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Chapters 5.1 – 5.6. Review of Factoring. Factors. Factors are numbers or variables that are multiplied in a multiplication problem. Factor an expression means to write the expression as a product of its factors a · b = c, then a and b are said to be factors of c Examples:

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## Chapters 5.1 – 5.6

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**Chapters 5.1 – 5.6**Review of Factoring**Factors**• Factors are numbers or variables that are multiplied in a multiplication problem. • Factor an expression means to write the expression as a product of its factors • a· b = c, then a and b are said to be factors of c • Examples: • 3 · 5 = 15, then 3 and 5 are factors of the product 15 • x3 · x4 = x7, then x3 and x4 are factors of x7 • x(x + 2) = x2 + 2x, then x and x + 2 are factors of x2 + 2x • (x – 1) (x + 3) = x2 + 2x – 3, then x – 1 and x + 3 are factors of x2 + 2x – 3**Prime Numbers and Composite Numbers**• Prime Number is an integer greater than 1 that has exactly two factors, itself and one. • The first 15 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 • Composite is a positive integer (other than 1) that is not a prime number. • The number 1 is neither prime nor composite, it is called a unit.**Greatest Common Factor**• Greatest Common Factor (GCF) of two or more numbers is the greatest number that divides into all the numbers. • GCF (two or more numbers) – Write each number as a product of prime factors. Determine the prime factors common to all numbers. Multiply the common factors. • Example: Find the GCF of 40 and 140 40 = 2 · 20 = 2 · 2 · 10 = 2 · 2 · 2· 5 = 23· 5 140 = 2 · 70 = 2 · 2 · 35 = 2 · 2 · 5· 7 = 22· 5 · 7 GCF = 22· 5 = 4· 5 = 20**Greatest Common Factor**• GCF (two or more terms) – Take each factor the largest number of times that it appears in all of the terms. • Example: Find the GCF of 18y2, 15 y3, and 27y5 18y2 = 2 · 9 · y · y = 2 · 3 · 3 · y · y 15 y3 = = 3 · 5 · y · y· y 27y5 = 3 · 9 · y · y · y · y· y = 3 · 3 · 3 · y · y · y · y· y GCF = 3 · y · y = 3 y2**Factor a Monomial from a Polynomial**• Determine the greatest common factor (GCF) of all the terms in the polynomial. • Write each term as the product of the GCF and its other factor. • Use the distributive property to factor out the GCF. a (b + c) = ab + ac • Example #9 pg 348: 7x – 35 GCF = 7 7(x – 5) • FOIL = First, Outer, Inner, Last (a + b) (c + d) First = (a)(c) Outer = (a)(d) Inner = (b)(c) Last = (b)(d)**Factor a Monomial from a Polynomial**• Determine the greatest common factor (GCF) of all the terms in the polynomial. • Write each term as the product of the GCF and its other factor. • Use the distributive property to factor out the GCF.Distributive property says the if a, b and c are all real numbers then a (b + c) = ab + ac • Monomial has one term: Exp: 5 , 4x, -6x2 • Binomial has two terms: Exp: x + 4 , x2 – 6 , 2x2 – 5x • Trinomial has three terms: Exp: x2 – 2x + 3 , 5x2 – 6x +7 • Polynomial has infinite number of terms: Exp: 2x4 – 4x2 – 6x + 3**Factor a Monomial from a Polynomial**• Example: 6a4 + 27a3 - 18a2 GCF = 3a2 3a2(2a2 + 9a – 6) • Example: x(5x-2) + 7(5x-2) (5x-2) is a factor (x+7)(5x-2)**Factor a Four-Term Polynomial by Grouping**• Determine if there is a common factor, if yes then factor out. • Arrange the four terms so that the first two terms and the last two terms have a common factor. • Use distributive property to factor each group of terms. (first two, last two) • Factor GCF from the results. • Example #27, pg 348 Example #22, pg 348: x2 + 3x – 2xy – 6y x2 – 5x + 4x – 20 x(x + 3) – 2y(x + 3) x(x – 5) + 4(x – 5) (x + 3) (x – 2y) (x – 5) (x + 4)**Signs**• If the 3rd term is positive the factor of last two terms will be positive. • If the 3rd term is negative the factor of the last two terms will be negative. • 2 positives – Factor positive x2 + b + c ( _ + _ ) ( _+ _ ) • b = Negative, c = Positive – Factor Negative x2 – b + c ( _ - _ ) ( _- _ ) • b = Positive, c = Negative – Factor Positive Negative x2 + b – c ( _ + _ ) ( _- _ ) • b = Negative, c = Negative – Factor Positive Negative x2 – b – c ( _ + _ ) ( _- _ )**Factoring Trinomials, a = 1**• In the form of ax2 + bx + c, where a = 1 • Find two numbers whose product equals the constant, c, and whose sum equals the coefficient of the x-term, b. • Use the two numbers found, including their signs, to write the trinomial in factored form . (x + one number)(x + second number) • Check using the FOIL method. • Example # 37, pg 348 Factors of c that add to b x2 + 11x + 18 8 = (2)(9) (x + 9) (x+ 2) 2 + 9 = 11**Factoring Trinomials**• Example # 40, pg 348 Factors of c (56)that add to b (-15) x2 – 15x + 56 56 = (-7)(-8) (x - 7) (x - 8) -7 + -8 = -15 • A prime polynomial is a polynomial that cannot be factored using only integer coefficients. • Example #36, pg 348 Factors of c (-15) that add to b (+4) x2 + 4x – 15 -15 = (-3)(5) PRIME -15 = (3)(-5) -3 + 5 ≠ 4 5 + -3 ≠ 4**Factoring Trinomials, a ≠ 1**• In the form of ax2 + bx + c, where a ≠1, by Trial and Error • Factor out any common factors to all three terms • Write all pairs of factors of the coefficient of the squared term, a • Write all pairs of factors of the constant term, c • Try various combinations of these factors until the correct middle term, bx, is found. • Example: Factors Possible Sum of 12 Factors Inner/Outer 3x2 + 20x + 12 (1)(12) (3x+1)(x+12) 36x + x = 37x (3x + 2) (x + 6) (2)(6) (3x+2)(x+6) 18x + 2x = 20x (3)(4) (3x+3)(x+4) 12x + 3x = 15x**Factoring Trinomials, a ≠ 1**• Example: 1. Common Factor = 2 2x2 + 2x – 12 2. Factors of -6 that add to 1 2(x2 + x – 6) (-1)(6) add -1 + 6 = 5 2x(x + 3) (x – 2) (-2)(3) add -2 + 3 = 1 Remember you can check with FOIL 2(x + 3) (x – 2) 2(x2 – 2x + 3x – 6) 2(x2 + x – 6) 2x2 + 2x – 12**Factoring Trinomials**• In the form of ax2 + bx + c, where a ≠1, by Grouping • Factor out any common factors to all three terms • Fine two numbers whose product is equal to the product of a times c, and whose sum is equal to b • Rewrite the middle term, bx, as the sum or difference of two terms using the numbers found • Factor by grouping. • HINT: If no factors of (a)(c) add up to (b) then cannot factor. • Example: No common factors 3x2 + 20x + 12 a = 3 b = 20 c = 12 3x2 + 18x + 2x + 12 (a)(c) = (3)(12) = 36 3x(x + 6) + 2(x + 6) factors of 36 that add to 20 (3x + 2) (x + 6) (1)(36) 1 + 36 = 37 (2)(18) 2 + 18 = 20 (3)(12) 3 + 12 = 15**Difference of Two Squares**• a2 – b2 = (a + b) (a – b) • Example #64, pg 349 Example # 70, page 349 x2 – 36 64x6 – 49y6 x2 – 62 (8x 3) 2 – (7y3) 2 (x + 6) (x – 6) (8x 3 + 7y3 )(8x 3 – 7y3 )**Sum of Two Cubes**• a3 + b3 = (a + b) (a2 – ab + b2) • Example #74, pg 349 Example # 77, page 349 x3 + 8 125a3 + b3 x3 + 23 a = x, b = 2 5a3 + b3 a = 5a, b = b (x + 2) (x 2 – 2x + 22) (5a + b) ((5a) 2 – 5ab + b2) (x + 2) (x 2 – 2x + 4) (5a + b) (25a 2 – 5ab + b2)**Difference of Two Cubes**• a3 – b3 = (a - b) (a2 + ab + b2) • Example #76, pg 349 Example # 73, page 349 b3 – 64 x3 – 1 b3 – 43 a = b , b = 4 x3 – 13 a = x , b = 1 (b – 4) (b2 + 4b + 42) (x – 1)(x2 + 1x + 12) (b – 4)(b2 + 4b + 16) (x – 1)(x2 + x + 1)**General Procedure for Factoring a Polynomial**• Factor any GCF of all terms • If a two term polynomial determine if it is a special factor. If so factor using the formula • If three term polynomial, factor according to methods discussed for a = 1 or a ≠ 1. • If more than three terms try factoring by grouping • Determine if there are any common factors, and factor them out.**Quadratic Equation**• Standard Form: a + bx + c = 0 a, b and c are real numbers • Zero factor Property – if ab = 0, then a = 0 or b = 0 • Solve the quadratic Equation by factoring • Write the equation in standard form • Factor the side of the equation that is not 0 • Set each factor equal to 0 and solve • Check each solution.**Quadratic Equation**ExampleCheck (x = 1) x2 – 3x = -2 x2 – 3x = -2 x2 – 3x + 2 = 0 12 – 3(1) = -2 Factors of 2 that sup to -3 1 – 3 = -2 (-1)(-2) = 2 -2 = -2 True (-1) + (-2) = -3 Check (x = 2) (x – 1)(x – 2) x2 – 3x = -2 x – 1 = 0 and x – 2 = 0 22 – 3(2) = -2 x = 1 x = 2 4 – 6 = -2 -2 = -2 True**Homework – Review Factoring**• Page 348 – 349: #11, 13, 15, 35, 39, 41, 55, 63, 69, 85

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