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Welcome to Interactive Chalkboard Algebra 2 Interactive Chalkboard Copyright © by The McGraw-Hill Companies, Inc. Send all inquiries to: GLENCOE DIVISION Glencoe/McGraw-Hill 8787 Orion Place Columbus, Ohio 43240
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Splash Screen
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Contents Lesson 5-1Monomials Lesson 5-2Polynomials Lesson 5-3Dividing Polynomials Lesson 5-4Factoring Polynomials Lesson 5-5Roots of Real Numbers Lesson 5-6Radical Expressions Lesson 5-7Rational Exponents Lesson 5-8Radical Equations and Inequalities Lesson 5-9Complex Numbers
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Lesson 1 Contents Example 1Simplify Expressions with Multiplication Example 2Simplify Expressions with Division Example 3Simplify Expressions with Powers Example 4Simplify Expressions Using Several Properties Example 5Express Numbers in Scientific Notation Example 6Multiply Numbers in Scientific Notation Example 7Divide Numbers in Scientific Notation
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Multiplying Monomials: When multiplying monomials you must ADD exponents. Example: 2x 3 3x 5 2x x x 3x x x x x 6x 8
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Example 1-1a Commutative Property Answer:Definition of exponents
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Example 1-1b Answer:
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Try These Multiply the following monomials. 1.a 2 a 6 3. (-3b 3 c)(7b 2 c 2 ) 2.3x 2 7x 4 4. 2x 2 (6y 3 )(2x 2 y)
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Try These Multiply the following monomials. 1.a 2 a 6 3. (-3b 3 c)(4b 2 c 2 ) 2.3x 2 7x 4 4. 2x 2 (6y 3 )(2x 2 y) a8a8 21x 6 -12b 5 c 3 24x 4 y 4
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Dividing Monomials: When dividing monomials you must SUBTRACT exponents. Example: Cancel x’s 3xx 3x 2
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Example 1-2a Subtract exponents. Remember that a simplified expression cannot contain negative exponents. Answer:Simplify. 1 1 11
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Example 1-2b Answer:
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Try These Divide the following monomials. 1. 3. 2. 4.
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Try These Divide the following monomials. 1. 3. 2. 4. an -y 3 z 2
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Power to a Power: When raising a power to a power you must MULTIPLY exponents. Example: (x 3 ) 5 This means 5 groups of (x 3 ). (x 3 ) (x 3 ) (x 3 ) (x 3 ) (x 3 ) (xxx)(xxx)(xxx)(xxx)(xxx) x 15
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Product to a Power: When raising a product to a power you raise every number/variable to that power. Example: (2x 2 y 3 ) 6 (2x 2 y 3 ) (2x 2 y 3 ) (2x 2 y 3 ) (2x 2 y 3 ) (2x 2 y 3 ) (2x 2 y 3 ) which can be written as: (2xxyyy) (2xxyyy) (2xxyyy) (2xxyyy) (2xxyyy) (2xxyyy) 64x 12 y 18
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Quotient to a Power When raising a quotient to a power you raise the numerator & denominator to that power.
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Example 1-3a Power of a power Answer:
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Example 1-3b Power of a power Answer:
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Example 1-3c Power of a product Power of a quotient Answer:
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Example 1-3d Negative exponent Power of a quotient Answer:
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Simplify each expression. a. b. c. d. Example 1-3e Answer:
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Try These Simplify each monomial. 1. (n 4 ) 4 3. (-2r 2 s) 3 (3rs 2 ) 2. (2x) 4 4.
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Try These Simplify each monomial. 1. (n 4 ) 4 3. (-2r 2 s) 3 (3rs 2 ) 2. (2x) 4 4. n 16 16x 4 -24r 7 n 5
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Negative Exponents To make a negative exponent positive, move the number/variable that is being raised to that exponent from the numerator to the denominator or vice versa. Example: x -3
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Try These Simplify each monomial. 1. 3. (a 3 b 3 )(ab) -2 2. 4.
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Try These Simplify each monomial. 1. 3. (a 3 b 3 )(ab) -2 2. 4.
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Example 1-4a Method 1 Raise the numerator and the denominator to the fifth power before simplifying.
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Example 1-4a Answer:
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Example 1-4b Method 2 Simplify the fraction before raising to the fifth power. Answer:
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Example 1-4c Answer:
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Example 1-5a Express 4, 560, 000 in scientific notation. 4,560,000 Answer: Write 1, 000, 000 as a power of 10.
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Example 1-5b Express 0.000092 in scientific notation. Use a negative exponent. Answer:
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Express each number in scientific notation. a. 52, 000 b. 0.00012 Example 1-5c Answer:
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Express the result in scientific notation. Example 1-6a Associative and Commutative Properties Answer:
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Associative and Commutative Properties Example 1-6b Express the result in scientific notation. Answer:
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Evaluate. Express the result in scientific notation. a. b. Example 1-6c Answer:
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Example 1-7a Divide the number of red blood cells in the sample by the number of red blood cells in 1 milliliter of blood. Answer: There are about 1.66 milliliters of blood in the sample. Biology There are about red blood cells in one milliliter of blood. A certain blood sample contains red blood cells. About how many milliliters of blood are in the sample? number of red blood cells in sample number of red blood cells in 1 milliliter
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Example 1-7b Answer: Biology A petri dish started withgerms in it. A half hour later, there are How many times as great is the amount a half hour later?
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End of Lesson 1
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Assignment: Page 226 #26, 32, 36, 40
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Lesson 2 Contents Example 1Degree of a Polynomial Example 2Subtract and Simplify Example 3Multiply and Simplify Example 4Multiply Two Binomials Example 5Multiply Polynomials
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Polynomials Polynomial: The sum of terms such as 5x, 3x 2, 4xy, 5 Polynomial Terms have variable and whole number exponents. There are no square roots of exponents, no fractional powers, and no variables in the denominators.
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Polynomials 6x -2 Not a polynomial term Has a negative exponent 1/x 2 Not a polynomial term Has variable in the denominator. Not a polynomial term Has variable in radical. 4x 2 Is a polynomial term
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Example 2-1a Determine whether is a polynomial. If it is a polynomial, state the degree of the polynomial. Answer: This expression is not a polynomial because is not a monomial.
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Example 2-1b Answer: This expression is a polynomial because each term is a monomial. The degree of the first term is 5 and the degree of the second term is 2 + 7 or 9. The degree of the polynomial is 9. Determine whether is a polynomial. If it is a polynomial, state the degree of the polynomial.
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Determine whether each expression is a polynomial. If it is a polynomial, state the degree of the polynomial. a. b. Example 2-1c Answer: yes, 5 Answer: no
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Adding Polynomials Add: (2x 2 - 4) + (x 2 + 3x - 3) 1.Remove parentheses. 2.Identify like terms. 3.Add the like terms. (2x 2 - 4) + (x 2 + 3x - 3) = 2x 2 - 4 + x 2 + 3x - 3 = 3x 2 + 3x - 7
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Subtracting Polynomials Subtract: (2x 2 - 4) - (x 2 + 3x - 3) 1. Remove parentheses. 2.Change the signs of ALL of the terms being subtracted. 3.Change the subtraction sign to addition. 4.Follow the rules for adding signed numbers. (2x 2 - 4) - (x 2 + 3x - 3) (Change the signs of terms being subtracted) = (2x 2 - 4) + (-x 2 - 3x + 3) = 2x 2 - 4 + -x 2 - 3x + 3 = x 2 - 3x - 1
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Example 2-2a Simplify Distribute the –1. Group like terms. Combine like terms.Answer:
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Example 2-2b Simplify Answer:
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Try These Add or subtract as indicated. 1.(3x 2 – x – 2) + (x 2 + 4x – 9) 2.(5y + 3y 2 ) + (– 8y – 6y 2 ) 3.(9r 2 + 6r + 16) – (8r 2 + 7r + 10) 4. (10x 2 – 3xy + 4y 2 ) – (3x 2 + 5xy)
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Try These Add or subtract as indicated. 1.(3x 2 – x – 2) + (x 2 + 4x – 9) 2.(5y + 3y 2 ) + (– 8y – 6y 2 ) 3.(9r 2 + 6r + 16) – (8r 2 + 7r + 10) 4. (10x 2 – 3xy + 4y 2 ) – (3x 2 + 5xy) 4x 2 + 3x - 11 -3y – 3y 2 r 2 – r + 6 7x 2 -8xy + 4y 2
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Multiplying Polynomials Simply multiply each term from the first polynomial by each term of the second polynomial. Example: (x + 3)(x² + 2x + 4) = x³ + 2x² + 4x + 3x² + 6x + 12 = x³ + 2x² + 3x² + 4x + 6x + 12 = x³ + 5x² + 10x + 12
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Example 2-3a Distributive Property Answer: Multiply the monomials.
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Example 2-3b Answer:
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Example 2-4a Outer termsInner termsLast termsFirst terms Answer: Multiply monomials and add like terms. +++
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Example 2-4b Answer:
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Example 2-5a Distributive Property Multiply monomials. Answer: Combine like terms.
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Example 2-5b Answer:
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Try These Multiply the polynomials. 1.4b(cb – zd) 2.2xy(3xy 3 – 4xy + 2y 4 ) 3.(3x + 8)(2x + 6) 4. (x – 3y) 2 5. (x 2 + xy + y 2 )(x – y)
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Try These Multiply the polynomials. 1.4b(cb – zd) 2.2xy(3xy 3 – 4xy + 2y 4 ) 3.(3x + 8)(2x + 6) 4. (x – 3y) 2 5. (x 2 + xy + y 2 )(x – y) 4b 2 c – 4bdz 6x 2 y 4 – 8x 2 y 2 + 4xy 5 6x 2 + 34x + 48 x 2 – 6xy + 9y 2 x 3 – y 3
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Assignment: Page 231-232 #25, 26, 30, 42
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End of Lesson 2
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Lesson 4 Contents Example 1GCF Example 2Grouping Example 3Two or Three Terms Example 4Quotient of Two Trinomials
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Factoring Lesson #1 Greatest Common Factor Polynomials in the form x 2 + bx + c
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Greatest Common Factor The first thing you should always do when factoring is to take out a common factor. This is the simplest technique of factoring, but it is important even when you learn fancier techniques, because you will make your later work much easier if you always look for common factors first. Taking out common factors is using the distributive property backwards. The distributive property says: a(b+c)=ab+ac The idea behind taking out a common factor is to look for something that all terms have “in common.” Look at thr right side of the above equation. There is a common factor, a.
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Greatest Common Factor A good trick for finding the greatest common factor to factor polynomials is to find the greatest common factor of the numbers and the smaller power of the variable, so here the greatest common factor of the numbers is 4 and the smallest power of x is 3, so we can take out 4x 3 as a common factor. Example: The polynomial: 4x 5 +12x 4 -8x 3 Can be factored into: 4x 3 (x 2 +3x-8)
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Example 1: Factor the polynomial: 2x 2 + 6x 4 by taking out a common factor. Solution: Choose the common factor. 2x 2. 2x 2 (1 + ___ ) 2x 2 (1 + 3x 2 ) Now Check your work: 2x 2 (1 + 3x 2 ) Multiply back together: 2x 2 + 6x 4
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Example 2: Factor the polynomial: 15x 2 y – 10xy 2 by taking out a common factor. Solution: Choose the common factor: 5xy. 5xy (3x – ___ ) 5xy (3x – 2y) Now Check your work: 5xy (3x – 2y) Multiply back together: 15x 2 y – 10xy 2
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Example 3: Factor the polynomial: 16a 3 b 5 – 24a 2 b 4 – 8a 4 b 7 c by taking out a common factor. Solution: Choose the common factor: 8a 2 b 4. 8a 2 b 4 (2ab – ___ – ___ ) 8a 2 b 4 (2ab – 3 – ___ ) 8a 2 b 4 (2ab – 3 – a 2 b 3 c ) Now Check your work: 8a 2 b 4 (2ab – 3 – a 2 b 3 c ) Multiply back together: 16a 3 b 5 – 24a 2 b 4 – 8a 4 b 7 c
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Now Try These: Factor the following polynomials and check your work. a.6x 2 y 3 + 8x 2 y 5 Solution: 2x 2 y 3 (3 + 4y 2 ) b.12a 4 b 2 c 3 – 18ab 2 c 4 + 24a 5 b 3 c 4 Solution: 6ab 2 c 3 (2a 3 – 3c + 4a 4 bc)
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Factoring Polynomials in the form x 2 + bx + c (General Quadratics) Examples of these “General Quadratics” are: a.x 2 + 7x + 10 b. x 2 + 13x - 30 c. x 2 - 8x + 15d. x 2 - 8x - 20
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Rules for Factoring General Quadratics If the constant term is positive: -Choose factors of the constant term whose SUM is the middle term. -Use the same signs – the sign of the middle term. Example: x 2 + 10x + 16 ( x )( x ) Choose factors of 16 whose sum is 10 (8 and 2) ( x 8 )( x 2 ) Use the same signs – sign of middle term (+) ( x + 8 )( x + 2 )
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Rules for Factoring General Quadratics If the constant term is negative: -Choose factors of the constant term whose DIFFERENCE is the middle term. -Use different signs – the larger factor gets the sign of the middle term. Example: x 2 - 2x - 24 ( x )( x ) Choose factors of 24 whose difference is 2 (6 and 4) ( x 6 )( x 4 ) Use different signs – the six gets the sign of middle term (-) ( x - 6 )( x + 4 )
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Now Try These: Factor the following polynomials and check your work. a.x 2 + 7x + 10 b. x 2 + 13x - 30 Answer: (x + 5)(x + 2)Answer: (x + 15)(x – 2) c. x 2 - 8x + 15d. x 2 - 8x – 20 Answer: (x - 5)(x – 3)Answer: (x – 10)(x + 2)
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End of Lesson 4 Part I
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Warm Up – Section 5-4 #1 Factor and check. 1. 10x 2 y + 15x 3 y 2 2. 16a 2 b 4 c 5 + 48a 3 bc 2 – 12ab 4 c 3 3.y 2 + 11y + 24 4.y 2 - 15y + 36 5.y 2 + 7y - 30 6.y 2 - 4y - 45
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Factoring Lesson #2 Polynomials in the form ax 2 + bx + c
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Factoring Polynomials in the form ax 2 + bx + c (Trial and Error) Examples of these “Trial and Error” Quadratics are: a.4x 2 - 8x - 45 b. 12x 2 + 13x - 14 c. 15x 2 - 26x + 7d. 25x 2 +15x + 2
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Rules for Factoring General Quadratics in the form ax 2 + bx + c - List all of the possible factors of the first term and the last term. -Choose the combination that will allow you to get the correct middle term. -Check your work!!! Example: 4x 2 - 24x + 35 4 x 15 x 7 2 x 235 x 1 Choose factors whose combination will give you the middle term (-24x). You may have to try different combinations before finding the one that works. ( 2x )( 2x ) Use the same signs – sign of middle term (-) ( 2x - 7 )( 2x - 5 )
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Now Try These: Factor the following polynomials and check your work. a.2x 2 + 7x + 6 b. 3x 2 + 10x + 3 Answer: (2x + 3)(x + 2)Answer: (3x + 1)(x + 3) c. 15x 2 - 38x + 7d. 10x 2 - 3x – 27 Answer: (5x - 1)(3x – 7)Answer: (5x – 9)(2x + 3)
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End of Lesson 4 Part II
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Warm Up – Section 5-4 #2 Factor using trial and error or the junk method and check your work. 1. 2x 2 + 11x + 14 2. 14y 2 – 19y – 3 3.3a 2 – 22a + 24 Factor. 4. 25r 2 s 4 t + 100rs 2 t 3 5. x 2 – 11x + 24 6. x 2 + 2x – 35
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Factoring Lesson #3 Difference of two perfect squares x 2 - y 2 Factoring by grouping Factoring Completely
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Factoring the difference of two perfect squares Examples of these polynomials are: a.4x 2 – 9 b. 16x 2 – 36 c. x 2 – 4d. 25x 2 – 16y 2
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Rules for Factoring the difference of two perfect squares -The square root of the first term becomes the first term of each binomial. -The square root of the second term becomes the second term of each binomial. - Use different signs. Example: x 2 - 64 Since the square root of x 2 is x, x is the first term of each binomial. ( x )( x ) Since the square root of 64 is 8, 8 is the second term of each binomial. ( x 8 )( x 8 ) Use different signs. ( x + 8 )( x - 8 )
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Now Try These: Factor the following polynomials and check your work. a. 4x 2 – 9 b. 16x 2 – 36 Answer: (2x + 3)(2x - 3)Answer: (4x + 6)(4x - 6) c.x 2 – 4 : d. 25x 2 – 16y 2 Answer: (x - 2)(x + 2)Answer: (5x – 4y)(5x + 4y)
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Factoring by grouping Examples of polynomials that are factored by grouping are: a.6x 2 + 3xy + 2xz + yz b. 6x 2 + 2xy – 3xz – yz Note: You will see 4 terms when using the grouping method.
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Rules for Factoring by grouping: - Group terms so that there is a common factor in each group. - Take out the common factor in both groups. -Combine like groups. Example: 10a 2 + 2ab + 5ad + bd I will group the first two terms and the last two terms since both of those groups contain a common factor. Note: I am adding these groups. (10a 2 + 2ab) + (5ad + bd) Take out a common factor. 2a(5a + b) + d(5a + b) Combine like groups: ( 2a + d )( 5a + b )
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Examples of factoring by grouping: Factor the polynomial: a 3 – 4a 2 + 3a – 12 Group: (a 3 – 4a 2 )+ (3a – 12) Factor: a 2 (a – 4) + 3(a – 4) Combine: (a 2 + 3)(a – 4) Factor the polynomial: 7ac 2 + 2bc 2 – 7ad 2 – 2bd 2 Group: (7ac 2 + 2bc 2 ) + (– 7ad 2 – 2bd 2 ) Factor: c 2 (7a + 2b) + d 2 (– 7a – 2b) Factor a negative out of second group: c 2 (7a + 2b) - d 2 ( 7a + 2b) Now groups match – so, Combine: (c 2 - d 2) ( 7a + 2b)
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Now Try These: Factor the following polynomials and check your work. a. 6x 2 + 3xy + 2xz + yz Answer: (3x + z)(2x + y) b.6x 2 + 2xy – 3xz – yz Answer: (2x - z)(3x + y)
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Factoring Completely Some polynomials can be factored more than once. This may not be apparent from the beginning. Just as integers can be factored into primes, polynomials can too, and it may take more than one step.
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Rules for Factoring Completely - Factor a polynomial using the appropriate method. - Check each factor to see if you can factor it again. -If so, do it until all polynomials are prime. Example: 3x 2 – 21x + 30 Here, I notice that I have a common factor of 3, so take it out. 3(x 2 – 7x + 10) Now x 2 – 7x + 10 can be factored. 3(x – 5)(x – 2) Now all terms are prime.
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Now Try These: Factor the following polynomials and check your work. a.2x 2 + 12x + 18 b. 3x 2 – 21x – 54 Answer: 2(x + 3)(x + 3)Answer: 3(x + 2)(x - 9) c.5x 2 – 20 : d. 25x 2 – 100y 2 Answer: 5(x - 2)(x + 2)Answer: 25(x – 2y)(x + 2y)
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End of Lesson 4 Part III
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Warm Up Section 5-4 #3 Factor. 1.7c 3 – 28c 2 d + 35cd 3 2.x 2 – 5x – 14 3.x 2 – 15x + 54 4.3x 2 – 22x + 35 5.64x 2 – 81 6. 3r + 3s + 5r 3 s + 5r 2 s 2
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Example 4-1a The GCF is 5ab. Answer: Distributive Property Factor
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Example 4-1b Answer: Factor
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Example 4-2a Group to find the GCF. Factor the GCF of each binomial. Factor Answer: Distributive Property
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Example 4-2b Answer: Factor
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Example 4-3a Rewrite the expression using –5y and 3y in place of –2y and factor by grouping. To find the coefficient of the y terms, you must find two numbers whose product is 3(–5) or –15 and whose sum is –2. The two coefficients must be 3 and –5 since and. Substitute – 5 y + 3 y for – 2 y. Factor
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Example 4-3b Factor out the GCF of each group. Answer: Distributive Property Associative Property
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Example 4-3c Answer: p 2 – 9 is the difference of two squares. Factor out the GCF. Factor
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Example 4-3d This is the sum of two cubes. Answer: Simplify. Sum of two cubes formula with and Factor
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Example 4-3e This polynomial could be considered the difference of two squares or the difference of two cubes. The difference of two squares should always be done before the difference of two cubes. Difference of two squares Answer: Sum and difference of two cubes Factor
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Factor each polynomial. a. b. c. d. Example 4-3f Answer:
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Example 4-4a Factor the numerator and the denominator. Divide. Assume a –5, –2. Answer: Therefore, Simplify
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Example 4-4b Simplify Answer:
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End of Lesson 4
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Lesson 3 Contents Example 1Divide a Polynomial by a Monomial Example 2Division Algorithm Example 3Quotient with Remainder Example 4Synthetic Division Example 5Divisor with First Coefficient Other than 1
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Steps for Dividing a Polynomial by a Monomial 1. Divide each term of the polynomial by the monomial. a) Divide numbers b) Subtract exponents 2. Remember to write the appropriate sign in between the terms. Example: Answer:
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Example 3-1a Sum of quotients Divide. Answer:
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Example 3-1b Answer:
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Try These Divide the following polynomials. 1. 2.
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Try These Divide the following polynomials. 1. 2.
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Example 3-2a Use factoring to find Answer:
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Example 3-2b Answer: x + 2 Use factoring to find
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Try These Divide the following polynomials by factoring. 1. 2. 3. 4.
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Try These Divide the following polynomials by factoring. 1. 2. 3. 4.
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End of Lesson 3
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Warm Up Section 5-5 Covering lessons 5.1-5.4 1.Simplify: 5x 2 y(4x 4 y 3 ) 2.Simplify: (4a 5 b 4 c 2 ) 3 3.Simplify: 4.Multiply: (3x + 7)(x – 4) 5.Factor: x 2 – 11x + 18 6. Factor: 3x 2 + 7x – 20
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Warm Up Section 5-5 Covering lessons 5.1-5.4 1.Simplify: 5x 2 y(4x 4 y 3 ) 20x 6 y 4 2.Simplify: (4a 5 b 4 c 2 ) 3 64a 15 b 12 c 6 3.Simplify: 4.Multiply: (3x + 7)(x – 4) 3x 2 – 5x – 28 5.Factor: x 2 – 11x + 18 (x – 9)(x – 2) 6. Factor: 3x 2 + 7x – 20 (3x – 5)(x + 4)
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Lesson 5 Contents Example 1Find Roots Example 2Simplify Using Absolute Value Example 3Approximate a Square Root
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Simplifying Radicals When working with the simplification of radicals you must remember some basic information about perfect square numbers.
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Perfect Squares 1= 1 x 1 4= 2 x 2 9= 3 x 3 16 = 4 x 4 25 = 5 x 5 36 = 6 x 6 49 = 7 x 7 64 = 8 x 8 81 = 9 x 9 100 = 10 x 10
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Perfect Squares Containing Variables a 2 = a x a a 4 = a 2 x a 2 a 6 = a 3 x a 3 a 8 = a 4 x a 4 a 10 = a 5 x a 5 So, a variable is a “perfect square” if it has an even exponent. To take the square root, just divide the exponent by 2.
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Simplifying Radical Expressions To simplify means to find another expression with the same value. It does not mean to find a decimal approximation. Example: and, although it is equivalent to 5.65, we do not use the decimal value since the radical value is exact and the decimal is an estimate.
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To simplify (or reduce) a radical: 1. Find the largest perfect square which will divide evenly into the number under your radical sign. This means that when you divide, you get no remainders, no decimals, no fractions. 2. Write the number appearing under your radical as the product (multiplication) of the perfect square and your answer from dividing. 3. Give each number in the product its own radical sign. 4. Reduce the "perfect" radical which you have now created.
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Example: Reduce : the largest perfect square that divides evenly into 48 is 16. Find the largest perfect square which will divide evenly into 48. Give each number in the product its own radical sign.
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Example Continued Reduce the "perfect" radical which you have now created.
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Example 5-1a Answer: The square roots of 16x 6 are 4x 3. Simplify
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Example 5-1b Simplify
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Example 5-1c Answer: The fifth root is 3a 2 b 3. Simplify
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Example 5-1d Answer: You cannot take the square root of a negative number. Thus, is not a real number. Simplify
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Example 5-1e Answer: 3x 4 Answer: 2xy 2 Answer: not a real number Answer: Simplify. a. b. c. d.
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Try These
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15Not real #-3 1/40.5z2z2 25g 2 13x 4 y 2 4x - y
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Example 5-2a Note that t is a sixth root of t 6. The index is even, so the principal root is nonnegative. Since t could be negative, you must take the absolute value of t to identify the principal root. Answer: Simplify
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Example 5-2b Since the index is odd, you do not need absolute value. Answer: Simplify
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Example 5-2c Answer: Simplify. a. b. Answer:
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Try These
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-1345 5/130.9z3z3 16x 4 6x 3 y 5 3x+6
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End of Lesson 5
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Assignment: P248 #40, 42, 46, 50
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Lesson 6 Contents Example 1Square Root of a Product Example 2Simplify Quotients Example 3Multiply Radicals Example 4Add and Subtract Radicals Example 5Multiply Radicals Example 6Use a Conjugate to Rationalize a Denominator
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Example 6-1a Factor into squares where possible. Product Property of Radicals Answer: Simplify. Simplify
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Example 6-1b Answer: Simplify
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Example 6-2a Simplify Quotient Property Factor into squares. Product Property
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Example 6-2b Rationalize the denominator. Answer:
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Example 6-2c Quotient Property Rationalize the denominator. Product Property Simplify
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Example 6-2d Multiply. Answer:
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Example 6-2e Answer: Simplify each expression. a. b. Answer:
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Try These
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Warm Up 5-6 Simplify. 1. 2. 3. 4. 5. 6.
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Warm Up 5-6 Simplify. 1. 2. 3. 4. 5. 6.
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Example 6-3a Factor into cubes. Product Property of Radicals Answer:Multiply. Simplify Product Property of Radicals
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Example 6-3b Answer: 24a Simplify
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Try These 1. 2.
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Try These 1. 2.
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Example 6-4a Product Property Multiply. Combine like radicals. Simplify Factor using squares. Answer:
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Example 6-4b Answer: Simplify
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Try These 1. 2.
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Try These 1. 2.
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Example 6-5a FOIL Product Property Answer: Simplify
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Example 6-5b FOIL Multiply. Answer: Subtract. Simplify
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Simplify each expression. a. b. Example 6-5c Answer: 41 Answer:
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Example 6-6a Multiply by since is the conjugate of FOIL Difference of squares Simplify
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Example 6-6b Multiply. Answer: Combine like terms.
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Example 6-6c Answer: Simplify
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End of Lesson 6
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Assignment P 254 #16-46 even
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Lesson 7 Contents Example 1Radical Form Example 2Exponential Form Example 3Evaluate Expressions with Rational Exponents Example 4Rational Exponent with Numerator Other Than 1 Example 5Simplify Expressions with Rational Exponents Example 6Simplify Radical Expressions
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Example 7-1a Write in radical form. Answer: Definition of
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Example 7-1b Write in radical form. Answer: Definition of
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Write each expression in radical form. a. b. Example 7-1c Answer:
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Example 7-2a Write using rational exponents. Definition of Answer:
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Example 7-2b Write using rational exponents. Definition of Answer:
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Write each radical using rational exponents. a. b. Example 7-2c Answer:
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Example 7-3a Evaluate Method 1 Answer: Simplify.
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Example 7-3b Multiply exponents. Method 2 Answer: Power of a Power
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Example 7-3c Answer: The root is 4. Evaluate. Method 1Factor. Power of a Power Expand the square. Find the fifth root.
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Example 7-3d Answer: The root is 4. Power of a Power Multiply exponents. Method 2
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Evaluate each expression. a. b. Example 7-3e Answer: 8 Answer:
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Example 7-4a According to the formula, what is the maximum that U.S. Weightlifter Oscar Chaplin III can lift if he weighs 77 kilograms? Answer: The formula predicts that he can lift at most 372 kg. Weight Lifting The formula can be used to estimate the maximum total mass that a weight lifter of mass B kilograms can lift in two lifts, the snatch and the clean and jerk, combined. Original formula Use a calculator.
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Example 7-4b Oscar Chaplin’s total in the 2000 Olympics was 355 kg. Compare this to the value predicted by the formula. Answer: The formula prediction is somewhat higher than his actual total. Weight Lifting The formula can be used to estimate the maximum total mass that a weight lifter of mass B kilograms can lift in two lifts, the snatch and the clean and jerk, combined.
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Example 7-4c Answer: 380 kg Answer: The formula prediction is slightly higher than his actual total. Weight Lifting Use the formula where M is the maximum total mass that a weight lifter of mass B kilograms can lift. a. According to the formula, what is the maximum that a weight lifter can lift if he weighs 80 kilograms? b. If he actually lifted 379 kg, compare this to the value predicted by the formula.
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Example 7-5a Simplify. Multiply powers. Answer: Add exponents.
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Example 7-5b Simplify. Multiply by
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Example 7-5c Answer:
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Simplify each expression. a. b. Example 7-5d Answer:
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Example 7-6a Simplify. Rational exponents Power of a Power
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Example 7-6b Quotient of Powers Answer:Simplify.
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Example 7-6c Simplify. Rational exponents Power of a Power Answer:Simplify. Multiply.
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Example 7-6d Answer:Multiply. Simplify. is the conjugate of
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Example 7-6e Answer: 1 Simplify each expression. a. b. c. Answer:
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End of Lesson 7
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Lesson 8 Contents Example 1Solve a Radical Equation Example 2Extraneous Solution Example 3Cube Root Equation Example 4Radical Inequality
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Solving Radical Equations 1.Isolate the radical 2.Raise each side to the appropriate power to eliminate the radical. 3.Solve for the variable. Example: Solve for x. 1. Isolate radical by adding 2 to both sides. 2. Square both sides. 3. So, x = 95
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Example 8-1a Solve Original equation Add 1 to each side to isolate the radical. Square each side to eliminate the radical. Find the squares. Add 2 to each side.
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Example 8-1b Check Answer: The solution checks. The solution is 38. Replace y with 38. Original equation Simplify.
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Example 8-1c Answer : 67 Solve
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Try These Solve each equation.
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25144 1-11
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Example 8-2a Solve Original equation Square each side. Find the squares. Isolate the radical. Divide each side by –4.
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Answer: The solution does not check, so there is no real solution. Example 8-2b Check Square each side. Evaluate the squares. Original equation Evaluate the square roots. Replace x with 16. Simplify.
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Example 8-2c Solve. Answer: no real solution
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Example 8-3a Solve In order to remove thepower, or cube root, you must first isolate it and then raise each side of the equation to the third power. Original equation Subtract 5 from each side. Cube each side. Evaluate the cubes.
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Example 8-3b Answer: The solution is –42. Divide each side by 3. Check Original equation Add. Replace y with –42. Simplify. The cube root of –125 is –5. Subtract 1 from each side.
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Example 8-3c Answer: 13 Solve
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Try These Solve each equation.
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495 9-20
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End of Lesson 8
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Assignment P 266 #16, 19, 22, 24
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Lesson 9 Contents Example 1Square Roots of Negative Numbers Example 2Multiply Pure Imaginary Numbers Example 3Simplify a Power of i Example 4Equation with Imaginary Solutions Example 5Equate Complex Numbers Example 6Add and Subtract Complex Numbers Example 7Multiply Complex Numbers Example 8Divide Complex Numbers
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Keep in Mind: The square root of a negative number does not exist. Example: is not 5 or -5 since 5 x 5 = 25 and -5 x -5 = 25. So up until now, we could not simplify.
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i i is defined to have the property that: i 2 = -1 therefore, we could say that square root of -1 is i. This allows us to simplify the square roots of negative numbers such as.
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Examples 1. Simplify: Since is 5 and is i, our answer is 5i. 2.Simplify 6ix 2
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Example 9-1a Simplify. Answer:
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Example 9-1b Simplify. Answer:
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Example 9-1c Simplify. a. b. Answer:
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Example 9-2a Answer: = 6 Simplify.
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Example 9-2b Answer: Simplify.
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Example 9-2c Answer: –15 Answer: Simplify. a. b.
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Example 9-3a Simplify Multiplying powers Power of a Power Answer:
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Example 9-3b Answer: i Simplify.
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Example 9-4a Solve Answer: Original equation Subtract 20 from each side. Divide each side by 5. Take the square root of each side.
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Example 9-4b Solve Answer:
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Find the values of x and y that make the equation true. Example 9-5a Set the real parts equal to each other and the imaginary parts equal to each other. Real parts Divide each side by 2. Imaginary parts Answer:
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Find the values of x and y that make the equation true. Example 9-5b Answer:
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Simplify. Example 9-6a Answer: Commutative and Associative Properties
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Simplify. Example 9-6b Commutative and Associative Properties Answer:
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Simplify. a. b. Example 9-6c Answer:
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Answer: The voltage isvolts. Example 9-7a Electricity In an AC circuit, the voltage E, current I, and impedance Z are related by the formula Find the voltage in a circuit with current 1 + 4 j amps and impedance 3 – 6 j ohms. Electricity formula FOIL Multiply. Add.
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Example 9-7b Electricity In an AC circuit, the voltage E, current I, and impedance Z are related by the formula E = I Z. Find the voltage in a circuit with current 1 – 3 j amps and impedance 3 + 2 j ohms. Answer: 9 – 7 j
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and are conjugates. Example 9-8a Multiply. Answer: Standard form Simplify.
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Example 9-8b Simplify. Multiply. Answer: Standard form Multiply by
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Simplify. a. b. Example 9-8c Answer:
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End of Lesson 9
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Algebra2.com Explore online information about the information introduced in this chapter. Click on the Connect button to launch your browser and go to the Algebra 2 Web site. At this site, you will find extra examples for each lesson in the Student Edition of your textbook. When you finish exploring, exit the browser program to return to this presentation. If you experience difficulty connecting to the Web site, manually launch your Web browser and go to www.algebra2.com/extra_examples.
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