REVIEW OF FACTORING Chapters 5.1 – 5.6

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  x 3 · x 4 = x 7, then x 3 and x 4 are factors of x 7  x(x + 2) = x 2 + 2x, then x and x + 2 are factors of x 2 + 2x  (x – 1) (x + 3) = x 2 + 2x – 3, then x – 1 and x + 3 are factors of x 2 + 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 = 2 · 20 = 2 · 2 · 10 = 2 · 2 · 2 · 5 = 2 3 · = 2 · 70 = 2 · 2 · 35 = 2 · 2 · 5 · 7 = 2 2 · 5 · 7 GCF = 2 2 · 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 18y 2, 15 y 3, and 27y 5 18y 2 = 2 · 9 · y · y = 2 · 3 · 3 · y · y 15 y 3 = = 3 · 5 · y · y · y 27y 5 = 3 · 9 · y · y · y · y · y = 3 · 3 · 3 · y · y · y · y · y GCF = 3 · y · y = 3 y 2

Factor a Monomial from a Polynomial 1. Determine the greatest common factor (GCF) of all the terms in the polynomial. 2. Write each term as the product of the GCF and its other factor. 3. 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 1. Determine the greatest common factor (GCF) of all the terms in the polynomial. 2. Write each term as the product of the GCF and its other factor. 3. 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, -6x 2 Binomial has two terms: Exp: x + 4, x 2 – 6, 2x 2 – 5x Trinomial has three terms: Exp: x 2 – 2x + 3, 5x 2 – 6x +7 Polynomial has infinite number of terms: Exp: 2x 4 – 4x 2 – 6x + 3

Factor a Monomial from a Polynomial Example: 6a a a 2 GCF = 3a 2 3a 2 (2a 2 + 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 1. Determine if there is a common factor, if yes then factor out. 2. Arrange the four terms so that the first two terms and the last two terms have a common factor. 3. Use distributive property to factor each group of terms. (first two, last two) 4. Factor GCF from the results. Example #27, pg 348Example #22, pg 348: x 2 + 3x – 2xy – 6yx 2 – 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 3 rd term is positive the factor of last two terms will be positive. If the 3 rd term is negative the factor of the last two terms will be negative. 2 positives – Factor positive x 2 + b + c ( _ + _ ) ( _+ _ ) b = Negative, c = Positive – Factor Negative x 2 – b + c ( _ - _ ) ( _- _ ) b = Positive, c = Negative – Factor Positive Negative x 2 + b – c ( _ + _ ) ( _- _ ) b = Negative, c = Negative – Factor Positive Negative x 2 – b – c ( _ + _ ) ( _- _ )

Factoring Trinomials, a = 1 In the form of ax 2 + bx + c, where a = 1 1. Find two numbers whose product equals the constant, c, and whose sum equals the coefficient of the x-term, b. 2. Use the two numbers found, including their signs, to write the trinomial in factored form. (x + one number)(x + second number) 3. Check using the FOIL method. Example # 37, pg 348Factors of c that add to b x x = (2)(9) (x + 9) (x+ 2)2 + 9 = 11

Factoring Trinomials Example # 40, pg 348 Factors of c (56)that add to b (-15) x 2 – 15x = (-7)(-8) (x - 7) (x - 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) x 2 + 4x – = (-3)(5) PRIME-15 = (3)(-5) ≠ ≠ 4

Factoring Trinomials, a ≠ 1 In the form of ax 2 + bx + c, where a ≠1, by Trial and Error 1. Factor out any common factors to all three terms 2. Write all pairs of factors of the coefficient of the squared term, a 3. Write all pairs of factors of the constant term, c 4. Try various combinations of these factors until the correct middle term, bx, is found. Example: Factors Possible Sum of 12 Factors Inner/Outer 3x x + 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 2x 2 + 2x – Factors of -6 that add to 1 2(x 2 + x – 6) (-1)(6) add = 5 2x(x + 3) (x – 2) (-2)(3) add = 1 Remember you can check with FOIL 2(x + 3) (x – 2) 2(x 2 – 2x + 3x – 6) 2(x 2 + x – 6) 2x 2 + 2x – 12

Factoring Trinomials In the form of ax 2 + bx + c, where a ≠1, by Grouping 1. Factor out any common factors to all three terms 2. Fine two numbers whose product is equal to the product of a times c, and whose sum is equal to b 3. Rewrite the middle term, bx, as the sum or difference of two terms using the numbers found 4. Factor by grouping. 5. HINT: If no factors of (a)(c) add up to (b) then cannot factor. Example:No common factors 3x x + 12 a = 3 b = 20 c = 12 3x x + 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) = 37 (2)(18) = 20 (3)(12) = 15

Difference of Two Squares a 2 – b 2 = (a + b) (a – b) Example #64, pg 349Example # 70, page 349 x 2 – 36 64x 6 – 49y 6 x 2 – 6 2 (8x 3 ) 2 – (7y 3 ) 2 (x + 6) (x – 6) (8x 3 + 7y 3 )(8x 3 – 7y 3 )

Sum of Two Cubes a 3 + b 3 = (a + b) (a 2 – ab + b 2 ) Example #74, pg 349Example # 77, page 349 x a 3 + b 3 x a = x, b = 25a 3 + b 3 a = 5a, b = b (x + 2) (x 2 – 2x ) (5a + b) ((5a) 2 – 5ab + b 2 ) (x + 2) (x 2 – 2x + 4)(5a + b) (25a 2 – 5ab + b 2 )

Difference of Two Cubes a 3 – b 3 = (a - b) (a 2 + ab + b 2 ) Example #76, pg 349Example # 73, page 349 b 3 – 64 x 3 – 1 b 3 – 4 3 a = b, b = 4 x 3 – 1 3 a = x, b = 1 (b – 4) (b 2 + 4b )(x – 1)(x 2 + 1x ) (b – 4)(b 2 + 4b + 16)(x – 1)(x 2 + x + 1)

General Procedure for Factoring a Polynomial 1. Factor any GCF of all terms 2. If a two term polynomial determine if it is a special factor. If so factor using the formula 3. If three term polynomial, factor according to methods discussed for a = 1 or a ≠ If more than three terms try factoring by grouping 5. 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 1. Write the equation in standard form 2. Factor the side of the equation that is not 0 3. Set each factor equal to 0 and solve 4. Check each solution.

Quadratic Equation Example Check (x = 1) x 2 – 3x = -2 x 2 – 3x + 2 = – 3(1) = -2 Factors of 2 that sup to -31 – 3 = -2 (-1)(-2) = 2 -2 = -2 True (-1) + (-2) = -3 Check (x = 2) (x – 1)(x – 2)x 2 – 3x = -2 x – 1 = 0 and x – 2 = 02 2 – 3(2) = -2 x = 1x = 24 – 6 = = -2True

Homework – Review Factoring Page 348 – 349: #11, 13, 15, 35, 39, 41, 55, 63, 69, 85