1. Polynomials Interpret the Structure of an Expression (MCC9-12.A.SSE.1a.b) Perform Arithmetic Operations on Polynomials (MCC9-12.A.APR.1)

2. Adding, Subtracting, and Multiplying Polynomials

3. A monomial is the product of non-negative integer powers of variables. Consequently, a monomial has NO variable in its denominator. It has one term. (mono implies one) 13, 3x, -57, x², 4y², -2xy, or 520x²y² (notice: no negative exponents, no fractional exponents) A binomial is the sum of two monomials. It has two unlike terms. (bi implies two) 3x + 1, x² - 4x, 2x + y, or y - y² A trinomial is the sum of three monomials. It has three unlike terms. (tri implies three) x 2 + 2x + 1, 3x² - 4x + 10, 2x + 3y + 2

4. A polynomial is the sum of one or more terms. (poly implies many) x 2 + 2x, 3x 3 + x² + 5x + 6, 4x - 6y + 8 Polynomials are in simplest form when they contain no like terms. x 2 + 2x + 1 + 3x² - 4x when simplified becomes 4x 2 - 2x + 1 Polynomials are generally written in descending order. Descending: 4x 2 - 2x + 1 exponents of variables decrease from left to right

5. The degree of a monomial is the sum of the exponents of the variables in the monomial. The degree of a polynomial is the greatest degree of its terms. When a polynomial is written so that the exponents of a variable decrease from left to right, the coefficient of the first term is called the leading coefficient. Example: Write so that the exponents decrease from left to right. Identify the degree and the leading coefficient of the polynomial. Degree of the polynomial is 5. The leading coefficient is 13. Degree is 3 Degree is 1 Degree is 5

6. Monomials that have the same variable factors are called like terms. The simplest "like terms" are numbers (constants). Remember the "variable" in a constant term is represented as x 0, which is 1. 5 = 5x 0 = 51 = 5 The terms 4x² and 5x² are like terms because each term has x² as a common factor. Because these values are like terms, they can be combined by addition or subtraction. Simply add or subtract the numbers in front of the terms (called coefficients). 4x² + 5x² = 9x²

7. Add like terms by adding the numerical portion of the terms, following the rules for adding signed numbers. (The numerical portion of an expression is called the coefficient.) Example: Add: (2x 2 - 4) + (x 2 + 3x - 3) Below are several different ways to attack this example:

8. 1. Using a horizontal method to add like terms: Remove parentheses. Identify like terms. Group the like terms together. Add the like terms. (2x 2 - 4) + (x 2 + 3x - 3) = 2x 2 - 4 + x 2 + 3x - 3.. identify like terms = 2x 2 + x 2 + 3x - 4 - 3.. group the like terms together = 3x 2 + 3x - 7... add the like terms

9. 2. Using a vertical method to add like terms: Arrange the like terms so that they are lined up under one another in vertical columns, adding 0 place holders if necessary. Add the like terms in each column following the rules for adding signed numbers. 2x 2 + 0x - 4 + x 2 + 3x - 3 3x 2 + 3x - 7

10. 3. Using the Distributive Property to add like terms: When you are adding the coefficients (the numbers in front of the variables) of like terms together, you are actually using the distributive property in reverse. 4x 2 + 5x 2 = (4 + 5) x 2 = 9x 2

11. Subtract like terms by changing the signs of the terms being subtracted, and following the rules for adding polynomials. Example: Simplify: (2x 2 - 4) - (x 2 + 3x - 3) Below are several different ways to attack this example:

12. 1. Using a horizontal method to subtract like terms: Change the signs of ALL of the terms being subtracted. Change the subtraction sign to addition. Follow the rules for adding signed numbers. (2x 2 - 4) - (x 2 + 3x - 3) = (2x 2 - 4) + (-x 2 - 3x + 3)... change signs of terms being subtracted and change subtraction to addition. = 2x 2 - 4 + -x 2 - 3x + 3... identify like terms = 2x 2 - x 2 - 3x - 4 + 3... group the like terms = x 2 - 3x - 1... add the like terms

13. 2. Using the vertical method to subtract like terms: 2x² + 0x – 4 -(x²+ 3x - 3) Now, change signs of all terms being subtracted and follow rules for adding. 2x² + 0x - 4 -x² - 3x + 3 (signs changed) x² - 3x - 1

14. 3.Using the Distributive Property to subtract like terms: When you are subtracting the coefficients (the numbers in front of the variables) of like terms, you are actually using the distributive property in reverse. 4x 2 - 5x 2 = (4 - 5) x 2 = -x 2

15. Monomials can be multiplied times other monomials, times binomials, times trinomials, and times polynomials in general. Multiplying by a monomial may involve: 1. applying rules for dealing with exponents 2. using the distributive property 3. remembering rules for multiplying signed numbers

16. Tools for Dealing with Monomials Distributive Property a(b + c) = ab + ac Multiplying Signed Numbers (+) (+) = (+) (+) (-) = (-) (-) (+) = (-) (-) (-) = (+)

17. Rules for Exponents

18. Example 1: monomial monomial (4x 3 ) (3x 2 ) = (4 3) (x 3 x 2 ) = 12 x 5 Notice that the factors were regrouped and then multiplied. Also, multiply powers with the same base by adding the exponents.

19. Example 2: monomial binomial x(x + 4) = x 2 + 4x This problem requires the distributive property. You need to multiply each term in the parentheses by the monomial (distribute the x across the parentheses).

20. Example 3: monomial trinomial 2x(x 2 + 3x + 4) = 2x(x 2 ) + 2x(3x) + 2x (4) = 2x 3 + 6x 2 + 8x Notice the distributive property at work again. Example 4: monomial polynomial 3x 2 (x 3 - 3x 2 + 6x – 5) = 3x 2 (x 3 )- 3x 2 (3x 2 )+ 3x 2 (6x) – 3x 2 (5) = 3x 5 - 9x 4 + 18x 3 - 15x 2 Again, the distributive property is needed along with the rule for multiplying powers.

21. WARM-UP Add or subtract, then simplify by combining like terms. Remember: Like terms have the same variable raised to the same power. (8x 2 + 3x + 3) – (2x 2 + 7x – 5) (3x 2 – 6x + 2) + (-5x 2 + 2x – 8)

22. Practice – Multiplying A Polynomial by a Monomial Simplify. 1. 5(2a + 3b) 2. 3x(4x – 2y)3. - 4(a 2 – b 2 ) 4. 5. 6. 7. 3(x + 5) – 6 = 18

23. Warm-Up Complete page 4 of the packet you were given last Thursday. – For examples 1-6 in the middle of the page also write the degree of the polynomial and the leading coefficient of each. You will have a Formative Assessment tomorrow on: – Simplifying polynomials – Degrees of polynomials / leading coefficients – Adding and subtracting polynomials – Multiplying a monomial by a polynomial – Using GCF to factor polynomials – Review on Coolmath.com/algebra

24. Try These - Distribute each problem:

25. Using GCF to Factor Polynomials

26. GCF Method is just distributing backwards!!

27. What is the GCF of 25a 2 and 15a? 5a Let’s go one step further… 1) FACTOR 25a 2 + 15a. Find the GCF and divide each term 25a 2 + 15a = 5a( ___ + ___ ) Check your answer by distributing. 5a3

28. 2) Factor 18x 2 - 12x 3. Find the GCF 6x 2 Divide each term by the GCF 18x 2 - 12x 3 = 6x 2 ( ___ - ___ ) Check your answer by distributing. 32x

29. Warm-Up Use GCF to factor each polynomial. -8x(3x-1) or 8x(-3x+1)

30. 3) Factor 28a 2 b + 56abc 2. GCF = 28ab Divide each term by the GCF 28a 2 b + 56abc 2 = 28ab ( ___ + ___ ) Check your answer by distributing. 28ab(a + 2c 2 ) a2c 2

31. Factor 20x 2 - 24xy 1.x(20 – 24y) 2.2x(10x – 12y) 3.4(5x 2 – 6xy) 4.4x(5x – 6y)

32. Factor out the GCF for each polynomial: Factor out means you need the GCF times the remaining parts. a)2x + 4y b)5a – 5b c)18xy 2 – 6y d)2m 2 + 6m 3 n e)5x 2 y – 10xy 2(x + 2y) 6y(3xy – 1) 5(a – b) 5xy(x - 2) 2m 2 (1 + 3mn) Greatest Common Factors aka GCF’s How can you check?

33. 5) Factor 28a 2 + 21b - 35b 2 c 2 GCF = 7 Divide each term by the GCF 28a 2 + 21b - 35b 2 c 2 = 7 ( ___ + ___ - ____ ) Check your answer by distributing. 7(4a 2 + 3b – 5b 2 c 2 ) 4a 2 5b 2 c 2 3b

34. Factor 16xy 2 - 24y 2 z + 40y 2 1.2y 2 (8x – 12z + 20) 2.4y 2 (4x – 6z + 10) 3.8y 2 (2x - 3z + 5) 4.8xy 2 z(2 – 3 + 5)

35. Ex 1 15x 2 – 5x GCF = 5x 5x(3x - 1)

36. Ex 2 8x 2 – x GCF = x x(8x - 1)

37. Ex 3 8x 2 y 4 + 2x 3 y 5 - 12x 4 y 3 GCF = 2x 2 y 3 2x 2 y 3 (4y + xy 2 – 6x 2 )

38. Factoring Trinomials Writing Quadratic Trinomials as a Product of Two Binomials

39. The standard form of any quadratic trinomial is Standard Form a=3 b=-4 c=1

40. Now you try. a =b =c = a =b = c = a =b =c =

41. Factoring Trinomials when a=1 and c > 0 First list all the factor pairs of c. Then find the factors with a sum of b These numbers are used to make the factored expression. 1, 12 2, 6 3, 4

42. Now you try. Factors of c: 1 15 3 5 Factors of c: 1 21 3 7 Binomial Factors ( x + ) Circle the factors of c with the sum of b Binomial Factors ( x + ) 3 5 37

43. Example 1:x 2 + 11x + 24 Their sum equals the middle term of the trinomial. When factoring these trinomials the factors will be two binomials: (x + )(x + ) We know that the first terms of each binomial must be x because the first term of the trinomial is x 2 and x  x = x 2. The challenge is to find the last term of each binomial. They must be chosen so that they will cause the coefficient of the middle term and the last term of the trinomial to work out. (That’s 11 and 24 in this case.) += 11 Their product of those same numbers equals the last term of the trinomial.  = 24

44. List the factors of 24: Example 1: x 2 + 11x + 24 124 212 38 46 124SUM = 25 212SUM = 14 38SUM = 11 46SUM = 10 It is the factors 3 and 8 that produce a sum of 11 AND a product of 24 so they must be the last terms of each binomial. (x + 3)(x + 8)

45. Example 2:a 2 + 16a + 28 Factors of 28: 128 214 47 SUM = 16 a  a = a 2 so they are the first terms of each binomial and the factors 2 and 14 make a sum of 16 AND a product of 28 so the are the last terms of each binomial. = (a + 2)(a + 14)

46. Example 3:y 2 + 2y + 1 Factors of 1: 11 y 2 + 2y + 1 = (y + 1)(y + 1) Factors of 1: 11 Sometimes there is only 1 pair of factors to consider. Example 4:m 2 + 3m + 1 Factors of 1: 11 In this example the factors available do not make a sum of 3 which means that the trinomial can’t be factored. SUM = 2 SUM = 3

47. Example 5:p 2 + 23p + 120 Factors of 120: 1120 260 340 430 524 620 815 1012 1120 260 340 430 524 620 815 1012 = (p + 8)(p + 15) In this example there are many pairs of factors to consider. Most examples will have fewer than these. The trick is in being able to quickly find all of the factors of c. p 2 + 23p + 120 SUM = 23 Example 6:x 2 + 5x + 6 Factors of 6: 16 23 SUM = 5 Factors of 6: 16 23 = (x + 2)(x + 3)

48. Factoring when c >0 and b < 0. c is positive and b is negative. Since a negative number times a negative number produces a positive answer, we can use the same method as before but… The binomial factors will have subtraction instead of addition.

49. Example 7: 112 26 34 We need a sum of -13 Make sure both values are negative! First list the factors of 12

50. Now you try. (x – 3)(x – 4) (x – 2)(x – 7) (x – 6)(x – 7)

51. Factoring when c < 0. We still look for the factors of c. However, in this case, one factor should be positive and the other negative in order to get a negative value for c Remember that the only way we can multiply two numbers and come up with a negative answer, is if one is number is positive and the other is negative!

52. Example 11: 112 26 34 In this case, one factor should be positive and the other negative. We need a sum of -1 + -

53. Example 12: x 2 + 5x - 6 Factors of -6: +6 -2+3 SUM = 5 When looking for the factors of a negative number, one must be positive and the other negative. If at the same time their sum is positive, then the factor that is bigger must be the positive one. +6 -2+3 = (x - 1)(x + 6)

54. Example 13: 118 29 36 List the factors of 18. We need a sum of 3 What factors and signs will we use?

55. Now you try. 14. 15. 16. 17.

56. REVIEW OF RULES FOR SIGNS Sign of bigger number (+) + (+) = (+) (+) + (-) = (-) + (+) = (-) + (-) = (-) ( ) ADDITION (+)(+) = (+) (+)(-) = (-) (-)(+) = (-) (-)(-) = (+) MULTIPLICATION Example 18: x 2 - 5x - 6 Factors of -6: SUM = -5 +1-6 +2-3 = (x + 1)(x - 6) When both the product and sum are negative, the factors have opposite signs but this time the bigger factor will be negative.

57. Example 19: x 2 - 5x + 6 Factors of 6: SUM = -5 -6 -2-3 = (x - 2)(x - 3) When looking for factors of a positive number when the sum is negative, both factors will be negative. Example 20: x 2 - 5x - 36 Factors of -36: 1-36 2-18 3-12 4-9 6-6 SUM = -5 1-36 2-18 3-12 4-9 6-6 = (x + 4)(x - 9)

58. Prime Trinomials Sometimes you will find a quadratic trinomial that is not factorable. You will know this when you cannot get b from the list of factors. When you encounter this write not factorable or prime.

59. Here is an example… 118 29 36 Since none of the pairs adds to 3, this trinomial is prime.

60. Now you try. factorableprime factorableprime factorableprime

61. Factor each trinomial, if possible. Watch out for signs!! 1) t 2 – 4t – 21 2) x 2 + 12x + 32 3) x 2 –10x + 24 4) x 2 + 3x – 18 Factor These Trinomials!

62. Solution #1: t 2 – 4t – 21 1) Factors of -21: 1 -21, -1 21 3 -7, -3 7 2) Which pair adds to (- 4)? 3) Write the factors. t 2 – 4t – 21 = (t + 3)(t - 7)

63. Solution #2: x 2 + 12x + 32 1) Factors of 32: 1 32 2 16 4 8 2) Which pair adds to 12 ? 3) Write the factors. x 2 + 12x + 32 = (x + 4)(x + 8)

64. Solution #3: x 2 - 10x + 24 1) Factors of 24: 1 24 2 12 3 8 4 6 2) Which pair adds to -10 ? 3) Write the factors. x 2 - 10x + 24 = (x - 4)(x - 6) None of them adds to (-10). For the numbers to multiply to +24 and add to -10, they must both be negative! -1 -24 -2 -12 -3 -8 -4 -6

65. Solution #4: x 2 + 3x - 18 1) Factors of -18: 1 -18, -1 18 2 -9, -2 9 3 -6, -3 6 2) Which pair adds to 3 ? 3) Write the factors. x 2 + 3x - 18 = (x - 3)(x + 6)

66. Factor x 2 + 3x + 2 1.(x + 2)(x + 1) 2.(x – 2)(x + 1) 3.(x + 2)(x – 1) 4.(x – 2)(x – 1)

67. When a ≠ 1. Instead of finding the factors of c: Multiply a times c. Then find the factors of this product. 170 235 514 710

68. 170 235 514 710 We still determine the factors that add to b. So now we have But we’re not finished yet….

69. Since we multiplied in the beginning, we need to divide in the end. Divide each constant by a. Simplify, if possible. Clear the fraction in each binomial factor

70. Divide each constant by a. Simplify, if possible. Clear the fractions. Multiply a times c. List factors. Look for sum of b Write 2 binomials using the factors with sum of b

71. Factor 2x 2 + 9x + 10 1.(2x + 10)(x + 1) 2.(2x + 5)(x + 2) 3.(2x + 2)(x + 5) 4.(2x + 1)(x + 10)

72. Now you try. (2x + 3)(2x – 1) (x - 3)(3x + 4) (3x - 1)(2x – 7)

73. Sometimes there is a GCF. If so, factor it out first. Then use the previous methods to factor the trinomial

74. Now you try. 1. 2.

75. First factor out the GCF. Then factor the remaining trinomial.

76. Factor each trinomial, if possible. Watch out for signs!! 1) 2x 2 + x – 21 2) 3x 2 + 11x + 10 Factor These Trinomials!

77. Solution #1: 2x 2 + x - 21 1) Multiply 2 (-21) = - 42; list factors of - 42. 1 -42, -1 42 2 -21, -2 21 3 -14, -3 14 6 -7, -6 7 2) Which pair adds to 1 ? 3) Write the temporary factors. 2x 2 + x - 21 = (x - 3)(2x + 7) ( x - 6)( x + 7) 4) Put “2” underneath. 2 2 5) Reduce (if possible). ( x - 6)( x + 7) 2 2 3 6) Move denominator(s)in front of “x”. ( x - 3)( 2x + 7)

78. Solution #2: 3x 2 + 11x + 10 1) Multiply 3 10 = 30; list factors of 30. 1 30 2 15 3 10 5 6 2) Which pair adds to 11 ? 3) Write the temporary factors. 3x 2 + 11x + 10 = (3x + 5)(x + 2) ( x + 5)( x + 6) 4) Put “3” underneath. 3 3 5) Reduce (if possible). ( x + 5)( x + 6) 3 3 2 6) Move denominator(s)in front of “x”. ( 3x + 5)( x + 2)

79. 1. 2.

80. Warm-Up Factor each trinomial. 2x² + 15x + 25 x² + 5x + 4 2x² + 9x - 81

81. 2x² + 15x + 25 (2x + 5) (x + 5) x² + 5x + 4 (x + 1) (x + 4) 2x² + 9x - 81 (x + 9) (2x - 9)

82. The Product of Two Binomials is a Trinomial Using the FOIL method and box method

83. The FOIL method is ONLY used when you multiply 2 binomials. It is an acronym and tells you which terms to multiply. Use the FOIL method to multiply the following binomials: (y + 3)(y + 7).

84. (y + 3)(y + 7). F tells you to multiply the FIRST terms of each binomial. y2y2

85. (y + 3)(y + 7). O tells you to multiply the OUTER terms of each binomial. y 2 + 7y

86. (y + 3)(y + 7). I tells you to multiply the INNER terms of each binomial. y 2 + 7y + 3y

87. (y + 3)(y + 7). L tells you to multiply the LAST terms of each binomial. y 2 + 7y + 3y + 21 Combine like terms. y 2 + 10y + 21

88. Remember, FOIL reminds you to multiply the: F irst terms O uter terms I nner terms L ast terms

94. F.O.I.L F irst O uter I nner L ast ( x + 2 ) ( x + 5 )

95. EXAMPLES ( x + 4 ) ( x + 8 ) = ( x + 5 ) ( x – 6) = x2x2 + 8x+ 4x+ 32 x2x2 − 6x+ 5x− 30 x 2 + 12x + 32 x 2 + 12x + 32 x 2 − x − 30

96. PRACTICE ( x − 7 ) ( x − 4 ) = ( x + 10 ) ( x + 3 ) = x 2 + 3x + 10x + 30 x 2 + 13x + 30 x 2 + 3x + 10x + 30 x 2 + 13x + 30 x 2 − 4x − 7x + 28 x 2 − 11x + 28 x 2 − 4x − 7x + 28 x 2 − 11x + 28

97. You Try!!! 1.(x + 5)(x + 7) 2. (y + 9)(y – 6) 3. (n – 8)(n + 5) 4.(x - 4)(x - 3)

98. ( 2x 2 + 4 ) ( 3x − 5 ) = ( 3x 2 − 6x) (4x + 2) = EXAMPLES 6x 3 − 10x 2 + 12x− 20 12x 3 + 6x 2 − 24x 2 − 12x 12x 3 − 18x 2 − 12x

99. ( 4x 2 + 2 ) ( 2x + 8 ) = 8x 3 + 32x 2 + 4x + 16 (3x 2 − 7x ) ( x − 9 ) = 3x 3 − 27x 2 − 7x 2 + 63x 3x 3 −34x 2 + 63x 3x 3 − 27x 2 − 7x 2 + 63x 3x 3 −34x 2 + 63x PRACTICE

100. QUICK REVIEW o When multiplying binomials we can think of it as using the distributive property twice or by using the FOIL method. First Outer Inner Last

101. EXIT SLIP On a piece of paper answer the following questions: – How does the distributive property related to the FOIL method? – What does FOIL stand for? – ( x + 3 ) ( x − 8 ) = – ( 2x 2 − 7 ) ( 4x − 2 ) =

102. I can show multiplying polynomials with the BOX METHOD. OBJECTIVE

103. The 2nd method is the Box Method. This method works for every problem! (3x – 5)(5x + 2) Here’s how you do it: (3x – 5)(5x + 2) Draw a box. Write a polynomial on the top & side of a box. It does not matter where. This will be modeled in the next problem along with FOIL. 3x-5 5x +2

104. 3) Multiply (3x - 5)(5x + 2) First terms: Outer terms: Inner terms: Last terms: Combine like terms. 15x 2 - 19x – 10 3x-5 5x +2 15x 2 +6x -25x -10 You have 2 techniques. Pick the one you like the best! 15x 2 +6x -25x -10

105. 4) Multiply (7p - 2)(3p - 4) First terms: Outer terms: Inner terms: Last terms: Combine like terms. 21p 2 – 34p + 8 7p-2 3p -4 21p 2 -28p -6p +8 21p 2 -28p -6p +8

106. BOX Method YOU TRY

110. Multiply (y + 4)(y – 3) 1.y 2 + y – 12 2.y 2 – y – 12 3.y 2 + 7y – 12 4.y 2 – 7y – 12 5.y 2 + y + 12 6.y 2 – y + 12 7.y 2 + 7y + 12 8.y 2 – 7y + 12

111. Multiply (2a – 3b)(2a + 4b) 1.4a 2 + 14ab – 12b 2 2.4a 2 – 14ab – 12b 2 3.4a 2 + 8ab – 6ba – 12b 2 4.4a 2 + 2ab – 12b 2 5.4a 2 – 2ab – 12b 2

112. So, what if you have a binomial x trinomial? Please copy: (x + 3)(x² + 2x + 4)

113. Horizontal Method: (x + 3)(x² + 2x + 4) = x³ + 2x² + 4x + 3x² + 6x + 12 Write terms in descending order… = x³ + 2x² + 3x² + 4x + 6x + 12 Combine like terms = x³ + 5x² + 10x + 12

114. Vertical Method: x² + 2x + 4 x + 3 x³ + 2x² + 4x 3x² + 6x + 12 x³ + 5x² + 10x + 12

115. OK, your turn… 1.(x+1)(x²+2x+1) 2.(b 2 + 6b –7)(3b – 4) 3.(2x 2 + 5x –1)(4x – 3) 4.(2y – 3)(3y 2 – y + 5) 5.(x 2 + 2x + 1)(x + 2)

116. Try These - Distribute each problem:

117. Special Polynomials  Difference of squares? a 2 – b 2 = (a – b)(a + b)  Perfect-square trinomial? a 2 + 2ab + b 2 = (a + b) 2 a 2 – 2ab + b 2 = (a – b) 2

118. Ex: Factor x 2 – 4 Notice the terms are both perfect squares x 2 = (x) 2 4 = (2) 2  x 2 – 4 = (x) 2 – (2) 2 a 2 – b 2 and we have a difference = (x – 2)(x + 2)  difference of squares = (a – b)(a + b) factors as Difference of Squares

119. Ex: Factor 9p 2 – 16 Notice the terms are both perfect squares 9p 2 = (3p) 2 16 = (4) 2  9a 2 – 16 = (3p) 2 – (4) 2 a 2 – b 2 and we have a difference = (3p – 4)(3p + 4)  difference of squares = (a – b)(a + b) factors as Difference of Squares

120. Ex: Factor y 6 – 25 Notice the terms are both perfect squares y 6 = (y 3 ) 2 25 = (5) 2  y 6 – 25 = (y 3 ) 2 – (5) 2 a 2 – b 2 and we have a difference = (y 3 – 5)(y 3 + 5)  difference of squares = (a – b)(a + b) factors as Difference of Squares

121. Ex: Factor 81 – x 2 y 2 Notice the terms are both perfect squares 81 = (9) 2 x 2 y 2 = (xy) 2  81 – x 2 y 2 = (9) 2 – (xy) 2 a 2 – b 2 and we have a difference = (9 – xy)(9 + xy)  difference of squares = (a – b)(a + b) factors as Difference of Squares

122. Perfect Square Trinomials The Square of a Binomial (a + b) 2 The square of a binomial comes up so often that you should be able to write the final product immediately. It will turn out to be a very specific trinomial. To see that, let us square (a + b): (a + b) 2 = (a + b)(a + b) = a 2 + 2ab + b 2 Why?… the outers plus the inners will be ab + ba = 2ab.outers plus the inners The order of factors does not matter.

123. Perfect Square Trinomials The Square of a Binomial The square of any binomial produces the following three terms: 1. The square of the first term of the binomial: a 2 2. Twice the product of the two terms: 2ab 3. The square of the second term: b 2

124. Perfect Square Trinomials The Square of a Binomial Example 1. Square the binomial (x + 6). Solution. (x + 6) 2 = x 2 + 12x + 36 x 2 is the square of x. 12x is twice the product of x with 6. (x · 6 = 6x. Twice that is 12x.) 36 is the square of 6. x 2 + 12x + 36 is called a perfect square trinomial -- which is the square of a binomial.

125. Perfect Square Trinomials The Square of a Binomial Example 2. Square the binomial (3x − 4). Solution. (3x − 4) 2 = 9x 2 − 24x + 16 9x 2 is the square of 3x. −24x is twice the product of 3x· −4. (3x· −4 = −12x. Twice that is −24x.) 16 is the square of −4.

126. Perfect Square Trinomials The Square of a Binomial Note: If the binomial has a minus sign, then the minus sign appears only in the middle term of the trinomial. Therefore, using the double sign ± ("plus or minus"), we can state the rule as follows:

127. Perfect Square Trinomials The Square of a Binomial Example 3. Is this a perfect square trinomial: x 2 + 14x + 49 ? Solution. Yes. It is the square of (x + 7). x 2 is the square of x. 49 is the square of 7. And 14x is twice the product of x · 7. In other words, x 2 + 14x + 49 could be factored as (x + 7) 2 Note: If the coefficient of x had been any number but 14, this would not have been a perfect square trinomial.

128. Perfect Square Trinomials The Square of a Binomial Example 4 Is this a perfect square trinomial: x 2 + 50x + 100 ? Answer. No, it is not. Although x 2 is the square of x, and 100 is the square of 10, 50x is not twice the product of x· 10. (Twice their product is 20x.)