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5-4 Factoring Quadratic Expressions M11.A.1.2.1: Find the Greatest Common Factor and/or the Least Common Multiple for sets of monomials M11.D.2.1.5: Solve quadratic equations using factoring M11.D.2.2.2: Factor algebraic expressions, including differences of squares and trinomials Objectives Finding Common Binomial Factors Factoring Special Expressions Vocabulary Factoring is rewriting an expression as the product of its factors. The greatest common factor (GCF) of an expression is a common factor of the terms of the expression. Finding Common Factors Factor each expression. a. 15x2 + 25x + 100 15x2 + 25x + 100 = 5(3x2) + 5(5x) + 5(20) = 5(3x2 + 5x + 20) Factor out the GCF, 5 Rewrite using the Distributive Property. b. 8m2 + 4m 8m2 + 4m = 4m(2m) + 4m(1) = 4m(2m + 1) Factor out the GCF, 4m Rewrite using the Distributive Property. Factoring when ac > 0 and b > 0 Factor x2 + 10x + 24. Step 1: Find factors with product ac and sum b. Since ac = 24 and b = 10, find positive factors with product 24 and sum 11. Factors of 24 Sum of factors 1, 24 25 2, 12 14 3, 8 11 6, 4 10 } } Step 2: Rewrite the term bx using the factors you found. Group the remaining terms and find the common factors for each group. x2 + 10x + 24 x2 + 4x + 6x + 24 Rewrite bx : 10x = 4x + 6x. x(x + 4) + 6(x + 4) Find common factors. Continued (continued) Step 3: Rewrite the expression as a product of two binominals. x(x + 4) + 6(x + 4) (x + 6)(x + 4) Rewrite using the Distributive Property. Check: (x + 6)(x + 4) = x2 + 4x + 6x + 24 = x2 + 10x + 24 Factoring when ac > 0 and b < 0 Factor x2 – 14x + 33. Step 1: Find factors with product ac and sum b. Since ac = 33 and b = –14, find negative factors with product 33 and sum b. Factors of 33 Sum of factors –1, –33 –34 –3, –11 –14 Step 2: Rewrite the term bx using the factors you found. Then find common factors and rewrite the expression as a product of two binomials. x2 + 14x + 33 x2 – 3x – 11x + 33Rewrite bx. } } x(x – 3) – 11(x – 3) Find common factors. (x – 11)(x – 3) Rewrite using the Distributive Property. Factoring When ac < 0 Factor x2 + 3x –28. Step 1: Find factors with product ac and sum b. Since ac = –28 and b = 3, find factors 2 with product –28 and sum 3. Factors of –28 Sum of factors 1, –28 –27 –1, 28 27 2, –14 –12 –2, 14 12 4, –7 –3 –4, 7 3 Step 2: Since a = 1, you can write binomials using the factors you found. x2 + 3x – 28 (x – 4)(x + 7) Use the factors you found. Factoring When a ≠ 1 and ac > 0 Factor 6x2 – 31x + 35. Step 1: Find factors with product ac and sum b. Since ac = 210 and b = –31, find negative factors with product 210 and sum –31. Factors of 210 Sum of factors –1, –210 –2, –105 –3, –70 –5, –42 –10, –21 –211 –107 –73 –47 –31 Step 2: Rewrite the term bx using the factors you found. Then find common factors and rewrite the expression as the product of two binomials. 6x2 – 31x + 35 2x(3x – 5) – 7(3x – 5) Find common factors. (2x – 7)(3x – 5) Rewrite using the Distributive Property. } Rewrite bx. } 6x2 – 10x – 21x + 35 Factoring When a ≠ 1 and ac < 0 Factor 6x2 + 11x – 35. Step 1: Find factors with product ac and sum b. Since ac = 210 and b = 11, find factors with product –210 and sum 11. Factors of –210 –1, –210 Sum of factors –209 Factors of –210 Sum of factors –3, 70 67 –1, 210 209 5, –42 –37 2, –105 –2, 105 –103 103 –5, 42 37 10, –21 –11 3, –70 –67 –10, 21 11 Step 2: Rewrite the term bx using the factors you found. Then find common factors and rewrite the expression as the product of two binomials. 6x2 + 11x + 35 6x2 – 10x + 21x – 35 Rewrite bx. 2x(3x – 5) + 7(3x – 5) Find common factors. (2x + 7)(3x – 5) Rewrite using the Distributive Property. Vocabulary A perfect square trinomial is the product you obtain when you square a binomial. a² + 2ab + b² = (a + b)² a² - 2ab + b² = (a – b)² Factoring a Perfect Square Trinomial Factor 100x2 + 180x + 81. 100x2 + 180x + 81 = (10x)2 + 180 + (9)2 Rewrite the first and third terms as squares. = (10x)2 + 180 + (9)2 Rewrite the middle term to verify the perfect square trinomial pattern. = (10x + 9)2 a2 + 2ab + b2 = (a + b)2 Real World Example A square photo is enclosed in a square frame, as shown in the diagram. Express the area of the frame (the shaded area) in completely factored form. Relate: frame area equals the outer area minus the inner area Define: Let x = length of side of frame. Write: area = x2 – (7)2 = (x + 7)(x – 7) The area of the frame in factored form is (x + 7)(x – 7) in2. Homework Pg 263 # 1, 2, 7, 13, 19, 25, 31 Pg 264 # 37, 38, 46