Simplify. x3 – x2 – 6x + 18 Combine like terms. Holt Algebra 1 7-7 Multiplying Polynomials Check It Out! Example 4b Multiply. (3x + 2)(x2 – 2x + 5) (3x + 2)(x2 – 2x + 5) x2 – 2x + 5 3x + 2 2x2 – 4x + 10 + 3x3 – 6x2 + 15x 3x3 – 4x2 + 11x + 10 Holt Algebra 1 Multiply each term in the top polynomial by 2. Multiply each term in the top polynomial by 3x, and align like terms. Combine like terms by adding vertically. 7-7 Multiplying Polynomials Example 5: Application The width of a rectangular prism is 3 feet less than the height, and the length of the prism is 4 feet more than the height. a. Write a polynomial that represents the area of the base of the prism. A = lw Write the formula for the area of a rectangle. A = lw Substitute h – 3 for w A = (h + 4)(h – 3) and h + 4 for l. A = h2 + 4h – 3h – 12 Multiply. A = h2 + h – 12 Combine like terms. The area is represented by h2 + h – 12. Holt Algebra 1 7-7 Multiplying Polynomials Example 5: Application The width of a rectangular prism is 3 feet less than the height, and the length of the prism is 4 feet more than the height. b. Find the area of the base when the height is 5 ft. A = h2 + h – 12 A = h2 + h – 12 Write the formula for the area the base of the prism. A = 52 + 5 – 12 Substitute 5 for h. A = 25 + 5 – 12 Simplify. A = 18 Combine terms. The area is 18 square feet. Holt Algebra 1 7-7 Multiplying Polynomials Check It Out! Example 5 The length of a rectangle is 4 meters shorter than its width. a. Write a polynomial that represents the area of the rectangle. Write the formula for the A = lw area of a rectangle. A = lw A = x(x – 4) A = x2 – 4x Substitute x – 4 for l and x for w. Multiply. The area is represented by x2 – 4x. Holt Algebra 1 7-7 Multiplying Polynomials Check It Out! Example 5 The length of a rectangle is 4 meters shorter than its width. b. Find the area of a rectangle when the width is 6 meters. A = x2 – 4x Write the formula for the area of a rectangle whose length is 4 A = x2 – 4x meters shorter than width . Substitute 6 for x. A = 62 – 4 6 A = 36 – 24 Simplify. A = 12 Combine terms. The area is 12 square meters. Holt Algebra 1 7-7 Multiplying Polynomials Lesson Quiz: Part I Multiply. 1. (6s2t2)(3st) 18s3t3 2. 4xy2(x + y) 4x2y2 + 4xy3 3. (x + 2)(x – 8) x2 – 6x – 16 4. (2x – 7)(x2 + 3x – 4) 2x3 – x2 – 29x + 28 5. 6mn(m2 + 10mn – 2) 6m3n + 60m2n2 – 12mn 6. (2x – 5y)(3x + y) 6x2 – 13xy – 5y2 Holt Algebra 1 7-7 Multiplying Polynomials Lesson Quiz: Part II 7. A triangle has a base that is 4cm longer than its height. a. Write a polynomial that represents the area of the triangle. 1 2 h + 2h 2 b. Find the area when the height is 8 cm. 48 cm2 Holt Algebra 1 7-9 Special Products of Binomials Warm Up Simplify. 1. 42 16 3. (–2)2 4 5. –(5y2) –25y2 4. (x)2 x2 7. 2(6xy) 12xy 8. 2(8x2) 16x2 Holt McDougal Algebra 1 2. 72 49 6. (m2)2 m4 7-9 Special Products of Binomials Objective Find special products of binomials. Holt McDougal Algebra 1 7-9 Special Products of Binomials Vocabulary perfect-square trinomial difference of two squares Holt McDougal Algebra 1 7-9 Special Products of Binomials Imagine a square with sides of length (a + b): The area of this square is (a + b)(a + b) or (a + b)2. The area of this square can also be found by adding the areas of the smaller squares and the rectangles inside. The sum of the areas inside is a2 + ab + ab + b2. Holt McDougal Algebra 1 7-9 Special Products of Binomials This means that (a + b)2 = a2+ 2ab + b2. You can use the FOIL method to verify this: F L (a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2 I = a2 + 2ab + b2 O A trinomial of the form a2 + 2ab + b2 is called a perfect-square trinomial. A perfect-square trinomial is a trinomial that is the result of squaring a binomial. Holt McDougal Algebra 1 7-9 Special Products of Binomials Example 1: Finding Products in the Form (a + b)2 Multiply. A. (x +3)2 (a + b)2 = a2 + 2ab + b2 (x + 3)2 = x2 + 2(x)(3) + 32 = x2 + 6x + 9 Use the rule for (a + b)2. Identify a and b: a = x and b = 3. Simplify. B. (4s + 3t)2 (a + b)2 = a2 + 2ab + b2 Use the rule for (a + b)2. Identify a and b: a = 4s and b = 3t. 2 2 2 (4s + 3t) = (4s) + 2(4s)(3t) + (3t) = 16s2 + 24st + 9t2 Holt McDougal Algebra 1 Simplify. 7-9 Special Products of Binomials Example 1C: Finding Products in the Form (a + b)2 Multiply. C. (5 + m2)2 (a + b)2 = a2 + 2ab + b2 Use the rule for (a + b)2. Identify a and b: a = 5 and b = m2. (5 + m2)2 = 52 + 2(5)(m2) + (m2)2 = 25 + 10m2 + m4 Holt McDougal Algebra 1 Simplify. 7-9 Special Products of Binomials Check It Out! Example 1 Multiply. A. (x + 6)2 (a + b)2 = a2 + 2ab + b2 (x + 6)2 = x2 + 2(x)(6) + 62 = x2 + 12x + 36 B. (5a + b)2 Use the rule for (a + b)2. Identify a and b: a = x and b = 6. Simplify. Use the rule for (a + b)2. (a + b)2 = a2 + 2ab + b2 Identify a and b: a = 5a and b = b. (5a + b)2 = (5a)2 + 2(5a)(b) + b2 = 25a2 + 10ab + b2 Holt McDougal Algebra 1 Simplify. 7-9 Special Products of Binomials Check It Out! Example 1C Multiply. (1 + c3)2 (a + b)2 = a2 + 2ab + b2 Use the rule for (a + b)2. Identify a and b: a = 1 and b = c3. (1 + c3)2 = 12 + 2(1)(c3) + (c3)2 = 1 + 2c3 + c6 Holt McDougal Algebra 1 Simplify. 7-9 Special Products of Binomials You can use the FOIL method to find products in the form of (a – b)2. F L (a – b)2 = (a – b)(a – b) = a2 – ab – ab + b2 I O = a2 – 2ab + b2 A trinomial of the form a2 – ab + b2 is also a perfect-square trinomial because it is the result of squaring the binomial (a – b). Holt McDougal Algebra 1 7-9 Special Products of Binomials Example 2: Finding Products in the Form (a – b)2 Multiply. A. (x – 6)2 (a – b)2 = a2 – 2ab + b2 (x – 6)2 = x2 – 2x(6) + (6)2 = x2 – 12x + 36 Use the rule for (a – b)2. Identify a and b: a = x and b = 6. Simplify. B. (4m – 10)2 Use the rule for (a – b)2. Identify a and b: a = 4m (a – b)2 = a2 – 2ab + b2 and b = 10. (4m – 10)2 = (4m)2 – 2(4m)(10) + (10)2 = 16m2 – 80m + 100 Holt McDougal Algebra 1 Simplify. 7-9 Special Products of Binomials Example 2: Finding Products in the Form (a – b)2 Multiply. C. (2x – 5y)2 Use the rule for (a – b)2. (a – b)2 = a2 – 2ab + b2 Identify a and b: a = 2x and b = 5y. 2 2 2 (2x – 5y) = (2x) – 2(2x)(5y) + (5y) = 4x2 – 20xy +25y2 D. (7 – r3)2 (a – b)2 = a2 – 2ab + b2 Simplify. Use the rule for (a – b)2. Identify a and b: a = 7 and 3. b = r 2 (7 – r3)2 = 72 – 2(7)(r3) + (r3) Simplify. = 49 – 14r3 + r6 Holt McDougal Algebra 1 7-9 Special Products of Binomials Check It Out! Example 2 Multiply. a. (x – 7)2 (a – b)2 = a2 – 2ab + b2 (x – 7)2 = x2 – 2(x)(7) + (7)2 = x2 – 14x + 49 b. (3b – 2c)2 Use the rule for (a – b)2. Identify a and b: a = x and b = 7. Simplify. Use the rule for (a – b)2. (a – b)2 = a2 – 2ab + b2 Identify a and b: a = 3b and b = 2c. 2 2 2 (3b – 2c) = (3b) – 2(3b)(2c) + (2c) = 9b2 – 12bc + 4c2 Holt McDougal Algebra 1 Simplify. 7-9 Special Products of Binomials Check It Out! Example 2c Multiply. (a2 – 4)2 Use the rule for (a – b)2. (a – b)2 = a2 – 2ab + b2 Identify a and b: a = a2 (a2 – 4)2 = (a2)2 – 2(a2)(4) + (4)2 and b = 4. = a4 – 8a2 + 16 Holt McDougal Algebra 1 Simplify. 7-9 Special Products of Binomials You can use an area model to see that (a + b)(a–b)= a2 – b2. Begin with a square with area a2. Remove a square with area b2. The area of the new figure is a2 – b2. So (a + b)(a – b) a2 – b2 is called a Holt McDougal Algebra 1 Remove the rectangle on the bottom. Turn it and slide it up next to the top rectangle. The new arrangement is a rectangle with length a + b and width a – b. Its area is (a + b)(a – b). = a2 – b2. A binomial of the form difference of two squares. 7-9 Special Products of Binomials Example 3: Finding Products in the Form (a + b)(a – b) Multiply. A. (x + 4)(x – 4) (a + b)(a – b) = a2 – b2 (x + 4)(x – 4) = x2 – 42 = x2 – 16 B. (p2 + 8q)(p2 – 8q) Use the rule for (a + b)(a – b). Identify a and b: a = x and b = 4. Simplify. Use the rule for (a + b)(a – b). (a + b)(a – b) = a2 – b2 (p2 + 8q)(p2 – 8q) = (p2)2 – (8q)2 Identify a and b: a = p2 and b = 8q. = p4 – 64q2 Simplify. Holt McDougal Algebra 1 7-9 Special Products of Binomials Example 3: Finding Products in the Form (a + b)(a – b) Multiply. C. (10 + b)(10 – b) Use the rule for (a + b)(a – b). (a + b)(a – b) = a2 – b2 (10 + b)(10 – b) = 102 – b2 = 100 – b2 Holt McDougal Algebra 1 Identify a and b: a = 10 and b = b. Simplify. 7-9 Special Products of Binomials Check It Out! Example 3 Multiply. a. (x + 8)(x – 8) Use the rule for (a + b)(a – b). (a + b)(a – b) = a2 – b2 (x + 8)(x – 8) = x2 – 82 = x2 – 64 b. (3 + 2y2)(3 – 2y2) (a + b)(a – b) = a2 – b2 (3 + 2y2)(3 – 2y2) = 32 – (2y2)2 = 9 – 4y4 Holt McDougal Algebra 1 Identify a and b: a = x and b = 8. Simplify. Use the rule for (a + b)(a – b). Identify a and b: a = 3 and b = 2y2. Simplify. 7-9 Special Products of Binomials Check It Out! Example 3 Multiply. c. (9 + r)(9 – r) (a + b)(a – b) = a2 – b2 (9 + r)(9 – r) = 92 – r2 = 81 – r2 Holt McDougal Algebra 1 Use the rule for (a + b)(a – b). Identify a and b: a = 9 and b = r. Simplify. 7-9 Special Products of Binomials Holt McDougal Algebra 1 7-9 Special Products of Binomials Example 4: Problem-Solving Application Write a polynomial that represents the area of the yard around the pool shown below. Holt McDougal Algebra 1 7-9 Special Products of Binomials Example 4 Continued 1 Understand the Problem The answer will be an expression that shows the area of the yard less the area of the pool. List important information: • The yard is a square with a side length of x + 5. • The pool has side lengths of x + 2 and x – 2. Holt McDougal Algebra 1 7-9 Special Products of Binomials Example 4 Continued 2 Make a Plan The area of the yard is (x + 5)2. The area of the pool is (x + 2) (x – 2). You can subtract the area of the pool from the yard to find the area of the yard surrounding the pool. Holt McDougal Algebra 1 7-9 Special Products of Binomials Example 4 Continued 3 Solve Step 1 Find the total area. (x +5)2 = x2 + 2(x)(5) + 52 = x2 + 10x + 25 Use the rule for (a + b)2: a = x and b = 5. Step 2 Find the area of the pool. (x + 2)(x – 2) = x2 – 2x + 2x – 4 Use the rule for (a + b)(a – b): a = x = x2 – 4 and b = 2. Holt McDougal Algebra 1 7-9 Special Products of Binomials Example 4 Continued 3 Solve Step 3 Find the area of the yard around the pool. Area of yard = a total area – area of pool = x2 + 10x + 25 – (x2 – 4) Identify like = + 10x + 25 – +4 terms. = (x2 – x2) + 10x + ( 25 + 4) Group like terms = 10x + 29 together x2 x2 The area of the yard around the pool is 10x + 29. Holt McDougal Algebra 1 7-9 Special Products of Binomials Example 4 Continued 4 Look Back Suppose that x = 20. Then the total area in the back yard would be 252 or 625. The area of the pool would be 22 18 or 396. The area of the yard around the pool would be 625 – 396 = 229. According to the solution, the area of the yard around the pool is 10x + 29. If x = 20, then 10x +29 = 10(20) + 29 = 229. Holt McDougal Algebra 1 7-9 Special Products of Binomials Remember! To subtract a polynomial, add the opposite of each term. Holt McDougal Algebra 1 7-9 Special Products of Binomials Check It Out! Example 4 Write an expression that represents the area of the swimming pool. Holt McDougal Algebra 1 7-9 Special Products of Binomials Check It Out! Example 4 Continued 1 Understand the Problem The answer will be an expression that shows the area of the two rectangles combined. List important information: • The upper rectangle has side lengths of 5 + x and 5 – x . • The lower rectangle is a square with side length of x. Holt McDougal Algebra 1 7-9 Special Products of Binomials Check It Out! Example 4 Continued 2 Make a Plan The area of the upper rectangle is (5 + x)(5 – x). The area of the lower square is x2. Added together they give the total area of the pool. Holt McDougal Algebra 1 7-9 Special Products of Binomials Check It Out! Example 4 Continued 3 Solve Step 1 Find the area of the upper rectangle. (5 + x)(5 – x) = 25 – 5x + 5x – x2 Use the rule for (a + b) (a – b): a = 5 and b = x. = –x2 + 25 Step 2 Find the area of the lower square. = xx = x2 Holt McDougal Algebra 1 7-9 Special Products of Binomials Check It Out! Example 4 Continued 3 Solve Step 3 Find the area of the pool. Area of pool = rectangle area + square area –x2 + 25 a = = –x2 + 25 + x2 = (x2 – x2) + 25 = 25 The area of the pool is 25. Holt McDougal Algebra 1 + x2 Identify like terms. Group like terms together 7-9 Special Products of Binomials Check It Out! Example 4 Continued 4 Look Back Suppose that x = 2. Then the area of the upper rectangle would be 21. The area of the lower square would be 4. The area of the pool would be 21 + 4 = 25. According to the solution, the area of the pool is 25. Holt McDougal Algebra 1 7-9 Special Products of Binomials Lesson Quiz: Part I Multiply. 1. (x + 7)2 x2 + 14x + 49 2. (x – 2)2 x2 – 4x + 4 3. (5x + 2y)2 25x2 + 20xy + 4y2 4. (2x – 9y)2 4x2 – 36xy + 81y2 5. (4x + 5y)(4x – 5y) 16x2 – 25y2 6. (m2 + 2n)(m2 – 2n) Holt McDougal Algebra 1 m4 – 4n2 7-9 Special Products of Binomials Lesson Quiz: Part II 7. Write a polynomial that represents the shaded area of the figure below. x+6 x–7 x–6 x–7 14x – 85 Holt McDougal Algebra 1