Factoring Quadratic Expressions Lesson 5-4 Additional Examples 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.

Factoring Quadratic Expressions Lesson 5-4 Additional Examples Factor x2 + 10x + 24. Step 1: Find factors with product ac and sum b. Factors of 24 Sum of factors 1, 24 25 2, 12 14 3, 8 11 6, 4 10 Since ac = 24 and b = 10, find positive factors with product 24 and sum 11. 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.

Factoring Quadratic Expressions Lesson 5-4 Additional Examples (continued) Step 3: Rewrite the expression as a product of two binominals. (x + 6)(x + 4) Rewrite using the Distributive Property. x(x + 4) + 6(x + 4) Check: (x + 6)(x + 4) = x2 + 4x + 6x + 24 = x2 + 10x + 24

Factoring Quadratic Expressions Lesson 5-4 Additional Examples Factor x2 – 14x + 33. Step 1: Find factors with product ac and sum b. Factors of 33 Sum of factors –1, –33 –34 –3, –11 –14 Since ac = 33 and b = –14, find negative factors with product 33 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. } x2 + 14x + 33 x2 – 3x – 11x + 33 Rewrite bx. x(x – 3) – 11(x – 3) Find common factors. (x – 11)(x – 3) Rewrite using the Distributive Property.

Factoring Quadratic Expressions Lesson 5-4 Additional Examples Factor x2 + 3x –28. Step 1: Find factors with product ac and sum b. Factors of –28 Sum of factors 1, –28 –27 –1, 28 27 2, –14 –12 –2, 14 12 4, –7 –3 –4, 7 3 Since ac = –28 and b = 3, find factors 2 with product –28 and sum 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 Quadratic Expressions Lesson 5-4 Additional Examples Factor 6x2 – 31x + 35. Step 1: Find factors with product ac and sum b. Factors of 210 Sum of factors –1, –210 –211 –2, –105 –107 –3, –70 –73 –5, –42 –47 –10, –21 –31 Since ac = 210 and b = –31, find negative factors with product 210 and sum –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 6x2 – 10x – 21x + 35 Rewrite bx. } 2x(3x – 5) – 7(3x – 5) Find common factors. (2x – 7)(3x – 5) Rewrite using the Distributive Property.

Factoring Quadratic Expressions Lesson 5-4 Additional Examples Factor 6x2 + 11x – 35. Step 1: Find factors with product ac and sum b. Factors of –210 Sum of factors –1, –210 –209 –1, 210 209 2, –105 –103 –2, 105 103 3, –70 –67 Since ac = 210 and b = 11, find factors with product –210 and sum 11. –3, 70 67 5, –42 –37 –5, 42 37 10, –21 –11 –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.

Factoring Quadratic Expressions Lesson 5-4 Additional Examples 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

Factoring Quadratic Expressions Lesson 5-4 Additional Examples 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.