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## 5-4 Factoring Quadratic Expressions

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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) Factor out the GCF, 5 = 5(3x2 + 5x + 20) Rewrite using the Distributive Property. b. 8m2 + 4m 8m2 + 4m = 4m(2m) + 4m(1) Factor out the GCF, 4m = 4m(2m + 1) Rewrite using the Distributive Property.**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 x2 + 10x + 24 x2 + 4x + 6x + 24 Rewrite bx : 10x = 4x + 6x. x(x + 4) + 6(x + 4) Find common factors. } } Factoring when ac > 0 and b > 0 Factor x2 + 10x + 24. Step 1:Find factors with product ac and sum b. Step 2: Rewrite the term bx using the factors you found. Group the remaining terms and find the common factors for each group.**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 Continued (continued) Step 3:Rewrite the expression as a product of two binominals.**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 x2 + 14x + 33 x2 – 3x – 11x + 33Rewrite bx. } } Factoring when ac > 0 and b < 0 Factor x2 – 14x + 33. Step 1: Find factors with product ac and sum b. 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. x(x – 3) – 11(x – 3)Find common factors. (x – 11)(x – 3) Rewrite using the Distributive Property.**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 Factoring When ac < 0 Factor x2 + 3x –28. Step 1: Find factors with product ac and sum b. 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.**Since ac = 210 and b = –31, find negative factors with**product 210 and sum –31. Factors of 210 Sum of factors –1, –210 –211 –2, –105 –107 –3, –70 –73 –5, –42 –47 –10, –21 –31 6x2 – 31x + 35 } } 6x2 – 10x – 21x + 35 Rewrite bx. Factoring When a ≠ 1 and ac > 0 Factor 6x2 – 31x + 35. Step 1: Find factors with product ac and sum b. 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. 2x(3x – 5) – 7(3x – 5) Find common factors. (2x – 7)(3x – 5) Rewrite using the Distributive Property.**Since ac = 210 and b = 11, find factors with product –210**and sum 11. Factors of –210 Sum of factors Factors of –210 Sum of factors –1, –210 –209 –3, 70 67 –1, 210 209 5, –42 –37 2, –105 –103 –5, 42 37 –2, 105 103 10, –21 –11 –10, 21 11 3, –70 –67 6x2 + 11x + 35 6x2 – 10x + 21x – 35 Rewrite bx. Factoring When a ≠ 1 and ac < 0 Factor 6x2 + 11x – 35. Step 1:Find factors with product ac and sum b. 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. 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)2Rewrite the first and third terms as squares. = (10x)2 + 180 + (9)2Rewrite the middle term to verify the perfect square trinomial pattern. = (10x + 9)2a2 + 2ab + b2 = (a + b)2**Relate: frame area equals the outer area minus the inner**area Write: area = x2 – (7)2 = (x + 7)(x – 7) 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. Define: Let x = length of side of frame. 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