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Five-Minute Check (over Lesson 4–2) CCSS Then/Now New Vocabulary Key Concept: FOIL Method for Multiplying Binomials Example 1: Translate Sentences into Equations Concept Summary: Zero Product Property Example 2: Factor GCF Example 3: Perfect Squares and Differences of Squares Example 4: Factor Trinomials Example 5: Real-World Example: Solve Equations by Factoring Over Lesson 4–2 Use the related graph of y = x2 – 4 to determine its solutions. A. 4, –4 B. 3, –2 C. 2, 0 D. 2, –2 Over Lesson 4–2 Use the related graph of y = –x2 – 2x + 3 to determine its solutions. A. –3, 1 B. –3, 3 C. –1, 3 D. 3, 1 Over Lesson 4–2 Solve –2x2 + 5x = 0. If exact roots cannot be found, state the consecutive integers between which the roots are located. A. 0 B. 0, between 2 and 3 C. between 1 and 2 D. 2, –2 Over Lesson 4–2 Which term is not another name for a solution to a quadratic equation? A. zero B. x-intercept C. root D. vertex Content Standards A.SSE.2 Use the structure of an expression to identify ways to rewrite it. F.IF.8.a Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context. Mathematical Practices 2 Reason Abstractly and quantitatively. You found the greatest common factors of sets of numbers. • Write quadratic equations in intercept form. • Solve quadratic equations by factoring. • factored form • FOIL method Translate Sentences into Equations (x – p)(x – q) = 0 Write the pattern. Replace p with and q with –5. Simplify. Use FOIL. Translate Sentences into Equations Multiply each side by 2 so b and c are integers. Answer: A. ans B. ans C. ans D. ans Factor GCF A. Solve 9y 2 + 3y = 0. 9y 2 + 3y = 0 3y(3y) + 3y(1) = 0 3y(3y + 1) = 0 3y = 0 3y + 1 = 0 y=0 Answer: Original equation Factor the GCF. Distributive Property Zero Product Property Solve each equation. Factor GCF B. Solve 5a2 – 20a = 0. 5a 2 – 20a = 0 5a(a) – 5a(4) = 0 5a(a – 4) = 0 5a = 0 a–4=0 a=0 a=4 Answer: 0, 4 Original equation Factor the GCF. Distributive Property Zero Product Property Solve each equation. Solve 12x – 4x2 = 0. A. 3, 12 B. 3, –4 C. –3, 0 D. 3, 0 Perfect Squares and Differences of Squares A. Solve x 2 – 6x + 9 = 0. x 2 = (x)2; 9 = (3)2 First and last terms are perfect squares. 6x = 2(x)(3) Middle term equals 2ab. x 2 – 6x + 9 is a perfect square trinomial. x 2 + 6x + 9 = 0 (x – 3)2 = 0 x–3 =0 x =3 Answer: 3 Original equation Factor using the pattern. Take the square root of each side. Add 3 to each side. Perfect Squares and Differences of Squares B. Solve y 2 = 36. y 2 = 36 y2 – 36 = 0 Subtract 36 from each side. y2 – (6)2 = 0 Write in the form a2 – b2. (y + 6)(y – 6) = 0 y+6=0 Factor the difference of squares. y–6=0 y = –6 Answer: –6, 6 Original equation y=6 Zero Product Property Solve each equation. Solve x 2 – 16x + 64 = 0. A. 8, –8 B. 8, 0 C. 8 D. –8 Factor Trinomials A. Solve x 2 – 2x – 15 = 0. ac = –15 a = 1, c = –15 Factor Trinomials x 2 – 2x – 15 = 0 Original equation x2 + mx + px – 15 = 0 Write the pattern. x 2 + 3x – 5x – 15 = 0 m = 3 and p = –5 (x 2 + 3x) – (5x + 15) = 0 Group terms with common factors. x(x + 3) – 5(x + 3) = 0 (x – 5)(x + 3) = 0 x–5=0 x+3 =0 x=5 Answer: 5, –3 x = –3 Factor the GCF from each grouping. Distributive Property Zero Product Property Solve each equation. Factor Trinomials B. Solve 5x 2 + 34x + 24 = 0. ac = 120 a = 5, c = 24 Factor Trinomials 5x 2 + 34x + 24 = 0 Original equation 5x2 + mx + px + 24 = 0 Write the pattern. 5x 2 + 4x + 30x + 24 = 0 m = 4 and p = 30 (5x 2 + 4x) + (30x + 24) = 0 Group terms with common factors. x(5x + 4) + 6(5x + 4) = 0 Factor the GCF from each grouping. (x + 6)(5x + 4) = 0 x+6=0 x = –6 5x + 4 = 0 Distributive Property Zero Product Property Solve each equation. Factor Trinomials Answer: A. Solve 6x 2 – 5x – 4 = 0. A. B. C. D. B. Factor 3s 2 – 11s – 4. A. (3s + 1)(s – 4) B. (s + 1)(3s – 4) C. (3s + 4)(s – 1) D. (s – 1)(3s + 4) Page 242 #18 - 42