Survey
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project
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 Lesson 5-1 Monomials Lesson 5-2 Polynomials Lesson 5-3 Dividing Polynomials Lesson 5-4 Factoring Polynomials Lesson 5-5 Roots of Real Numbers Lesson 5-6 Radical Expressions Lesson 5-7 Rational Exponents Lesson 5-8 Radical Equations and Inequalities Lesson 5-9 Complex Numbers Example 1 Simplify Expressions with Multiplication Example 2 Simplify Expressions with Division Example 3 Simplify Expressions with Powers Example 4 Simplify Expressions Using Several Properties Example 5 Express Numbers in Scientific Notation Example 6 Multiply Numbers in Scientific Notation Example 7 Divide Numbers in Scientific Notation Multiplying Monomials: • When multiplying monomials you must ADD exponents. • Example: 2x3 3x5 2x x x 3x x x x x 6x8 Definition of exponents Commutative Property Answer: Definition of exponents Answer: Try These Multiply the following monomials. 1. a2 • a6 2. 3x2 • 7x4 3. (-3b3c)(7b2c2) 4. 2x2(6y3)(2x2y) Try These Multiply the following monomials. 1. a2 • a6 a8 2. 3x2 • 7x4 21x6 3. (-3b3c)(4b2c2) 4. 2x2(6y3)(2x2y) -12b5c3 24x4y4 Dividing Monomials: • When dividing monomials you must SUBTRACT exponents. 7 6x • Example: 5 2 x • • Cancel x’s 6 xxxxxxx 2 xxxxx 3xx 3x2 Subtract exponents. Remember that a simplified expression cannot contain negative exponents. 1 1 Answer: 1 1 Simplify. Answer: Try These Divide the following monomials. 1. 3. 6 2. 5 3 3 4. 2 a n 5 an 3a b c 3 7 9a b c y z 2 5 y z 5 7 3 2 5 2c d (3c d ) 4 2 30c d Try These Divide the following monomials. 1. 3. 2 2. 6 a n 5 an 5 3 3 3a b c 3 7 9a b c an 4. y z 2 5 y z 5 7 3 -y3z2 2 5 2c d (3c d ) 4 2 30c d Power to a Power: • When raising a power to a power you must MULTIPLY exponents. • Example: (x3)5 This means 5 groups of (x3). (x3) (x3) (x3) (x3) (x3) • (xxx)(xxx)(xxx)(xxx)(xxx) • x15 Product to a Power: • When raising a product to a power you raise every number/variable to that power. • Example: (2x2y3) 6 (2x2y3) (2x2y3) (2x2y3) (2x2y3) (2x2y3) (2x2y3) which can be written as: (2xxyyy) (2xxyyy) (2xxyyy) (2xxyyy) (2xxyyy) (2xxyyy) 64x12y18 Quotient to a Power • When raising a quotient to a power you raise the numerator & denominator to that power. 3 x x x 3 125 5 5 3 3 Power of a power Answer: Answer: Power of a power Power of a quotient Power of a product Answer: Negative exponent Power of a quotient Answer: Simplify each expression. a. Answer: b. Answer: c. Answer: d. Answer: Try These Simplify each monomial. 1. (n4)4 2. (2x)4 3. (-2r2s) 3 (3rs2) 4. 6x2 y4 3 ( 4 3) 3x y Try These Simplify each monomial. 1. (n4)4 2. (2x)4 n16 3. (-2r2s) 3 (3rs2) -24r7n5 16x4 4. 6x2 y4 3 ( 4 3) 3x y 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 1 x3 Try These Simplify each monomial. 2 1. 15 x y 4 5 x 2 y 6 3. (a3b3)(ab)-2 2. 15 x 2 y 4 5 x 2 y 6 4. xy 2 ( 2) y Try These Simplify each monomial. 1. 28 x 4 6 x 2 3. (a3b3)(ab)-2 2. 15 x 2 y 4 5 x 2 y 6 4. xy 2 ( 2) y Method 1 Raise the numerator and the denominator to the fifth power before simplifying. Answer: Method 2 Simplify the fraction before raising to the fifth power. Answer: Answer: Express 4,560,000 in scientific notation. 4,560,000 Answer: Write 1,000,000 as a power of 10. Express 0.000092 in scientific notation. Answer: Use a negative exponent. Express each number in scientific notation. a. 52,000 Answer: b. 0.00012 Answer: Express the result in scientific notation. Associative and Commutative Properties Answer: Express the result in scientific notation. Associative and Commutative Properties Answer: Evaluate. Express the result in scientific notation. a. Answer: b. Answer: 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? Divide the number of red blood cells in the sample by the number of red blood cells in 1 milliliter of blood. number of red blood cells in sample number of red blood cells in 1 milliliter Answer: There are about 1.66 milliliters of blood in the sample. Biology A petri dish started with germs in it. A half hour later, there are How many times as great is the amount a half hour later? Answer: Assignment: Page 226 #26, 32, 36, 40 Example 1 Degree of a Polynomial Example 2 Subtract and Simplify Example 3 Multiply and Simplify Example 4 Multiply Two Binomials Example 5 Multiply Polynomials Polynomials • Polynomial: The sum of terms such as 5x, 3x2, 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. Polynomials 6x-2 Not a Has a negative polynomial term exponent 1/x2 Not a Has variable in polynomial term the denominator. Not a Has variable in polynomial term radical. 4x2 Is a polynomial term 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. Determine whether is a polynomial. If it is a polynomial, state the degree of the polynomial. 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 each expression is a polynomial. If it is a polynomial, state the degree of the polynomial. a. Answer: yes, 5 b. Answer: no Adding Polynomials • 1. 2. 3. • • • Add: (2x2 - 4) + (x2 + 3x - 3) Remove parentheses. Identify like terms. Add the like terms. (2x2 - 4) + (x2 + 3x - 3) = 2x2 - 4 + x2 + 3x - 3 = 3x2 + 3x - 7 Subtracting Polynomials • 1. 2. 3. 4. Subtract: (2x2 - 4) - (x2 + 3x - 3) Remove parentheses. Change the signs of ALL of the terms being subtracted. Change the subtraction sign to addition. Follow the rules for adding signed numbers. • • • • • (2x2 - 4) - (x2 + 3x - 3) (Change the signs of terms being subtracted) = (2x2 - 4) + (-x2 - 3x + 3) = 2x2 - 4 + -x2 - 3x + 3 = x2 - 3x - 1 Simplify Distribute the –1. Group like terms. Answer: Combine like terms. Simplify Answer: Try These Add or subtract as indicated. 1. (3x2 – x – 2) + (x2 + 4x – 9) 2. (5y + 3y2) + (– 8y – 6y2) 3. (9r2 + 6r + 16) – (8r2 + 7r + 10) 4. (10x2 – 3xy + 4y2) – (3x2 + 5xy) Try These Add or subtract as indicated. 1. (3x2 – x – 2) + (x2 + 4x – 9) 2. (5y + 3y2) + (– 8y – 6y2) 4x2 + 3x - 11 -3y – 3y2 3. (9r2 + 6r + 16) – (8r2 + 7r + 10) r2 – r + 6 4. (10x2 – 3xy + 4y2) – (3x2 + 5xy) 7x2 -8xy + 4y2 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 Distributive Property Answer: Multiply the monomials. Answer: + + + First terms Outer terms Inner terms Last terms Answer: Multiply monomials and add like terms. Answer: Distributive Property Distributive Property Multiply monomials. Answer: Combine like terms. Answer: Try These Multiply the polynomials. 1. 4b(cb – zd) 2. 2xy(3xy3 – 4xy + 2y4) 3. (3x + 8)(2x + 6) 4. (x – 3y)2 5. (x2 + xy + y2)(x – y) Try These Multiply the polynomials. 1. 4b(cb – zd) 4b2c – 4bdz 2. 2xy(3xy3 – 4xy + 2y4) 6x2y4 – 8x2y2 + 4xy5 3. (3x + 8)(2x + 6) 6x2 + 34x + 48 4. (x – 3y)2 x2 – 6xy + 9y2 5. (x2 + xy + y2)(x – y) x 3 – y3 Assignment: Page 231-232 #25, 26, 30, 42 Example 1 GCF Example 2 Grouping Example 3 Two or Three Terms Example 4 Quotient of Two Trinomials Factoring Lesson #1 • Greatest Common Factor • Polynomials in the form x2 + bx + c 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. 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 4x3 as a common factor. Example: The polynomial: 4x5+12x4-8x3 Can be factored into: 4x3(x2+3x-8) Example 1: Factor the polynomial: 2x2 + 6x4 Now Check your work: 2x2 (1 + 3x2) by taking out a common factor. Multiply back together: 2x2 + 6x4 Solution: Choose the common factor. 2x2. 2x2 (1 + ___ ) 2x2 (1 + 3x2) Example 2: Factor the polynomial: 15x2y – 10xy2 Now Check your work: 5xy (3x – 2y) by taking out a common factor. Multiply back together: 15x2y Solution: Choose the common factor: 5xy. 5xy (3x – ___ ) 5xy (3x – 2y) – 10xy2 Example 3: Factor the polynomial: 16a3b 5 – 24a2b4 – 8a4b7c Now Check your work: 8a2b4 (2ab – 3 – a2b3c ) by taking out a common factor. Multiply back together: 16a3b 5 – 24a2b4 – 8a4b7c Solution: Choose the common factor: 8a2b4. 8a2b4 (2ab – ___ – ___ ) 8a2b4 (2ab – 3 – ___ ) 8a2b4 (2ab – 3 – a2b3c ) Now Try These: Factor the following polynomials and check your work. a. 6x2y3 + 8x2y5 Solution: 2x2y3 (3 + 4y2) b. 12a4b2c3 – 18ab2c4 + 24a5b3c4 Solution: 6ab2c3 (2a3 – 3c + 4a4bc) Factoring Polynomials in the form x2 + bx + c (General Quadratics) Examples of these “General Quadratics” are: a. x2 + 7x + 10 b. x2 + 13x - 30 c. x2 - 8x + 15 d. x2 - 8x - 20 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: x2 + 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 ) 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: x2 - 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 ) Now Try These: Factor the following polynomials and check your work. a. x2 + 7x + 10 b. x2 + 13x - 30 Answer: (x + 5)(x + 2) Answer: (x + 15)(x – 2) c. x2 - 8x + 15 d. x2 - 8x – 20 Answer: (x - 5)(x – 3) Answer: (x – 10)(x + 2) Part I Warm Up – Section 5-4 #1 Factor and check. 1. 10x2y + 15x3y2 2. 16a2b4c5 + 48a3bc2 – 12ab4c3 3. y2 + 11y + 24 4. y2 - 15y + 36 5. y2 + 7y - 30 6. y2 - 4y - 45 Factoring Lesson #2 • Polynomials in the form ax2 + bx + c Factoring Polynomials in the form ax2 + bx + c (Trial and Error) Examples of these “Trial and Error” Quadratics are: a. 4x2 - 8x - 45 b. 12x2 + 13x - 14 c. 15x2 - 26x + 7 d. 25x2 +15x + 2 Rules for Factoring General Quadratics in the form ax2 + 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. • Example: 4x2 - 24x + 35 4x1 2x2 5x7 35 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 (-) - Check your work!!! ( 2x - 7 )( 2x - 5 ) Now Try These: Factor the following polynomials and check your work. a. 2x2 + 7x + 6 b. 3x2 + 10x + 3 Answer: (2x + 3)(x + 2) Answer: (3x + 1)(x + 3) c. 15x2 - 38x + 7 d. 10x2 - 3x – 27 Answer: (5x - 1)(3x – 7) Answer: (5x – 9)(2x + 3) Part II Warm Up – Section 5-4 #2 Factor using trial and error or the junk method and check your work. 1. 2x2 + 11x + 14 2. 14y2 – 19y – 3 3. 3a2 – 22a + 24 Factor. 4. 25r2s4t + 100rs2t3 5. x2 – 11x + 24 6. x2 + 2x – 35 Factoring Lesson #3 • Difference of two perfect squares x2 - y2 • Factoring by grouping • Factoring Completely Factoring the difference of two perfect squares Examples of these polynomials are: a. 4x2 – 9 b. 16x2 – 36 c. x2 – 4 d. 25x2 – 16y2 Rules for Factoring the difference of two perfect squares - - - The square root of the first term becomes the first term of each binomial. Example: x2 - 64 The square root of the second term becomes the second term of each binomial. (x Use different signs. Since the square root of x2 is x, x is the first term of each binomial. )( 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 ) Now Try These: Factor the following polynomials and check your work. a. 4x2 – 9 b. 16x2 – 36 Answer: (2x + 3)(2x - 3) Answer: (4x + 6)(4x - 6) c. x2 – 4: d. 25x2 – 16y2 Answer: (x - 2)(x + 2) Answer: (5x – 4y)(5x + 4y) Factoring by grouping Examples of polynomials that are factored by grouping are: a. 6x2 + 3xy + 2xz + yz b. 6x2 + 2xy – 3xz – yz Note: You will see 4 terms when using the grouping method. 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: 10a2 + 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. (10a2 + 2ab) + (5ad + bd) Take out a common factor. 2a(5a + b) + d(5a + b) Combine like groups: ( 2a + d )( 5a + b ) Examples of factoring by grouping: Factor the polynomial: a3 – 4a2 + 3a – 12 Factor the polynomial: 7ac2 + 2bc2 – 7ad2 – 2bd2 Group: (a3 – 4a2)+ (3a – 12) Factor: a2 (a – 4) + 3(a – 4) Combine: (a2 + 3)(a – 4) Group: (7ac2 + 2bc2) + (– 7ad2 – 2bd2) Factor: c2(7a + 2b) + d2 (– 7a – 2b) Factor a negative out of second group: c2(7a + 2b) - d2 ( 7a + 2b) Now groups match – so, Combine: (c2 - d2) ( 7a + 2b) Now Try These: Factor the following polynomials and check your work. a. 6x2 + 3xy + 2xz + yz Answer: (3x + z)(2x + y) b. 6x2 + 2xy – 3xz – yz Answer: (2x - z)(3x + y) 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. 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: 3x2 – 21x + 30 Here, I notice that I have a common factor of 3, so take it out. 3(x2 – 7x + 10) Now x2 – 7x + 10 can be factored. 3(x – 5)(x – 2) Now all terms are prime. Now Try These: Factor the following polynomials and check your work. a. 2x2 + 12x + 18 b. 3x2 – 21x – 54 Answer: 2(x + 3)(x + 3) Answer: 3(x + 2)(x - 9) c. 5x2 – 20: d. 25x2 – 100y2 Answer: 5(x - 2)(x + 2) Answer: 25(x – 2y)(x + 2y) Part III Warm Up Section 5-4 #3 Factor. 1. 7c3 – 28c2d + 35cd3 2. x2 – 5x – 14 3. x2 – 15x + 54 4. 3x2 – 22x + 35 5. 64x2 – 81 6. 3r + 3s + 5r3s + 5r2s2 Factor The GCF is 5ab. Answer: Distributive Property Factor Answer: Factor Group to find the GCF. Factor the GCF of each binomial. Answer: Distributive Property Factor Answer: Factor 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 . Rewrite the expression using –5y and 3y in place of –2y and factor by grouping. Substitute –5y + 3y for –2y. Associative Property Factor out the GCF of each group. Answer: Distributive Property Factor Factor out the GCF. Answer: p2 – 9 is the difference of two squares. Factor This is the sum of two cubes. Sum of two cubes formula with and Answer: Simplify. Factor 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 each polynomial. a. Answer: b. Answer: c. Answer: d. Answer: Simplify Factor the numerator and the denominator. Divide. Assume a –5, –2. Answer: Therefore, Simplify Answer: Example 1 Divide a Polynomial by a Monomial Example 2 Division Algorithm Example 3 Quotient with Remainder Example 4 Synthetic Division Example 5 Divisor with First Coefficient Other than 1 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: Sum of quotients Divide. Answer: Answer: Try These Divide the following polynomials. 1. 2. Try These Divide the following polynomials. 1. x 2 xy 3x y 2 3 3 2. 2ab 3a b 5a 2 3 3 Use factoring to find Answer: Use factoring to find Answer: x + 2 Try These Divide the following polynomials by factoring. 1. 2. 3. 4. Try These Divide the following polynomials by factoring. 1. 2. x2 3. x2 4. x3 x 5 Warm Up Section 5-5 Covering lessons 5.1-5.4 1. Simplify: 5x2y(4x4y3) 2. Simplify: (4a5b4c2) 3 6 3 18 x y 3. Simplify: 4 8 12 x y 4. Multiply: (3x + 7)(x – 4) 5. Factor: x2 – 11x + 18 6. Factor: 3x2 + 7x – 20 Warm Up Section 5-5 Covering lessons 5.1-5.4 1. Simplify: 5x2y(4x4y3) 2. Simplify: (4a5b4c2) 3 6 3 18 x y 3. Simplify: 4 8 12 x y 20x6y4 64a15b12c6 3x 2 5 2y 4. Multiply: (3x + 7)(x – 4) 3x2 – 5x – 28 5. Factor: x2 – 11x + 18 (x – 9)(x – 2) 6. Factor: 3x2 + 7x – 20 (3x – 5)(x + 4) Example 1 Find Roots Example 2 Simplify Using Absolute Value Example 3 Approximate a Square Root Simplifying Radicals • When working with the simplification of radicals you must remember some basic information about perfect square numbers. Perfect Squares 1 =1x1 4 =2x2 9 =3x3 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 Perfect Squares Containing Variables a2 = a x a a4 = a2 x a 2 a6 = a3 x a 3 a8 = a4 x a 4 a10 = a5 x a5 So, a variable is a “perfect square” if it has an even exponent. To take the square root, just divide the exponent by 2. 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. 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. 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. Example Continued • Reduce the "perfect" radical which you have now created. Simplify Answer: The square roots of 16x6 are 4x3. Simplify Simplify Answer: The fifth root is 3a2b3. Simplify Answer: You cannot take the square root of a negative num Thus, is not a real number. Simplify. a. Answer: 3x4 b. Answer: c. Answer: 2xy2 d. Answer: not a real number Try These 225 1 16 (5 g ) (7) 2 8 169 x y 27 4 0.25 4 3 4 z 8 (4 x y ) 2 Try These 225 (7) 15 Not real # 1 16 1/4 (5 g ) 25g2 2 27 -3 4 0.25 z 8 z2 0.5 4 3 8 169 x y 13x4y2 4 (4 x y ) 4x - y 2 Simplify Note that t is a sixth root of t6. 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 Since the index is odd, you do not need absolute value. Answer: Simplify. a. Answer: b. Answer: Try These 169 25 169 (2 x) 8 2 3 0.81 4 (4) 6 36 x y 10 125 z 12 (3x 6) 2 Try These 169 -13 16x4 3 125 4 25 169 5/13 (2 x) (4) 2 5 0.81 8 0.9 6 36 x y 6x3y5 4 z 12 z3 10 (3x 6) 3x+6 2 Assignment: • P248 #40, 42, 46, 50 Example 1 Square Root of a Product Example 2 Simplify Quotients Example 3 Multiply Radicals Example 4 Add and Subtract Radicals Example 5 Multiply Radicals Example 6 Use a Conjugate to Rationalize a Denominator Simplify Factor into squares where possible. Product Property of Radicals Answer: Simplify. Simplify Answer: Simplify Quotient Property Factor into squares. Product Property Rationalize the denominator. Answer: Simplify Quotient Property Rationalize the denominator. Product Property Multiply. Answer: Simplify each expression. a. Answer: b. Answer: Try These 72 4 32 x y 5 1 6 7 wz 32 3 5 3 4 54 16 y 4r 8 9 t 3 3 96 4 5 2 24m n 20 x 5 10 y Try These 72 3 54 32 x y 5 3 2 4x2 y 2 2 y 5 1 6 7 wz 32 1 5 2 wz wz 2 3 16 y 3 96 24 6 3 6 2 4 4 3 4 5 2 24m n 2 y3 2 4r 8 4 9 2 r t t t 5 4mn3 3mn2 20 x 5 10 2 x 2 5 x y y5 Warm Up 5-6 Simplify. 1. 36 2. 81x 2 y 6 3. 81x 6 y 7 4. 40 x 4 y11 5. 6. 3 5 3 24a b 4 32s 4t 7 Warm Up 5-6 Simplify. 1. 36 6 6 9xy 6 7 3 4 11 2. 81x y 3. 81x y 4. 40 x y 5. 6. 3 9x y 2 3 y 5 10 y 3 2 2x y 3 5 2ab 3b 4 7 4 24a b 4 3 2 32s t 2st 2t 3 Simplify Product Property of Radicals Factor into cubes. Product Property of Radicals Answer: Multiply. Simplify Answer: 24a Try These 1. (3 12 )(2 21) 2. (3 24 )(5 20) Try These 1. (3 12 )(2 21) 36 7 2. (3 24 )(5 20) 60 30 Simplify Factor using squares. Product Property Multiply. Combine like radicals. Answer: Simplify Answer: Try These 1. 12 48 27 2. 5 20 24 180 7 54 Try These 1. 12 48 27 3 3 2. 5 20 24 180 7 54 4 5 23 6 Simplify F O I L Product Property Answer: Simplify FOIL Multiply. Answer: Subtract. Simplify each expression. a. Answer: b. Answer: 41 Simplify Multiply by since is the conjugate of FOIL Difference of squares Multiply. Answer: Combine like terms. Simplify Answer: Assignment • P 254 #16-46 even Example 1 Radical Form Example 2 Exponential Form Example 3 Evaluate Expressions with Rational Exponents Example 4 Rational Exponent with Numerator Other Than 1 Example 5 Simplify Expressions with Rational Exponents Example 6 Simplify Radical Expressions Write Answer: in radical form. Definition of Write Answer: in radical form. Definition of Write each expression in radical form. a. Answer: b. Answer: Write Answer: using rational exponents. Definition of Write Answer: using rational exponents. Definition of Write each radical using rational exponents. a. Answer: b. Answer: Evaluate Method 1 Answer: Simplify. Method 2 Power of a Power Multiply exponents. Answer: Evaluate . Method 1 Factor. Power of a Power Expand the square. Find the fifth root. Answer: The root is 4. Method 2 Power of a Power Multiply exponents. Answer: The root is 4. Evaluate each expression. a. Answer: b. Answer: 8 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. According to the formula, what is the maximum that U.S. Weightlifter Oscar Chaplin III can lift if he weighs 77 kilograms? Original formula Use a calculator. 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. 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 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? Answer: 380 kg b. If he actually lifted 379 kg, compare this to the value predicted by the formula. Answer: The formula prediction is slightly higher than his actual total. Simplify . Multiply powers. Answer: Add exponents. Simplify . Multiply by Answer: Simplify each expression. a. Answer: b. Answer: Simplify . Rational exponents Power of a Power Quotient of Powers Answer: Simplify. Simplify . Rational exponents Power of a Power Multiply. Answer: Simplify. Simplify . is the conjugate of Answer: Multiply. Simplify each expression. a. Answer: 1 b. Answer: c. Answer: Example 1 Solve a Radical Equation Example 2 Extraneous Solution Example 3 Cube Root Equation Example 4 Radical Inequality Solving Radical Equations 1. 2. 3. Isolate the radical Raise each side to the appropriate power to eliminate the radical. 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 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. Check Original equation Replace y with 38. Simplify. Answer: The solution checks. The solution is 38. Solve Answer: 67 Try These Solve each equation. Try These Solve each equation. 25 144 1 -11 Solve Original equation Square each side. Find the squares. Isolate the radical. Divide each side by –4. Square each side. Evaluate the squares. Check Original equation Replace x with 16. Simplify. Evaluate the square roots. Answer: The solution does not check, so there is no real solution. Solve Answer: no real solution . Solve In order to remove the power, 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. Subtract 1 from each side. Divide each side by 3. Check Original equation Replace y with –42. Simplify. The cube root of –125 is –5. Add. Answer: The solution is –42. Solve Answer: 13 Try These Solve each equation. Try These Solve each equation. 49 5 9 -20 Assignment P 266 #16, 19, 22, 24 Example 1 Square Roots of Negative Numbers Example 2 Multiply Pure Imaginary Numbers Example 3 Simplify a Power of i Example 4 Equation with Imaginary Solutions Example 5 Equate Complex Numbers Example 6 Add and Subtract Complex Numbers Example 7 Multiply Complex Numbers Example 8 Divide Complex Numbers 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 . i i is defined to have the property that: i2 = -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 . Examples 1. Simplify: 2. Simplify 6ix2 Since is 5 and is i, our answer is 5i. Simplify Answer: . Simplify Answer: . Simplify. a. Answer: b. Answer: Simplify Answer: . =6 Simplify Answer: . Simplify. a. Answer: –15 b. Answer: Simplify Multiplying powers Power of a Power Answer: Simplify Answer: i . Solve Original equation Subtract 20 from each side. Divide each side by 5. Take the square root of each side. Answer: Solve Answer: Find the values of x and y that make the equation true. 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: Find the values of x and y that make the equation true. Answer: Simplify . Commutative and Associative Properties Answer: Simplify . Commutative and Associative Properties Answer: Simplify. a. Answer: b. Answer: 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. Answer: The voltage is volts. 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 Simplify . and are conjugates. Multiply. Answer: Standard form Simplify . Multiply by Multiply. Answer: Standard form Simplify. a. Answer: b. Answer: 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. End of Custom Shows WARNING! Do Not Remove This slide is intentionally blank and is set to auto-advance to end custom shows and return to the main presentation.