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A LGEBRA I – CHAPTER 10 10.1 Adding and Subtracting Polynomials Objectives: 1. Know the anatomy of a polynomial 2. Add and subtract polynomials.

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Presentation on theme: "A LGEBRA I – CHAPTER 10 10.1 Adding and Subtracting Polynomials Objectives: 1. Know the anatomy of a polynomial 2. Add and subtract polynomials."— Presentation transcript:

1 A LGEBRA I – CHAPTER 10 10.1 Adding and Subtracting Polynomials Objectives: 1. Know the anatomy of a polynomial 2. Add and subtract polynomials

2 V OCABULARY – PAGE 576 Polynomial Standard form Degree Degree of polynomial Leading Coefficient Term Monomial Binomial Trinomial

3 T ERM VS. C OEFFICIENT TermCoefficient -12x³? x³y? -z? 2? Constant – If a term contains only a number like 2 it is called a constant.

4 T YPES OF P OLYNOMIALS Monomials 1 Term Binomials 2 Terms Trinomials 3 Terms ax²x + yx² + 4xy + y² -3x6y² - 2-x³ + 2x + 1 4½ z³ - 2z8y² - 2y -10 All other longer polynomials do not have a special name.

5 There are two ways to classify polynomials: 1.degree – (for each term) It is the exponent of the variable 2.number of terms – count the number of terms SEE EXAMPLE 2 on Page 576

6 P OLYNOMIAL C LASSIFICATIONS PolynomialDegree Classification # of Terms Classification 6 -2x + 1 x² + 2x - 5 x³ - 8x PLEASE NOTE: X ⁴ DEGREE IS QUARTIC

7 P OLYNOMIAL IN STANDARD FORM 2x³ + 5x² - 4x + 7 Degree of a polynomial – largest degree of any of its terms. Leading Coefficient Degree Constant Term

8 Anatomy of a Polynomial terms Total degree constant degree of polynomial standard form leading coefficient

9 Anatomy of a Polynomial terms degree constant degree of polynomial standard form leading coefficient

10 Like Terms: terms that have identical variables both in letters and degree Combining Like Terms: adding and subtracting like terms – changes only the coefficient Adding and Subtracting Polynomials: simply combining like terms 2 ways to add and subtract 1.Horizontal 2.Vertical

11 Adding & Subtracting Horizontally Steps: 1.Distribute all subtraction signs into polynomials. 2.Combine like terms (6x 2 – x + 3) + (-2x + x 2 – 7)

12 Adding & Subtracting Vertically Steps: 1.Distribute all subtraction signs into polynomials. 2.Combine like terms (6x 2 – x + 3) – (-2x + x 2 – 7)

13 Adding & Subtracting Steps: 1.Distribute all subtraction signs into polynomials. 2.Combine like terms (-8x 3 + x – 9x 2 + 2) + (8x 2 – 2x + 4) + (4x 2 – 1 – 3x 3 )

14 Adding & Subtracting Steps: 1.Put each polynomial on a level in standard form lining up like terms 2.Carefully add or subtract the coefficients (-6x 3 + 5x – 3) – (2x 3 + 4x 2 – 3x + 1)

15 Adding & Subtracting Steps: 1.Put each polynomial on a level in standard form lining up like terms 2.Carefully add or subtract the coefficients (4x 2 – 1) – (3x – 2x 2 )

16 Adding & Subtracting Steps: 1.Put each polynomial on a level in standard form lining up like terms 2.Carefully add or subtract the coefficients (-8x 3 + x – 9x 2 + 2) + (8x 2 – 2x + 4) – (4x 2 – 1 – 3x 3 )

17 Find the sum or difference: Steps: 1.Put each polynomial on a level in standard form lining up like terms 2.Carefully add or subtract the coefficients (2x 2 + 9x – 4) + (6x – 3x 2 + 1)

18 A LGEBRA I – L ESSON 10.1 Homework Worksheet 10.1 10.1 Quiz Tomorrow

19 A LGEBRA I – C HAPTER 10 10.2 Multiplying Polynomials Objectives: 1. Multiply 2 binomials 2. Multiply polynomials Vocabulary: none

20 M ULTIPLYING P OLYNOMIALS Multiplying variable terms: Multiply the coefficients together and then multiply the variables together. Then combine like terms, if necessary. 3x(6xy) = 18x²y 2x²(5y² + 7xy) =10x²y² + 14x³y

21 A DDING VS. M ULTIPLYING P OLYNOMIALS Variable exponents must be the same to perform operation. Coefficients must be the same to perform operation. Variable and exponents will stay the same after operation. Variable exponents change after operation Adding PolynomialsYesNoYesNo Multiplying PolynomialsNo No* *Unless: 7(3x²) = 21 x² Yes*

22 M ULTIPLYING B INOMIALS USING FOIL (3x + 4)(x + 5) = 3x² + 15x + 4x + 20 = 3x² + 19x + 20 First Terms Inner Terms Last Terms Outer Terms

23 M ULTIPLYING P OLYNOMIALS H ORIZONTALLY Find the product (x – 4)(5x + 9 - 2x²)

24 M ULTIPLYING P OLYNOMIALS V ERTICALLY Find the product (x + 4)(5x + 3 - 3x²) Align like terms (descending order) in columns. - 3x² + 5x + 3 * x + 4

25 Multiply these polynomials: (x + 9)(x + 7) (x + 5)(x + 5) (x + 7)(x – 7) (2x + 3)(2x +3)+ 3)

26 Multiply these polynomials: (3x 2 + 3)(5x 2 – 2x) (5x - 3)(5x 2 – 3x + 2)

27 A LGEBRA I – L ESSON 10.2 Homework Worksheet 10.2 10.2 Quiz Tomorrow

28 A LGEBRA I – C HAPTER 10 10.1-10.2 Quick Review (x 2 – x + 3) – (2x + x 2 – 3) (2x-1)(x – 7)

29 A LGEBRA I – C HAPTER 10 10.3 Special Products of Polynomials Objectives: 1. Multiply using the square of a binomials pattern 2. Multiply using the sum and difference pattern

30 F IND THE PATTERN : Use FOIL to find the products: ( x - 2)(x+2) (2n +3)(2n - 3) (4t – 1)(4t + 1) (x + y)(x – y) x² - 4 4n² - 9 16t² - 1 x² - y²

31 S UM AND D IFFERENCE P ATTERN (a + b)(a – b) = a² - b² (3x – 4) (3x + 4) = 9x² - 16 Ex. 1 (2x -5)(2x + 5) = Ex. 2 (x - 2y)(x + 2y) =

32 Multiply: (x + 3)(x - 3)

33 Multiply: (x - 4) (x + 4)

34 Multiply: (2x + 7) (2x - 7)

35 Multiply: (3x - 4) (3x + 4)

36 Multiply: (3x 3 + 11)(3x 3 - 11)

37 F IND THE PATTERN : Use FOIL to find the products: (x + 3)² (3m + 1)² (5s + 2)² (x + y)² x² + 6x + 9 9m² + 6m + 1 25s² + 20s + 4 x² + 2xy + y²

38 S QUARE OF A B INOMIAL P ATTERN (a + b)² = a² + 2ab + b² Ex: (x + 4)² = x² + 8x + 16 (a - b)² = a² - 2ab + b² Ex: (2x – 6)² = 4x² - 24x + 36 (3b -5)² = 9b² - 30b + 25

39 Multiply: (5x + 4) 2

40 Multiply: (2x 2 – 3y) 2

41 A LGEBRA I – L ESSON 10.3 Homework Worksheet 10.3 10.3 Quiz Next Monday 10.1-10.3 Test Next Wednesday

42 10.1-10.3 Review Complete Both Problems on your OWN! (6x + 2x ³ - 5x⁵) – (9x³ + 2x⁵ - 4) (6x + 2)(4x – 3)

43 A LGEBRA I –C HAPTER 10 10.4 Solving Polynomial Equations in Factored Form Objectives: 1. Solve a polynomial equation in factored form 2. Understand factors, solutions, and zeros Vocabulary: factored form, zero-product form

44 V OCABULARY We use evaluate, simplify, and find the product for expressions. In order to solve there must be an equation. Equations have the = sign. A polynomial in factored form is written as a product of two or more linear factors. The factored form of x² +4x + 4 is (x + 2)(x + 2).

45 Standard form: 2x 2 + 7x – 15 = 0 Factored form: (2x – 3)(x + 5) = 0

46 Z ERO -P RODUCT P ROPERTY Let a and b be real numbers. If ab = 0, then a = 0 or b = 0 Solve the equation (x – 4)(x + 1) = 0 By the zero product property: x – 4 = 0 or x + 1 = 0 x= 4 or x = -1 The solutions are 4 and -1. Check in original equation.

47 S OLVE THE FOLLOWING EQUATIONS. Solve (x + 8)² = 0 Solve (x – 4)(x + 3) = 0 Solve x(x+3)(2x – 3) = 0 (3x – 2)(4x + 3)(x + 4) = 0 -8 4, -3 0, -3, 3/2 2/3, -3/4, -4

48 (x – 9)(x + 6) = 0 (5x – 2)(2x + 7) = 0 (x + 8) 2 = 0 (3x – 2)(4x + 3)(x + 4) = 0 (x – 3)(5x + 19)(5x – 1) = 0 S OLVE THE FOLLOWING EQUATIONS.

49 F ACTORS, S OLUTIONS, AND X INTERCEPTS For any quadratic polynomial ax² + bx + c, if one statement is true, then all 3 statements are true. (x – 4) and (x + 4) are factors of x² -16 x = 4 and x = -4 are solutions of x² - 16 = 0 4 and -4 are x-intercepts of y = x² - 16

50 S KETCH THE GRAPH OF Y = ( X – 3)( X + 1) Solve (x – 3) (x + 1) = 0 x intercepts = 3, -1 Coordinates of the vertex: The x coordinate is the avg. of the x intercepts x = (3 + (-1)) / 2 = 1 Substitute to find y- coordinate y = (1 – 3)(1 + 1) y = (-2)(2) = -4 The coordinates of the vertex are (1, -4). (3,0) (-1,0) (1,-4)

51 S KETCH THE GRAPH OF Y = ( X – 4)( X + 2) Solve (x – 4) (x + 2) = 0 x intercepts = Coordinates of the vertex: The x coordinate is the avg. of the x intercepts Substitute to find y- coordinate The coordinates of the vertex are ____

52 A LGEBRA I – L ESSON 10.4 Homework Worksheet 10.4 10.4 Quiz Tomorrow

53 A LGEBRA I – CHAPTER 10 10.5 Factoring x 2 + bx + c Objectives: 1. Factor x 2 + bx + c 2. Solve x 2 + bx + c = 0 equations by factoring Vocabulary: none

54 F ACTORING X ² + BX + C Find 2 numbers whose product is c and whose sum is b. The factored form is: (x + one number) (x + other number)

55 F ACTORING WHEN B AND C ARE POSITIVE x² + 3x + 2 (x + ) Fill in numbers from table. (x + 1)(x + 2) Check by using FOIL: (x + 1)(x + 2) = x² + 1x + 2x + 2 = x² + 3x + 2 cb 23 1·21+2

56 F ACTOR : X ² + 10 X + 16 Check by using foil: (x + 2) (x + 8) = x² + 8x + 2x + 16 = x² + 10x + 16 cb + 1, 16 2, 8 4, 4 17 10 8 (x + 2) (x + 8) 1610

57 F ACTOR X ² - 5 X + 6 Check by foil: (x – 2) (x - 3) = x² - 2x - 3x + 6 = x² - 5x + 6 6-5 + -1, -6 -2, -3 -3, -2 -6, -1 -7 -5 -7 (x + -2) (x + -3) = (x - 2) (x - 3) Since c is a positive number we will look for 2 numbers that have the same signs. The 2 numbers will be - since b is -.

58 F ACTOR X ² - 2 X - 8 Check by foil: (x – 4) (x + 2) = x² - 4x + 2x – 8 = x² - 2x - 8 -8 -2 + -1, 8 -2, 4 -4, 2 -8, 1 7 2 -2 -7 (x + -4) (x +2) = (x – 4) (x + 2) Since c is a negative number we will look for 2 numbers that have opposite signs. The larger number will be - since b is -.

59 S IGN PATTERNS x² + bx + c x² - bx - c x² + bx - c x² - bx + c + means factors have the same sign- means factors have different signs Both numbers will be +Both numbers will be - Larger abs value will be -Larger abs value will be +

60 YOU TRY TO F ACTOR : X ² + 7 X - 18 7 + -18 Since c is a negative number we will look for 2 numbers that have opposite signs. The larger number will be + since b is +.

61 S OLVING A Q UADRATIC E QUATION x² - 3x = 10Equation x² - 3x – 10 = 0Write in Standard Form (x - 5)(x + 2) = 0Factor left side x–5=0 or x+2= 0Zero Product Property X = 5 or x = -2Solve for x The solutions are 5 and -2. Check by substituting values into original equation. 5² - 3(5) = 10 25 – 15 = 10 (-2)² - 3 (-2) = 10 4 – (-6) = 10

62 Solve: x 2 – 5x - 24 = 0 Steps to solve quadratic by factoring 1.Put into standard form 2.Factor quadratic 3.Set each factor to 0 and solve 4.Check! x 2 + 3x = 10

63 Solve: x 2 + 5x = -6 Steps to solve quadratic by factoring 1.Put into standard form 2.Factor quadratic 3.Set each factor to 0 and solve 4.Check! x 2 + 3x = 10

64 A LGEBRA I – L ESSON 10.5 Homework Worksheet 10.5 10.5 Quiz Tomorrow

65 10.4 Review Factor and Solve on your OWN! x² - 10x + 24 = 0

66 A LGEBRA I – CHAPTER 10 10.6 Factoring ax 2 + bx + c Objectives: 1. Factor ax 2 + bx + c 2. Solve ax 2 + bx + c = 0 equations by factoring Vocabulary: none

67 F ACTOR : 2 X ² + 11 X + 5 Step 1: Put in standard form and factor out GCF if possible. Step 2: Find factors of a c that add up to b: Step 3: Write as factors: (ax + )(ax + ): (2x + 1)(2x + 10) Step 4: Factor out and discard a. Leave whole numbers. (2x + 1)(2x + 10) = (2x + 1)(x + 5) a c 10 b 11 2,57 1,1011 a bc 22

68 F ACTOR : 3 X ² - 4 X - 7 Step 1: Put in standard form and factor out GCF if possible. Step 2: Find factors of a c that add up to b: Step 3: Write as factors: (ax + )(ax + ): (3x + 3)(3x - 7) Step 4: Factor out and discard a. Leave whole numbers. (3x + 3)(3x - 7) = (x + 1)(3x - 7) a c -21 b -4 1, -21-20 3, -7-4 a b c 33

69 F ACTOR : 6 X ² - 19 X + 15 Step 1: Put in standard form and factor out GCF if possible. Step 2: Find factors of ac that add up to b: Step 3: Write as factors: (ax + )(ax + ): Step 4: Factor out and discard a. Leave whole numbers. abc

70 Y OU TRY TO F ACTOR : 2 X ² + 3 X - 9 Step 1: Put in standard form and factor out GCF if possible. Step 2: Find factors of a c that add up to b: Step 3: Write as factors: (ax + )(ax + ): Step 4: Factor out and discard a. Leave whole numbers. a bc

71 T OGETHER - S OLVE FOR X : 6 X ² - 2 X - 8 Step 1: Put in standard form and factor out GCF if possible: 2(3x ² – x – 4) Step 2: Find factors of a c that add up to b: Step 3: Write as factors: (ax + )(ax + ): Step 4: Factor out and discard a c -12 b

72 Y OU T RY - F ACTOR : 15 X ² - 27 X – 6 = 0 Step 1: Put in standard form and factor out GCF if possible. Step 2: Find factors of a c that add up to b: Step 3: Write as factors: (ax + )(ax + ): Step 4: Factor out and discard a. Leave whole numbers.

73 A LGEBRA I – L ESSON 10.6 Homework Worksheet 10.6 10.6 Quiz Tomorrow

74 A LGEBRA I – CHAPTER 10 10.7 Factoring Special Products Objectives: 1. Factor out the GCF 2. Factor special products Vocabulary: GCF

75 D IFFERENCE OF 2 S QUARES a ² - b ² = (a + b)(a – b) 9x ² - 16 = (3x + 4)(3x – 4) Both binomial terms are perfect squares with a “–” between positive terms.

76 S OLVE FOR X : 2 X ² - 72 = 0 Put in standard form and factor out GCF: 2(x ² - 36) = 0 Check for pattern and factor: 2(x + 6)(x – 6) = 0 Use zero product property: x = -6, 6 Check in original equation: 2(6) ² - 72 = 0 2 (-6) ² - 72 = 0 a = xb = 6

77 S OLVE FOR X : 2 X ² - 72 = 0 If you forget formula, you can still factor using product-sum method by rewriting with 0x. After GCF: 2(x ² - 36) = 0 Put in standard form: 2(x ² + 0x – 36) = 0 Rewrite as factored form: 2(x + 6)(x – 6) = 0 Solve using zero product property: x = 6, -6 -360 6,-60 a b c acb

78 You Try Factor each Expression. 100x² - 121 x² - 49 98x² - 50 x² + 25 + 36x = -108

79 P ERFECT S QUARE T RINOMIALS a ² + 2ab + b ² = (a + b) ² x ² + 8x + 16 = (x + 4) ² a ² - 2ab + b ² = (a - b) ² x ² - 12x + 36 = (x – 6) ²

80 F ACTORING P ERFECT S QUARE T RINOMIALS First and last terms are perfect squares. Middle term is 2 √first term √last term. 1. x ² - 4x + 4 2. 16y ² + 24y + 9

81 You Try Factor each Expression. 16x² -12x + 9 3x² 3x² - 30x + 75 + 36x = -108

82 A LGEBRA I – L ESSON 10.7 Homework Worksheet 10.7 10.7 Quiz Monday after Spring Break

83 A LGEBRA I – CHAPTER 10 10.8 Factoring Using the Distributive Property Objectives: 1. Factor out the GCF 2. Solve equations with degrees > 2 Vocabulary: none

84 B INOMIALS : B OTH TERMS MUST BE PERFECT SQUARES SEPARATED WITH A – SIGN. a) m ² - 4 b) 4p ² - 25 c) 50 – 98x ² d) x ² + 36

85 YOUR TURN: A LWAYS F ACTOR GCF F IRST a) 33x - 121x ² b) 4p ² + 12p c) 18d ³ - 6d ² + 3d

86 F ACTORING BY G ROUPING x ³ + 2x ² + 3x + 6 (x ³ + 2x ² ) + (3x + 6) x ² (x + 2) + 3(x + 2) (x ² + 3)(x +2) No GCF Split to group terms Factor each group Distributive Property

87 TOGETHER F ACTOR BY G ROUPING x ³ + 2x ² - 36x – 72

88 YOUR TURN F ACTOR BY G ROUPING x ³ - 2x ² - 9x + 18

89 ONE MORE F ACTOR BY G ROUPING x ³ - 3x ² + x – 3

90 S UMMARY OF F ACTORING A P OLYNOMIAL Step 1: Factor out GCF Step 2: 2 Terms: Factor a ² - b ² 3 Terms: Factor x ² + bx + c Factor ax ² + bx + c 4 Terms: Factor by Grouping Step 3: Check by multiplying back together. Can it be factored further?

91 Homework Worksheet 10.8 10.8 Quiz Tomorrow 10.4-10.8 Test Friday A LGEBRA I – L ESSON 10.8


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