Algebra I. Book 2. Powered by...


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1 Algebra I Book 2 Powered by...
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3 ALGEBRA I Units 47 by The Algebra I Development Team
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5 ALGEBRA I UNIT 4 POWERS AND POLYNOMIALS Review Properties of Exponents Polynomials: Classifying and Simplifying Adding and Subtracting Polynomials Multiplying By a Monomial Factoring: Integers and Monomials Common Monomial Factors UNIT 5 MORE FACTORING Review Polynomial Products Factoring: x 2 + bx + c Factoring: ax 2 + bx + c and ax 2 + bxy + cy Factoring Special Types Factoring Combined Types Factoring By Grouping Solving Quadratic Equations by Factoring Consecutive Integer Problems UNIT 6 MULTIPLYING AND DIVIDING RATIONAL EXPRESSIONS Review Rational Expressions Rational Expressions: Simplest Form Using 1 in Factoring Dividing Powers Simplifying a Product Multiplying and Dividing Dividing by a Monomial Dividing by a Binomial UNIT 7 ADDING AND SUBTRACTING RATIONAL EXPRESSIONS Review Adding Rational Expressions with Monomial Denominators Adding Rational Expressions with Polynomial Denominators Adding Other Types of Rationals Subtracting Rational Expressions Simplifying by Factoring Out Complex Rational Expressions Complex Rational Expressions: Polynomial Denominators iii
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7 UNIT 4 POWERS AND POLYNOMIALS Page Lesson 4.0 Review Lesson 4.1 Properties of Exponents Lesson 4.2 Polynomials: Classifying and Simplifying Lesson 4.3 Adding and Subtracting Polynomials Lesson 4.4 Multiplying By a Monomial Lesson 4.5 Factoring: Integers and Monomials Lesson 4.6 Common Monomial Factors
8 Lesson 4.0 Review Unit 4 Terminology: An open sentence contains one or more variables and is neither true nor false until a replacement is made for the variable(s). A solution is any replacement which makes an open sentence true. Equivalent equations are those that have the same solution. Properties for Equations: Addition: If a = b, then a + c = b + c. (also a  c = b  c) Solving Equations: Multiplication: If a = b and c 0, then ac = bc (also a = b) c c 1. Simplify each side. 2. Use the addition property to get a variable term on one side and a constant on the other. 3. Use the multiplication property to get the variable alone. Example 1: x  4 = 12 x = x = 16 Example 2: 2 x = 12 3 Example 3:. Addition property x = Multiplication property x = 18 2(3x + 4) + x = 3x x 6x x = 2x  7. Simplify. 7x + 8 = 2x  7. Simplify. 7x + 82x = 2x  72x. Addition property. 5x + 8 = 7. Simplify. 5x = Addition property. 5x = 15. Simplify. 5x = 15. Multiplication property. 5 5 (Multiply by 1/5 or divide by 5.) x = 3. Answer. 2
9 Words to Symbols: Lesson 4.0 Understanding key words and punctuation marks allows word problems to be translated into equivalent algebraic equations. (See Lesson 3.6.) Solving Word Problems: 1. Choose a variable to represent what is to be found. 2. Plan and write the equation, using the facts in the problem. 3. Solve the equation and state the solution to the problem. 4. Check the solution with the facts of the problem. Review Problems Solve each equation. Check the solution: 1. 3 = x a + 24 = 10a x = x + 3 = 2x x = t = 6t x = 2(5) 12. 3m m = m x = x = 5x  55x (a  3) = 2(a  6) x  2x = (8 + 2x) = 3(1 + 4x) 7. 10p p = r  3(42r) = 6(r + 1) = 3m m Write an equation for each problem; then solve it: 17. Two less than five times a number is 23. Find the number. 18. Joe's age is increased by 5 times his age. The result is 12. What is Joe's age? 19. Separate 52 marbles into two groups so the second group has 2 less than 5 times the number in the first group. 20. What are the dimensions of a rectangle which is twice as long as it is wide if its perimeter is 48cm? 3
10 Lesson 4.1 Properties of Exponents Unit 4 Rules: In x4, x is the base and 4 is the exponent. This can be read "the fourth power of x" or "x to the fourth power" or "x to the fourth." Product of Powers Rule: When multiplying powers with the same base, add the exponents and keep the base. am. an = am+n Examples: 1. Simplify: y3. y4 =. Same base is "y". y 3+4 =. Keep the base and add exponents. y7 2. Simplify:. (Think: y. y. y. y. y. y. y = y7.) x6. x =. Same base. Remember x = x1. x6+1 =. Keep same base; add exponents. x7. Add Simplify: x3. y3. Does not have the same base so it cannot be simplified any further. Practice: Simplify: A. y2. y4 = B. y8. y3 = C. x. x7 = D. x. y5 = Rule: The Commutative and Associative Properties can be used to regroup a problem so the product of powers rule can be applied. 4
11 Example: 4. Simplify: (3y2)(2y4) = (3. 2)(y2. y4) =. Group like factors. 6. y2+4 =. Multiply like factors. 6y6. Product of Powers. Lesson 4.1 Practice: Simplify: E. (4x2)(2x5) F. (3x2y4)(4xy3) Example: 5. Simplify; then evaluate for x = 1: (x3)(3x2)(2x4) = (1x3)(3x2)(2x4) =. x3 = 1x3. ( )(x3. x 2. x4) =. Regroup. (6) (x3+2+4) =. Product of Powers. 6x9 =. Simplified answer. 6(1)9 =. Substitute 1 for x. 6(1) =. (1)9 = Evaluated answer. Rule: Power of a Power Rule: To simplify a power of a power, multiply the exponents. (am)n = am n. Example: 6. Simplify: (y3)4 = y3 4 = y12. Multiply the exponents.. (Think: y3. y3. y3. y3 = y12.) Practice: Simplify: G. (x4)6 = H. (c5)7 = Rule: Power of a Product Rule: To find the power of a product, apply the exponent outside the grouping symbols to each factor inside the grouping symbols. (ab)m = am. bm 5
12 Lesson 4.1 Examples: 7. Simplify: 8. Simplify: (2x)3 = (xy4)3 = 23. x3 = x3. (y4 ) 3 = 8x3 x3y12 Note: From the examples above, notice that each exponent inside the grouping symbols is multiplied by the exponent outside the grouping symbols. 9. Simplify: (4x2)4 = (4)1 4 x2 4 =. Remember 4 = (4)1. (4)4. x8 =. Multiply exponents. 256x8. Simplify: (4)4 = 256. Practice: Simplify: I. (3x4)2 = J. (2y3)3 = Example: 10. Simplify: (3a3b2c)4 = (31. a3. b2. c1)4 =. 3 = 31 and c = c1. (31)4(a3)4(b2)4(c1)4 = 34a12b8c4 =. Multiply each exponent by 4. 81a12b8c4. Simplify: 34 = 81. Practice: Simplify: K. (4x2y3z)2 = L. (22d2e3f7)2 = Example: 11. Simplify: (4a2b)2(ab3)3(2a2b) = (42a4b2)(a3b9)(2a2b) = (16a4b2)(a3b9)(2a2b) = (16. 2)(a4. a3. a2)(b2. b9. b) =. b = b1. 32a9b12 6
13 Lesson 4.1 HOMEWORK Simplify, if possible: 1. x2. x4 8. (6z2)(7z5) 2. c3. c5 9. (5a2)(6a) 3. r4. r 10. (6ab2)(2a3b5) 4. d3. d8 11. (bm5)(3b3m4) 5. (5x2)(3x3) 12. (5x2yz3)(4y3) 6. (4y3)(2y) 13. (3x)(6y2) 7. (7b2)(3b5) 14. (6x)(yz) Simplify; then evaluate for x = 3, y = 4: 15. (4x2)(3x) 16. (3x2)(2y) 17. (2xy2)(3y3) 18. (6x2)(3x3)(x) Simplify: 19. (6y2)3 25. (6a2b)2(ab2)3 20. (3x2y3)4 26. (2xy)2(x3y4)2 21. (3a3b2c8)2 27. (x)2(xy)2 22. (x2y3z9)7 28. (x2y)2(3x3y4)2 23. (5a2)2(a3)8 29. (de)2(2de) 24. (2x3)2(4x2)3 30. (a2y)x(ay5)x 7
14 Lesson 4.2 Polynomials: Classifying and Simplifying Unit 4 Rule: Examples: Rule: Examples: A monomial is a term that is either a constant, a variable, or a product of a constant and one or more variables. Types of monomials: is a constant 2. x is a variable 3. 3 x 4 is a product of a constant and a variable a2bc3 is a product of a constant and several variables. The degree of a monomial is determined by finding the sum of the exponents of the variable in the monomial. A nonzero constant has a degree of 0 and the constant 0 has no degree. Find the degree of each monomial: 5. 11y3 Degree: x2y3 Degree: a2n3r Degree: 6. (6 = ) 8. 7 Degree: 0 Practice: Rules: Find the degree of each monomial: A. 11xy2 Degree: B. 19 Degree: C. 6a2x3y Degree: D. 7rx5y Degree: A polynomial is a monomial or the sum or difference of two or more monomials. Polynomials with one, two, or three terms have special names: Monomial: A polynomial with one term: 4y3 Binomial: A polynomial with two terms: 2x2 + 7y Trinomial: A polynomial with three terms: 7x2 + 6x + 5 Practice: Label each polynomial as either a monomial, binomial, or trinomial: E. 3x2 + 2x  6: F. xy2 + 2x: G. 8a2bc3: 8
15 Lesson 4.2 Rule: The degree of a polynomial in simple form is the same as the highest degree of any of its terms. When simplifying a polynomial, combine all like terms and then write it in descending order of exponents (highest exponent first, next highest exponent second, etc.). Example: 9. Simplify: 6y28y3 + 2y2 + 4y  8y3 + 4 = 8y38y3 + 6y2 + 2y2 + 4y + 4 =. Regroup like terms. 16y3 + 8y2 + 4y + 4 =. Combine like terms. Degree: 3 HOMEWORK What is the degree of each polynomial? 1. 4x7 4. x2y3z8 2. 8x 5. 2x23xy x6 + 3x2y52y4 Classify each polynomial as a monomial, binomial, or trinomial: 7. 4x23y 10. 4b2 + 2x 8. 6a4 + 2x  3y 11. 6x23x d 12. 2(x2 + 3y3) Simplify; write answers in descending order of exponents: 13. 8x23x + 4x ax26y + 5ax24y 14. 3y3 + 2y  6y2 + 4y 21. 2b + 6ab23a2b + 7b 15. 3a + 4a26a + 2a a  3a2 + 2a  6ab 16. 7c + 6c2 + 5c33c a2b32ab + 6ab  4a2b r + 2r26r  2r xy2 + 2x2y  6x2y2 + 2x2y  4x2y2 18. ey2 + 2y3 + 6y  4y2 + 2y 25. 6x2y + 2xy  3x2y  xy  2x2y  x2y 19. 6a + 4a  3a2 + 2a  5a2 9
16 Lesson 4.3 Adding and Subtracting Polynomials Unit 4 Rule: Example: To add polynomials, group like terms in descending order of exponents; then combine like terms. 1. Add: (2x23x + 2) + (6x24x + 7) = 2x2 + 6x23x  4x =. Group like terms, descending order). 8x27x + 9. Combine like terms. Practice: Add: A. (3y3 + 4y  2) + (5y3 + 2y  8) B. (5a2 + 3a34a + 2) + (3a5 + 6a  5) Rule: Example: Practice: To subtract polynomials, add the opposite of each term of the polynomial that is to be subtracted. (This is the same as distributing 1.) 2. Subtract: (a2 + 3a  4)  (6a24a + 5) = Subtract: a2 + 3a  46a2 +4a  5 =. Add the opposite. a26a2 + 3a + 4a =. Group like terms. 5a2 + 7a  9. Combine like terms. C. (3y32y2 + 4y  5)  (3y2 + 6y + 4) D. (2a4 + 3a35a + 2a2)  (7a2 + 4a3  a + 7) 10
17 Add: HOMEWORK Lesson (7x23x + 2) + (8x2 + 2x + 6) 2. (b47b + 9) + (b5  b4 + 2b3) 3. (2y5  y4 + 3y3) + (y49y + 5) 4. (2c23c + 5) + (3c27c + 4) Subtract: 5. (4x2 + 3x  2)  (7x + 7) 6. (5y2 + 3y  4)  (6y + 8) 7. (2x43x2 + 2x)  (3x3 + 4x25) 8. (a52a + 4a25)  (6a53a3 + 2a27) 9. (7d3 + 5d  4)  (5d3 + 4d  4) 10. (7x3 + 3x x)  (2x + 3x x3) Perform the indicated operation: 11. (2x23x + 4) + (5x + 3) 12. (2x + 4) + (3x25x3 + 4x  2) 13. (6x34x  3)  (5x3 + 4x23x + 5) 14. (2y + 3y25) + (y2 + 3y  2) 15. (5a43a2 + 4a34)  (6a52a + 4) 16. (5c3 + 3)  (7c2 + 4c33c2 + c  8) 17. (13c2 + 7c + 5) + (7c  37c c2) 18. (8y2 + 2y  5)  (6y3 + 3y  2) 19. (9x + 4x23x3)  (6y3 + 3y  2) 20. (21x  3x2 + 7x) + (7x23x + 14) 11
18 Lesson 4.4 Multiplying by a Monomial Unit 4 Rule: Example: Practice: Some polynomials can be simplified by first using the Distributive Property and then combining like terms. 1. Simplify: x2 + 6x (x2 + 3x  4) = x2 + 6x x2 + 12x  16 =. Distribute the 4. x2 + 4x2 + 6x + 12x =. Group like terms. Simplify: 5x2 + 18x  8. Combine like terms. A. y2 + 4y + 3(2y2 + 3y  2) B. 3a2 + 4a + 5(a22a  5) Rule: Example: To simplify the product of a monomial and a polynomial, use the Distributive Property and the Product of Powers Property. 2. Multiply: 2a2(5a4 + 6a3 + 2a) = (2a2. 5a4) + (2a2. 6a3) + (2a2. 2a) =. Distribute the 2a2. 10a6 + 12a5 + 4a3 Practice: Multiply: C. x2(3x32x2 + 4) D. c3(4c5 + 6c33c + 5) Rule: Example: The same steps are followed when there are two or more variables. 3. Multiply: a2c3(3a2 + 4ac32a + 3c  5) (a2c3. 3a 2) + (a2c3. 4ac 3)  (a2c3. 2a) + (a 2c3. 3c)  (a 2c3. 5). Distribute the a2c3. 3a4c3 + 4a3c62a3c3 + 3a2c45a2c3. Group like terms. 12
19 Lesson Multiply: 2xy(3x24xy2)  3x(2x3y + 4x27) = (2xy. 3x 2)  (2xy. 4xy 2) + (3x. 2x 3y) + (3x. 4x 2)  (3x. 7) = 6x3y  8x2y36x4y  12x3 + 21x = 6x4y + 6x3y  12x38x2y3 + 21x Practice: Multiply: E. xy(3x4 + 4y  2xy2) F. 2s2t3(4st + 3s22t3 + 5) HOMEWORK Simplify: Multiply: 1. x2 + 3x  5(x2 + 5x + 3) 5. r2(3r42r2 + 4) 2. y3 + 3y2 + 2(y33y2 + 7y  4) 6. 2x(5x43x2 + 4) 3. p + p25(p2 + 3p  4) 7. a3(a43a2 + 6a  3) 4. c42c + 4(c3 + c24c + 2) 8. 3x3(3x3 + 2x2 + x  5) Simplify: 9. x2y3(3x2 + 4y23xy3) 10. ab2(3ab + 2a23ab34) 11. b3d2(3bd2 + 2b  3d  4d3) 12. xy3(3x3 + y24xy + 2) 13. 3a3c4(2a43c2 + 2ac + 6) xy5(2xy43xy + 4y3 + 2) 15. 3qr(4r2 + 3r  2qr3)  6qr(2r2 + 3qr35r) a3b2(3a4b  2a2b + 2)  4a3b2(2a4b + 3a2b  8) 17. xy2(3xy4 + 2x2y  5y)  xy2(2x2y  3y  6xy4) 18. m2r3(3mr + 2mr26m2r) + m2r3(3mr  2mr2 + 6m2r) 13
20 Lesson 4.5 Note: Example: Practice: Use a similar approach to find a missing factor with a numerical coefficient. 3. Find the missing factor: (6x3)(?) = 24x8 (6x3)(?) = 24. x8 6. (?). x3. x? = 24. x8 6 (4) = (4). x3. x5 = 24. x8 x3 x5 = x8 6x3 (4x5) = 24x8 Find the missing factor: E. (3y2)(?) = 9y7 F. (?) (5m3) = 35m8 HOMEWORK Factor each number into primes, if possible: Find the missing factor: 7. (y3)(?) = y4 14. (11r4)(?) = 44r8 8. (?)(a7) = a (?)(16y) = 16y2 9. (?)(c) = c5 16. (3ab)(?) = 3a2b 10. (r4)(?) = r (6xy)(?) = 24x2y4 11. (4x2)(?) = 8x3 18. (?)(5r2s3) = 15r7s (?)(6a3) = 24a5 19. (?)(4x2y5) = 28x5y7 13. (?)(5c4) = 20c9 20. (x8y2z5)(?) = 21x9y3z5 15
21 Lesson 4.5 Factoring: Integers and Monomials Unit 4 Rules: A prime number is a whole number greater than one whose only factors are 1 and the number itself. The prime factorization of a number is the result of rewriting the number as the product of its prime factors. There is only one prime factorization of any given number. Example: 1. Factor 36 into primes: 36 or Note: Notice that there is only one prime factorization of 36, , but it can be found in several different ways. Rule: Example: By using the Product of Powers Property, a missing factor can be found. 2. Find the missing factor: x4.? = x11 x4. x? = x11 x4+? = x11 x4+7 = x11 Thus, the missing factor is x7. Practice: Factor each number into primes: A. 24 B. 32 Find the missing factor: C. y4.? = y7 D.?. a9 = a13 14
22 Lesson 4.6 Common Monomial Factors Unit 4 Rule: Example: To factor a polynomial into the product of a monomial and a polynomial, the greatest common factor (GCF) must be found. Factor out the greatest common monomial: 1. 3x2 + 6x + 12 = 3(x2) + (3)(2)(x) + (3)(2)(2) =. 3 is the greatest common 3(x2 + 2x + 4) factor. 2. 3a2 + 12a  3 = 3(a2) + (2)(2)(3)(a) + 3(1) =. 3 = (a2 + 4a  1). 3 is the greatest common factor. Practice: Factor out the greatest common monomial: A. 4x2 + 16x + 8 B. 5y215y  5 Example: 3. Factor out the GCF: 15x245x + 30 = x x =. Factor 15, 45, and 30 15(x23x + 2) into primes.. The GCF is 3. 5 or Factor out the GCF: x35x2 + x = (x)(x)(x)  5(x)(x) + (x) =. The GCF is x. x(x25x + 1). Remember that 1. x = x. Practice: Example: Factor out the GCF: C. 10x2 + 30x  40 D. y3 + 3y2  y 5. Factor out the GCF: 3y4 + 12y315y2 = 3(y4 + 4y35y2) =. Find the greatest common whole number factor, 3. 3y2(y2 + 4y  5). Now look for the greatest common variable factor, y2. The GCF is 3y2. 16
23 Practice: Factor the GCF from each: Lesson 4.6 E. 4x512x3 + 6x2 F. 12y66y418y Note: Remember these steps when factoring out the greatest common monomial factor: 1. Factor out the Greatest Common Numerical Factor, if possible 2. Factor out the Greatest Common Variable Factor, if possible. HOMEWORK Factor out the Greatest Common Factor: 1. 3y2 + 6y x316x2 2. 4x26x r93r3 + 3r 3. 5y210y x9 + 45x675x3 4. 2a212a a535a 5. 4c218c a5 + 3a42a3 + a 6. 7x221x a520a y227y x320x2 + 24x 8. 8a2 + 16a t334t 9. 21d2 + 42d a43a3 + a y2 + 34y y2 + 7y x37x2 + 14x x x c5 + 21c34c x216x r8 + 6r64r x5112x b43b2 + b a2123a m77m4 + 21m r7155r
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25 UNIT 5 MORE FACTORING 5.0 Review Polynomial Products Factoring: x 2 + bx + c Factoring: ax 2 + bx + c and ax 2 + bxy + cy Factoring Special Types Factoring Combined Types Factoring By Grouping Solving Quadratic Equations by Factoring Consecutive Integer Problems
26 Lesson 5.0 Review Unit 5 Properties of Exponents: 1. Product of Powers : am. an = am+n Example : x2. x3 = x2+3 = x5 2. Power of a Power : (am)n = amn Example : (x2)3 = x2 3 = x6 3. Power of a Product : (ab)m = ambm Example : (3x)2 = (32)(x2) = 9x2 Terminology: Monomial: 4xy2 (1 term) Binomial: 2x2 + 7y (2 terms) Trinomial: 7x2 + 6x + 5 (3 terms) Polynomial: A monomial or the sum or difference of two or more monomials. Degree: 3x2y3 + 4x3y + 2xy. This is a 5th degree polynomial. 5th degree 4th degree 2nd degree term term term Adding and Subtracting: To add polynomials, group and combine like terms in descending order of exponents. To subtract a polynomial, add the opposite of each of its terms. Example: (7a2 + 3a  4)  (2a25a + 6) = 7a2 + 3a  42a2 + 5a  6 =. Add the opposite. 7a22a2 + 3a + 5a =. Group like terms. 5a2 + 8a Combine like terms. Multiplying by a Monomial: Use the Distributive Property and the properties of exponents. 2a2b(3a + 2a2b2 + 5b3) = 6a3b + 4a4b3 + 10a2b4 = 4a4b3 + 6a3b + 10a2b4 Common Monomial Factors: 1. Find the greatest common numerical factor. 2. Find the greatest common variable factor. 3. Use the Distributive Property to "undo" multiplication: 6a3 + 4a4b3 + 10a3b4 = 2a3 (3 + 2ab3 + 5b4) 20
27 Review Problems Lesson 5.0 Simplify, if possible: 1. x5. x4 4. (2x3)4 2. (2x3)(3x5) 5. (3x2y)(5x3y4)2 3. (x4)5 Simplify each polynomial; then tell its degree: 6. 6r + 2r26r b + 6a2b33a3b2 + 7b 8. 7a2b32ab + 6ab  4a2b3 Simplify: 9. (2x43x2 + 5x) + (5x33x2 + 2) 10. (6x34x  3)  (5x3 + 4x23x + 5) 11. (8x  3x3 + 4x4)  (6x3 + 3x  2) 12. m42m + 4(m3 + m24m + 2) 13. x3y2(3xy2 + 2x  3y  4y3) 14. 2ab(4b2 + 3b  2ab3)  6ab(2b2 + 3ab35b) Find the missing factor(s): 15. (4x2)(?) = 12x6 16. (?)(4a2b5) = 28a5b7 Factor completely: 17. 3x2 + 6x a512a3 + 6a x3 + 7x214x r816r6 + 64r4 21
28 Lesson 5.1 Polynomial Products Unit 5 Rule: The Distributive Property makes it possible to multiply two binomials in the same way a monomial is multiplied by a binomial. Examples: 1. Multiply: 2x(4x + 3) =. Monomial x Binomial. 2x. 4x + 2x. 3 =. Distribute 2x. 8x2 + 6x. Multiply. 2. Multiply: (2x + 1)(4x + 3) =. Binomial x Binomial. If a = 2x + 1, this problem becomes: a(4x + 3) =. Substitute "a" for 2x + 1. a(4x) + a(3) = 4ax + 3a In the same way, (2x + 1)(4x + 3) =. Distribute "a".. Multiply. (2x + 1)4x + (2x + 1)3 =. Distribute 2x + 1. (2x)4x + (1)4x + (2x)3 + (1)3 =. Distribute 4x and 3. 8x2 + 4x + 6x + 3 8x2 + 10x + 3 =. Multiply.. Combine like terms. Rule: The FOIL method makes multiplication easier. It suggests multiplying First terms, Outer terms, Inner terms, and Last terms. Examples: 3. Multiply: (2x + 1)(4x + 3) 2x. 4x = 8x2 2x. 3 = 6x. First terms.. Outer terms. 1. 4x = 4x. Inner terms = 3. Last terms. 8x2 + 10x + 3. Combine like terms. 22
29 Lesson Multiply: (5m  4p)(2m + 3p) 5m. 2m = 10m2. First terms. 5m. 3p = 15mp. Outer terms. 4p. 2m = 8mp. Inner terms. 4p. 3p = 12p2. Last terms. 10m2 + 7mp  12p2. Combine like terms. Practice: Multiply: A. (2x + 3)(3x + 4) B. (4x  1)(4x + 1) Rule: The FOIL method may be used to square a binomial. First, write the power in expanded form, then multiply using FOIL. Example: 5. Multiply: (3x + 2)2 (3x + 2) (3x + 2). Product of two binomials. 9x2 6x 6x. F. O. I 4. L 9x2 + 12x + 4. Combine like terms. Example: A quick method of squaring a binomial: 6. Multiply: (4x  5y)2 (4x)2 16x2 + 2(4x) (5y) 16x2 + 2(20xy) + (5y)2 16x240xy + 25y2. Square the first term.. Double product of the two terms.. Square the last term.. Simplify each term. 23
30 Lesson 5.1 Practice: Square each binomial: C. (x + 4)2 D. (2x  3)2 Rule: The Distributive Property allows multiplication of any two polynomials. In order, multiply the first polynomial by each term of the second polynomial. Examples: 7. Multiply: (2x  1)(3x22x + 5) (2x  1)(3x2) = 6x33x2. Multiply polynomial by 3x2. (2x  1)(2x) = 4x2 + 2x. Multiply polynomial by 2x. (2x  1)(5) = 10x  5. Multiply polynomial by 5. (6x33x2)+(4x2+2x)+(10x5) 6x37x2 + 12x Multiply: (5x + 3)(x2 + 4x  2). Add the resulting products.. Combine like terms. (5x + 3)(x2) = 5x3 + 3x2. Multiply polynomial by x2. (5x + 3)(4x) = 20x2 + 12x. Multiply polynomial by 4x. (5x + 3)(2) = 10x  6. Multiply polynomial by 2. (5x3+3x2)+(20x2+12x)+(10x6) 5x3 + 23x2 + 2x  6. Add the resulting products.. Combine like terms. Rule: For a simplified method of multiplying two polynomials, multiply each term of one polynomial by each term of the other polynomial; then combine like terms. Example: 9. Multiply: (5x + 3)(x2 + 4x  2). Same as Example 8. 5x(x2) + 5x(4x) + 5x(2) +. Multiply each term by 5x. 3(x2) + 3(4x) + 3(2) =. Multiply each term by 3. 5x3 + 20x2 + 3x210x + 12x  6 = 5x3 + 23x22x  6. Simplify.. Combine like terms. 24
31 Lesson 5.1 HOMEWORK Multiply: 1. (3x + 1)(x + 2) 11. (r  2p)2 2. (x + 4)(2x  3) 12. (3r  4x)2 3. (2a + 1)(3a + 5) 13. (2a  3b)(4a + 5b) 4. (r  3)(r + 3) 14. (2x + y)(3x  2y) 5. (4m + 5)(4m  5) 15. (2p  3)(p + 1) 6. (2a  1)(3a  2) 16. (2r  3)(2r  3) 7. (5m + 2)(5m + 2) 17. (2x  3y)(2x + 3y) 8. (2p + 1)2 18. (3a  1)(2a2 + 5a  7) 9. (3r  4)2 19. (2x + 3)(x24x  5) 10. (m  3)2 20. (3m + 2)2(2m  1) 25
32 Lesson 5.2 Factoring x 2 + bx + c Unit 5 Rule: Multiplying two binomials often results in a trinomial. Factoring such a trinomial reverses the process to obtain the two binomials. For example, since (x + 3)(x + 4) = x2 + 7x + 12, then the factored form of x2 + 7x + 12 is (x + 3)(x + 4). Example: 1. Factor: x2 + 5x + 6 x2 + 5x + 6 = (x )(x ). Factor the first term. x2 + 5x + 6 = (x +?)(x +?) 3x. Factor the last term.. Think of the integer pairs whose product is 6: 1. 6; ; 2. 3; Select the integer pair x2 + 5x + 6 = (x + 2)(x + 3) 2x whose sum is 5: = 5. Thus, x2 + 5x + 6 = (x + 2)(x + 3). 2. Factor: x25x + 6 x25x + 6 = (x )(x ). Factor the first term. x25x + 6 = (x +?)(x +?) 3x. Factor the last term. Think of the integer pairs whose product is 6: 1. 6; ; 2. 3; Select the integer pair x25x + 6 = (x  2)(x  3) 2x whose sum is 5: 2 + (3) = 5. Thus, x25x + 6 = (x  2)(x  3). Note: When the third term is positive, (+), the two factors must have the same sign as the middle term. Example: 3. Factor: x2 + 5x  24 x2 + 5x  24 = (x )(x ). Factor the first term. x2 + 5x  24 = (x +?)(x +?) 3x. Factor the last term. Think of the integer pairs whose product is 24: ; ; ; ; 8. 3; 8. 3; 6. 4; Select the integer pairs whose x2 + 5x  24 = (x + 8)(x  3) 8x sum is +5: 8 + (3 )= +5. Thus, x2 + 5x  24 = (x + 8)(x  3). 26
33 4. Factor: x25x  24 x25x  24 = (x )(x ). Factor the first term. x25x  24 = (x +?)(x +?) Lesson 5.2. Factor last terms. Write the integer pairs whose product is 24: ; ; ; ; 8. 3; 8. 3; 6. 4; 3x Select the integer pair whose x25x  24 = (x  8)(x + 3) 8x sum is 5: = 5. Thus, x25x  24 = (x  8)(x + 3). Note: When the third term is negative, (), the two factors must have opposite signs, and the factor with the larger absolute value will have the same sign as the middle term. Practice: A. Factor: m22m  15 HOMEWORK Factor each trinomial into two binomials: 1. x2 + 6x r25r a2 + 12a x2 + 10x m2 + 8m a2  a g2 + 4g b25b x24x x22x r212r m2 + 6m n26n x212x m2  m x29x a2 + 7a m211m x23x z212z c2 + 4c a26a m2 + 9m x2a + 10xa d215d
34 Lesson 5.3 Factoring ax 2 + bx + c and ax 2 + bxy + cy 2 Unit 5 Rule: The same trial and error process used to factor trinomials of the form x2 + bx + c (with an x2 coefficient of 1) is also used to factor trinomials in which the x2 coefficient is not 1: ax2 + bx + c, where "a" is a positive integer. Example: 1. Factor: 3x214x x214x + 15 = (3x )(x ) 3x214x + 15 = (3x?)(3x?) 3x214x + 15 = (3x  5)(x  3) Thus, 3x214x + 15 = (3x 5)(x  3). Practice: A. Factor: 5x213x + 6. Factor the first term. (3 is prime, so the only possible factors are 3x and x.). Factor the last term.. Write the integer pairs whose product is 15: 15. 1; ; 9x 5. 3; Select the integer pairs whose  5x Inner and Outer products have a sum of 14x: 5x + 9x = 14x. Rule: If the x2 coefficient is not prime, different combinations of it may also need to be tried. Example: 2. Factor: 6k2 + 11k Note negative third term. 6k2 + 11k  10 = (6k )(k ) 6k2 + 11k  10 = (6k?)(k?). Factor the first term. Possible factors are 6k.k or 3k. 2k. Try 6k. k first.. Factor the last term. Try (6k + 10)(k  1) factors of 10 whose Inner and (6k  1)(k + 10) Outer products have a sum of (6k  5)(k + 2) 11k. None of these works. (6k  2)(k + 5) Since the above combinations do not give a middle term of +11k and reversing the signs gives a negative middle term, try combinations using 3k and 2k as factors of 6k2. 28
35 6k2 + 11k  10 = (3k?)(2k?) Lesson 5.3 (3k 1)(2k +10). Again, try factors of 10 whose (3k+10)(2k 1) Inner and Outer products have (3k + 5)(3k  2) a sum of +11k. (3k  2)(2k + 5). This combination works. So, the factorization of 6k2 + 11k  10 is (3k  2)(2k + 5). Practice: B. Factor: 4x2 + 4x  15 Rule: With practice, the correct combination of factors may be found without trying every possibility. Think of all possible factors of first and last terms; the signs of the middle and last terms may eliminate half of the combinations. Example: 3. Factor: 18x29x  20 Some possible factors: Middle term: (18x  20)(x + 1) 2x (18x + 10)(x  2) 26x (9x  10)(2x + 2) 2x (9x + 5)(2x  4) 26x (6x  10)(3x + 2) 18x (6x  20)(3x + 1) 54x (6x + 4)(3x  5) 18x (6x + 5)(3x  4) 9x Correct term. Although other possible factors could be tried, since 9x is the correct middle term, 18x29x  20 = (6x + 5)(3x  4). 4. Factor: 16x2 + 14x  15 Some possible factors: Middle term: (16x  15)(x + 1) x (16x  1)(x + 15) 239x (16x  5)(x + 3) 43x (16x  3)(x + 5) 77x (8x + 15)(2x  1) 22x (8x  1)(2x + 15) 118x (8x  5)(2x + 3) 14x Correct term. So, 16x2 + 14x  15 = (8x  15)(2x  3). 29
36 Lesson 5.3 Rule: Factoring a polynomial with two variables is done in the same way a polynomial with one variable is factored: Factor out the GCF; then look for combinations of factors of the first and last terms that give the correct middle term. Examples: 5. Factor completely: 40r314r2t  12rt2 40r314r2t  12rt2 2r(20r27rt  6t2). 2 and r are both common factors.. 2r is the GCF. 2r(4r )(5r ). Try 4r and 5r as factors. 2r(4r  3t)(5r + 2t) 15rt. Try combinations of factors of 6t2: 3t and + 2t are correct.. 15rt + 8rt = 7rt, the correct middle term. + 8rt So, the complete factorization of 40r314r2t  12rt2 is 2r(4r  3t)(5r + 2t). Notice that to have a middle term with both variables, an "r" is kept with each factor of 20r2 and a "t" is kept with each factor of 6t2. 6. Factor completely: 12k275m2 12k275m2. 3 is a common factor. 3(4k225m2). 3 is the GCF. 3[(2k)2  (5m)2]. 4k225m2 is a difference of two squares. 3(2k + 5m)(2k  5m). A difference of two squares factors into a sum and a difference. +10km. 10km  10km = 0; Inner and Outer products must add to km So, the complete factorization of 12k275m2 is 3(2k + 5m)(2k  5m). 7. Factor completely: 405x580xy4 405x580xy4 5x(81x416y4) 5x[(9x2)2  (4y2)2] 5x(9x2 + 4y2)(9x24y2) 5x(9x2 + 4y2)(3x+ 2y)(3x 2y). 5 and x are common factors.. 5x is the GCF.. 81x4 and 16y4 are perfect squares.. A difference of two squares factors into a sum and a difference.. 9x24y2 is a difference of two squares, but 9x2 + 4y2 is not. So, the complete factorization of 405x580xy4 is 5x(9x2 + 4y2)(3x + y)(3x  y). 30
37 HOMEWORK Lesson 5.3 Factor each trinomial into two binomials: 1. 3x2 + 11x m2 + m x217x y28y r2 + 5r n213n p2 + 9p m215m m2  m a  5a2 Factor completely, if possible: 11. 3x2 + 3xy  18y2 12. m24p a2  r2 14. r212rx + 35x m2 + 20mp + 18p a216r x281y p216q k2 + 13km  5m a23ab  b g2 + 4gf  30f a29ab  15b cd324c3d m3n + 21m2n29mn a3b + 7a2b220ab3 31
38 Lesson 5.4 Factoring Special Types Unit 5 Rules: Some polynomials have a special pattern. Recognition of this pattern makes them easier to factor. This lesson shows two of these special cases: perfect trinomial squares and the difference of two squares. Trinomials of the type a2 + 2ab + b2 or a22ab + b2 are called perfect trinomial squares. Factoring a perfect trinomial square gives the square of a binomial. a2 + 2ab + b2 = (a + b)2 a22ab + b2 = (a  b)2 Examples: 1. Factor: x2 + 10x + 25 x2 + 2(5x) + 52 x2 + 2(5x) Can it be written in the a 2 + 2ab + b2 form?. First and last terms are perfect squares.. Middle term is twice the product of x and 5. (x + 5)2. Square of a binomial sum. So, x2 + 10x + 25 = (x + 5)2. 2. Factor: 9x x + 4 (3x) 2 + 2(3x)(2) (3x) 2 + 2(3x)(2) (3x + 2) 2. Answer. 3. Factor: 25x 230x + 9 (5x) 22(5x)(3) Write it in a2 + 2ab + b2 form.. First and last terms are perfect squares.. Middle term is twice the product of 3x and 2.. Rewrite. Note negative middle term. (5x  3) 2. Answer. Note negative sign. 4. Factor: 16x 240xy + 25y 2 (4x) 22(4x)(5y) + (5y) 2. Rewrite. Note negative middle term. (4x  5y) 2. Answer. Square of a difference. 32
39 Practice: A. x 214x + 49 B. 25x x + 81 Lesson 5.4 Rule: Factoring the difference of two squares gives a product of two binomials. The binomial is a sum and a difference of the square roots of the two squares: a2  b2 = (a + b)(a  b) Examples: 5. Multiply: (x + 2)(x  2). Use FOIL. x24. First and last products give x24. (x + 2) (x  2) 2x 2x. Inner and Outer products give 2x  2x = 0x. So, the product is a binomial, the difference of two squares: (x + 2)(x  2) = x Factor: x Factor: x236y2 (x )(x ). Factor first term. (x + )(x  ). Positive. negative = negative. (x + 2)(x  2). Square root of last term is 2. (x )(x ). Factor first term. (x + )(x  ). Positive. negative = negative. (x + 6y)(x  6y). Square root of last term is 6y. Practice: C. Factor: 16m29r2 D. Factor: 1.69a249b2 33
40 Lesson 5.4 Factor the perfect trinomials: 1. x214x x2 + 16x y26y a212a x216x x2 + 14x x2 + 2x t2 + 18t t26t n2 + 4nt + t d210d x4 + 2x y410y2x + x c2 + 16c + 4 HOMEWORK 34
41 Lesson 5.4 Factor the difference of two squares: 15. x r x m x2 20. x m r n x p n r21.21t x t m
42 Lesson 5.5 Factoring Combined Types Unit 5 Rules: In order to factor some polynomials completely, it may be necessary to use more than one type of factoring. To factor a polynomial completely: First, factor out the GCF other than 1, if any exists. Then factor the remaining polynomial, if possible, until it factors no further. Examples: 1. Factor completely: 3x29x x29x Look for common factors. 3(x23x  10). 3 is the GCF. 3(x )(x ). Factor x23x  10 into two binomials, if possible. 3(x  5)(x + 2). Try combinations of factors of 10: 5 and + 2 are correct.  5x +2x. 5x + 2x = 3x, the correct term. So, the complete factorization of 3x24x  30 is 3(x  5)(x + 2). 2. Factor completely: 30a3 + 22a228a 30a3 + 22a228a. Look for common factors. 2a(15a2 + 11a  14). 2a is the GCF. 2a( )( ). Factor the trinomial into two binomials, if possible. 2a(5a )(3a ). Try 5a and 3a as factors. 2a(5a + 7)(3a  2). Try combinations of factors of 14: +7 and 2 are correct. + 21a  10a. 21a  10a = 11a, the correct term. So, the complete factorization of 30a322a228a is 2a(5a + 7)(3a  2). Practice: A. Factor completely: 5r235r
43 Examples: 3. Factor completely: 5x3125x 5x3125x 5x(x225). Look for common factors.. 5x is the GCF. Lesson 5.5 5x( + )(  ). x225 factors into a sum and a difference. 5x(x + 5)(x  5). Completely factored form. 4. Factor completely: 1622y y4. Look for common factors. 2(81  y4). 2 is the GCF. 2[92  (y2)2] y4 is a difference of two squares: y4 is the square of y2. 2(9 + y2)(9  y2). A difference of two squares factors into a sum and a difference. 2(9 + y2)(3 + y)(3  y). 9  y2 is a difference of two squares, but 9 + y2 is not. So, the complete factorization of 1622y4 is 2(9 + y2)(3 + y)(3  y). Factor completely, if possible: HOMEWORK 1. 3x2 + 15x p24p 2. 2a2 + 6a m33m230m 3. 5r a3 + 26a26a 4. 2m34m26m a36a 5. 2c33c2  c 15. 4m2 + 7m x39x 16. 2k27k x32x240x 17. 5a320a 8. 3a3 + 15a2 + 12a x39x212x 9. x25x n384n2 + 39n 10. 4k316k2 + 12k r
44 Lesson 5.6 Factoring by Grouping Unit 5 Rule: Some polynomials may be factored by grouping two (or more) terms having a common factor and then factoring out all common factors, first from each group and then from the whole polynomial. Examples: 1. Factor: 2x + px. x is a common factor. (2 + p)x. Factored form; common factor on right. 2. Factor: 2(x + 1) + p(x + 1). (x + 1) is a common factor. (2 + p)(x + 1). Factored form; common factor on right. 3. Factor: a(x + 1)  b(x + 1). (x + 1) is a common factor. (a  b)(x + 1). Factored form; common factor on right. 4. Factor: (2x + 3)3m + (2x + 3)5r. (2x + 3) is a common factor. (2x + 3)(3m + 5r). Factored form; common factor on left. 5. Factor: (3x + 2y)2a  7b(3x + 2y). (3x + 2y) is a common factor. (3x + 2y)(2a  7b). Factored form; choose either side for common factor. 38
45 Practice: Factor out the common factor: A. 5(r  3) + m(r  3) Lesson 5.6 B. 2a(x + 7)  b(x + 7) Examples: 6. Factor: 2x px + p (2x + 2) + (px + p). Group so terms in each group have a common factor: 2 and p. 2(x + 1) + p(x + 1). Factor the GCF from each group; (x + 1) is now a common factor. (2 + p)(x + 1). (x + 1) is factored out; compare to Example 2 above. 7. Factor: 6mx + 10rx + 9m + 15r (6mx + 10rx) + (9m + 15r) 2x(3m + 5r) + 3(3m + 5r). Group terms with common factors.. Factor each group. (2x + 3)(3m + 5r). Factored form; (3m + 5r) is a common factor. 8. Factor: 6ax + 4ay  21bx  14by (6ax + 4ay) + (21bx  14by). Show addition of the groups when third term is negative. 2a(3x + 2y)  7b(3x + 2y). Negative common factor (7b) in second group. (2a  7b)(3x + 2y). Factored form; common factor is (3x + 2y). Here's another way to factor 6ax + 4ay  21bx  14by: (6ax  21bx) + (4ay  14by) 3x(2a  7b) + 2y(2a  7b). Rearrange so third term is positive; then group.. Extract common factors from each group. (2a  7b)(3x + 2y). Factored form; common factor is (2a  7b). 39
46 Lesson 5.6 Practice: Factor: C. 7ax + 3ay  7bx  3by Rule: After factorization by grouping, be sure to check each resulting factor for any possible further factorization. Example: 9. Factor: 4a2m + 4a225m  25 (4a2m + 4a2) + (25m  25). Write as a sum of two groups (binomials). 4a2(m + 1)  25(m + 1). Common factors are 4a2 and 25. (4a225)(m + 1). (m + 1) is a common factor. (2a + 5)(2a  5)(m + 1). 4a225 is a difference of two squares. Since no factor will factor further, the complete factorization of 4a2m + 4a225m  25 is (2a + 5)(2a  5)(m + 1). 40
47 Lesson 5.6 HOMEWORK Use grouping techniques to factor each polynomial completely, if possible: 1. (c + d)x + (c + d)y 2. 2(m  n) + p(m  n) 3. (x + 5)a2 + 3(x + 5) 4. (c225)9  d2(c225) 5. k(64  a2) + (64  a2)m 6. 81(r29)  m4(r29) 7. ak + am + bk + bm 8. xc  xd + 4c  4d 9. xy + 2x + ry + 2r 10. x2 + 5x + xy + 5y 11. r25r + 3r c22c + cd  2d 13. 2am + 5ak + 2bm + 5bk 14. cx + 6c + x2 + 6x 15. m2 + 3m  km  3k 16. r2c + r2d  9c  9d x g2x  g2 18. m34mn2 + 2m28n2 19. x2t + 2x2100t b481a2b24b a2 41
48 Lesson 5.7 Solving Quadratic Equations by Factoring Unit 5 Rules: A quadratic equation is an equation of the form ax2 + bx + c = 0. (Notice that one member is a second degree polynomial and the other member is 0.) Zero Product Property: If x. y = 0, then either x = 0, y = 0, or both x and y are 0. If the polynomial member can be factored, then the quadratic equation may be solved by first factoring it and then applying the Zero Product Property. Example: 1. Solve: x(x + 2) = 0. The product equals 0. x = 0 or x + 2 = 0. Set each factor equal to 0. x = 0 or x = 2. Solve each equation for x. Check: x(x + 2) = 0 x(x + 2) = 0 Let x = 0. Let x = 2. 0(0 + 2) = 02(2 + 2) = 0 0(2) = 02(0) = 0 0 = 0 0 = 0 TRUE, so 0 is a solution. TRUE, so 2 is a solution. The solutions of x(x + 2) = 0 are 0 and 2. Rule: For each quadratic, or second degree equation, two roots, or solutions, must be found. Example: 2. Solve: x212x + 35 = 0 (x  7)(x  5) = 0. Factor. x  7= 0 or x  5 = 0. Set each factor equal to 0. x= 7 or x = 5. Solve each equation. Check: x212x + 35 = 0 x212x + 35 = 0. Try (7) + 35= (5) + 35 = 0. Try = = 0 0= 0 0= 0 TRUE TRUE So, solutions are 7 and 5. Practice: A. Solve: x25x  14 = 0 42
49 Lesson 5.7 Note: Sometimes a double root or multiple root occurs in a quadratic, cubic, or higher degree equation. The degree of an equation tells the maximum number of roots of the equation. Example: 3. Solve: 4x340x x = 0. Third degree (cubic) equation. 4x(x210x + 25) = 0. GCF is 4x. 4x(x  5)(x  5) = 0. Completely factored form. 4x = 0 or x  5 = 0 or x  5 = 0. 4 cannot equal 0. x = 0 or x = 5 or x = 5. Solve each equation and check. Solutions are 0 and 5. Notice that 5 appears twice as a root. For the cubic equation, 4x340x = 0, the three roots are: 0 and the double root, 5. Practice: B. Solve: 6x3 + 16x26x = 0 Example: 4. Solve: 15x2 + 14x  8 = 0 (3x + 4)(5x  2) = 0. Factor the trinomial. 3x + 4 = 0 or 5x  2 = 0. Let each factor equal 0. 3x = 4 or 5x = 2. Solve each equation. x = 4/3 or x = 2/5. Solutions are rational. Both roots satisfy the original equation, so the solutions to 15x2 + 14x  8 = 0 are 4/3 and 2/5. Rule: Generally, a quadratic equation is easier to solve if it is first written in standard form. A quadratic equation is in standard form when written: ax2 + bx + c = 0, where: 1. The polynomial equals Terms of the polynomial are arranged in descending powers of the variable. 3. a, b, and c, are integers, with "a" positive. Example: 5. Write 3n = 64n2 in standard form. Since the n2 term is negative (4n2), make the right member 0. 4n2 + 3n = 6 4n2 + 3n  6 = 0. Add 4n2 to each side.. Subtract 6 on each side to obtain standard form. 43
50 Lesson 5.7 Practice: Example: Write each equation in standard form: C. 5x2 + 2 = 4x D. 3 = 4a  2a2 E. 3m = 5m Solve: 5x = 6 + x2 Since the x2 term is positive, make the left member 0. 0 = 6 + x25x. Subtract 5x from each side. 0 = x25x + 6. Arrange in descending powers of x. 0 = (x  3)(x  2). Factor. x  3 = 0 or x  2 = 0. Let each factor be 0. x = 3 or x = 2. Solve each equation. Check: 5x = 6 + x2 5x = 6 + x2 Try 3. 5(3) = 6 + (3)2 5(2) = 6 + (2)2 Try = = = 15; TRUE 10 = 10; TRUE So, the solutions are 3 and 2. Practice: Write in standard form; then solve each: F. 2 = x23x G. 7  x2 = 6x Example: 7. Solve: 49x = 2x2 2x29x + 4 = 0 (2x  1)(x  4) = 0. Factor.. Add 2x2 to each side and write in standard form. 2x  1 = 0 or x  4 = 0. Let each factor be 0. 2x = 1 or x = 4. Solve each equation. x = 1/2 or x = 4. Check; both equations are true. So, the solutions are 1/2 and 4. Rule: Some word problems result in a quadratic equation. The same fourstep method for problemsolving is used to solve such word problems. 44
51 Lesson 5.7 Example: 8. Eight less than 6 times a number is equal to the square of the number. Find the number: 8 less than 6x is x2. Choose a variable. Let x be the number. 6x  8 = x2 6x  8 = x2. Plan and write an equation.. Solve. 0 = x26x + 8. Write in standard form. 0 = (x  4)(x  2). Factor. x  4 = 0 or x  2 = 0. Let each factor be 0. x = 4 or x = 2. Solve each equation. Check both roots in the equation. Then check them with the conditions of the original problem. The roots of this equation satisfy the conditions stated, so 4 and 2 are both solutions to the problem. HOMEWORK Use the zero product property and solve each equation by factoring: 1. a24a = r27r = 0 2. x2 + 8x  9 = c2 + 13c  24 = t213t  10 = m2 + 5m  42 = x222x + 7 = x2 + 23x + 56 = a211a + 5 = r3245r = x2400 = g3  g220g = 0 Solve each equation: 13. x2 + 8 = 6x 18. 8y + y = x2 = 3x 19. 8x = 215x2 15. x2 = 9x 20. 9x + x2 = x2 = x 21. 4m = m x  6x2 = x  14 = 6x2 Write an equation; then solve each problem: 23. Five times a number, decreased by 6, is the same as the square of the number. Find the number. 24. The sum of 6 times a number and the square of the number is 16. Find the number. 25. Twelve less than 9 times a number is the same as the square of the number increased by two. Find the number. 45
52 Lesson 5.8 Consecutive Integer Problems Unit 5 Rule: Consecutive integers are obtained by starting with a given integer and counting by one's. Examples: List four consecutive integers beginning with: 1. 3 : 3, 4, 5, : 2, 1, 0, 1 3. x: x, x+1, x+2, x+3 Practice: List four consecutive integers beginning with: A. 19 : B. 7 : C. m : Rule: Even integers are divisible by 2. Consecutive even integers are obtained by starting with an even integer and counting by two's. Consecutive odd integers are not divisible by 2 but are also obtained by counting by two's starting with an odd integer. Examples: List four consecutive even integers beginning with: 4. 6: 6, 8, 10, 12. Add 2 each time : 4, 2, 0, 2. Notice: Zero is even. 6. x: x, x+2, x+4, x+6. If x is even. List four consecutive odd integers beginning with: : 11, 13, 15, 17. Add 2 each time : 1, 1, 3, 5. Add 2 each time. 9. x : x, x+2, x+4, x+6. If x is odd. 46
53 Lesson 5.8 Practice: List: D. 3 consecutive even integers, beginning with 28. E. 3 consecutive odd integers, beginning with c. F. 3 consecutive even integers, beginning with m. Rule: A consecutive integer problem may be solved using the four basic steps of problem solving. Example: 10. Find three consecutive integers whose sum is 66. Let x = the first Let x + 1 = the second Let x + 2 = the third. Choose a variable; then represent the integers. x + x x + 2 = 66. Plan and write an equation.. 3x + 3 = 66 3x = 63 x = 21 x + 1 = 22 x + 2 = 23. Solve the equation.. First integer.. Second integer.. Third integer = 66. Check: The roots of the equation satisfy the original problem. So, the solution is: 21, 22, 23. Rule: It is necessary to check the roots of an equation to see that they satisfy the conditions of the original problem. Consecutive integer problems may result in equations that give two sets of roots; however, one, none, or both sets of roots may be a solution to the problem. 47
54 Lesson 5.8 Examples: 11. Find two consecutive odd integers whose product is 195. Let x = first Let x + 2 = second. Choose a variable; then represent each integer. x(x + 2) = 195 x2 + 2x = 195. Plan and write the equation.. Solve the equation; quadratic may have 2 solutions. x2 + 2x = 0 Rewrite in standard form. (x  13)(x + 15) = 0 Factor. x  13 = 0 or x + 15 = 0 Let each factor be 0. x = 13 or x = 15 First integer: 13 or 15. x + 2 = 15 or x + 2 = 13 Second integer: 15 or 13. (13)(15) = 195. Check both pairs of (15)(13) (15)(13) = 195 consecutive odd integers. So, there are two solutions: 13 and 15; 15 and Find three consecutive even integers such that the square of the second decreased by 5 times the square of the first is 16 less than twice the third. Let x = first Let x + 2 = second Let x + 4 = third. Choose a variable; then represent each integer. (x+2)25x2 = 2(x+4) Plan and write the equation.. Solve the equation: x2 + 4x + 45x2 = 2x x2 + 4x + 4 = 2x  8 Simplify each side. 4x2 + 2x + 12 = 0 Add 2x and 8 to each side. 2x2  x  6 = 0 Divide each side by 2. (2x + 3)(x  2) = 0 Factor. 2x + 3 = 0 or x  2 = 0 Let each factor be 0. 2x = 3 or x = 2 Solve each equation. x = 3/2 or x = 23/2 is not an integer.. Check: Only these three consecutive Try: x = 2 = First x + 2 = 4 = Second even integers satisfy the x + 4 = 6 = Third problem. So, there is only one solution: 2, 4, and 6. 48
55 Lesson 5.8 HOMEWORK Write an equation and solve each problem: 1. Find three consecutive integers whose sum is Find three consecutive integers whose sum is Find three consecutive even integers whose sum is Find three consecutive odd integers whose sum is Find five consecutive integers such that four times the third, increased by 15, is 5 times the last. 6. Find four consecutive odd integers such that the first equals the product of the second and fourth. 7. Find two consecutive even integers whose product is Find two consecutive odd integers such that the square of the second, increased by the first is Find two consecutive even integers such that twice the square of the second increased by the first is Find two consecutive integers such that the sum of their squares is Find four consecutive integers such that the difference of the squares of the second and fourth is Find three consecutive odd integers such that the square of the third decreased by the square of the first is 8 times the second. 13. Find three consecutive integers such that the square of the first is 15 less than the square of the second. 14. Find three consecutive even integers such that the product of the first and third is two more than 5 times the second. 15. Find four consecutive odd integers such that the square of the sum of the first and fourth equals the square of the sum of the second and third. 49
56
57 UNIT 6 MULTIPLYING AND DIVIDING RATIONAL EXPRESSIONS 6.0 Review Rational Expressions Rational Expressions: Simplest Form Using 1 in Factoring Dividing Powers Simplifying a Product Multiplying and Dividing Dividing by a Monomial Dividing by a Binomial
58 Lesson 6.0 Review Unit 6 Multiplying Binomials: FOIL Method  Multiply First, Outer, Inner, and Last terms. first last 3 6x2 (3x + 1) (2x  3) = (3x + 1)(2x  3) = 6x27x  3 inner 2x outer 9x Multiplying Polynomials: Multiply each term of one by each term of the other. (2x + 3)(4x2 + 2x + 1) = 8x3 + 4x2 + 2x + 12x2 + 6x + 3 = 8x3 + 16x2 + 8x + 3 Factoring: Quadratic Trinomials: third term 3x214x + 15 = (3x  5)(x  3). Both negative. positive 3x2 + 14x + 15 = (3x + 5)(x + 3). Both positive. third 6x211x  10 = (2x  5)(3x + 2). Largest product is term negative. negative 6x2 + 11x  10 = (2x + 5)(3x  2). Largest product is positive. Special Cases: perfect a2 + 2ab + b2 = (a + b)2. square of trinomial a22ab + b2 = (a  b)2 a binomial square difference a2  b2 = (a + b)(a  b). binomial sum times of two binomial difference squares By Grouping: x2 + 3x  kx  3k (x2 + 3x)  (kx + 3k). Group. x(x + 3)  k(x + 3). Common factors are x and k. (x  k)(x + 3). Common factor is (x + 3). 52
59 Zero Product Property: Lesson 6.0 If x. y = 0, then x = 0, y = 0, or both x and y are 0. Solving Quadratic Equations: Solve: 14x = 612x2 12x214x  6 = 0. Write in standard form. 2(3x + 1)(2x  3) = 0. Factor nonzero member. 3x + 1 = 0 or 2x  3 = 0. Set each variable factor equal to zero. 3x = 1 2x = 3. Solve each linear equation formed. x = 1 x = 3. Check each solution. 3 2 Multiply: Review Problems 1. 5x(3x  4) 4. (2x  1)(2x  6) 2. (x + 4)(x  3) 5. (3x  4)2 3. (2x + 3)(3x + 4) 6. (3x + 2)(2x2  x + 3) Factor completely: 7. x2  x x2  y2 8. x2 + 9x x2y236p2 9. m26m a210ab + b m214m x4 + 2x r2 + 13r mr + 4mt  21pr  14pt Solve each equation: 17. 3x2 + 15x + 18 = r250 = x  2 = 4x x2 = 12x  30x3 53
60 Lesson 6.1 Rational Expressions Unit 6 Rule: A rational expression is of the form a where "a" and "b" are polynomials. b A rational expression a is undefined (or meaningless) for b any value of the variable which makes b = 0. Example: 1. Recall 16 = 8, a unique number, 2 because only 8. 2 = 16. But 16 = no unique number because 0 (no number). (0) = 16. And 0 = no unique number because 0 (any number). (0) = 0. Rule: To find the value(s) of the variable for which a rational expression is undefined, set the denominator equal to 0 and solve the resulting equation. Examples: 2. For what value of x is x  4 undefined? 3x  6 3x  6 = 0. Set the denominator equal to 0 3x = 6 and solve for x. x = 2. This value of x makes the denominator equal to 0. Thus, the fraction is undefined when x = 2. 54
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