1 / 33

Splash Screen

Splash Screen. 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

jemma
Download Presentation

Splash Screen

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Splash Screen

  2. 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 Lesson Menu

  3. Use the related graph of y = x2 – 4 to determine its solutions. A. 4, –4 B. 3, –2 C. 2, 0 D. 2, –2 5-Minute Check 1

  4. 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 5-Minute Check 2

  5. 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 5-Minute Check 3

  6. Use a quadratic equation to find two real numbers that have a sum of 5 and a product of –14. A. 10, –4 B. 5, –1 C. –2, 7 D. –5, 2 5-Minute Check 4

  7. Which term is not another name for a solution to a quadratic equation? A. zero B.x-intercept C. root D. vertex 5-Minute Check 5

  8. 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. CCSS

  9. You found the greatest common factors of sets of numbers. • Write quadratic equations in intercept form. • Solve quadratic equations by factoring. Then/Now

  10. factored form • FOIL method Vocabulary

  11. Concept

  12. Replace p withand q with –5. Translate Sentences into Equations (x – p)(x – q) = 0 Write the pattern. Simplify. Use FOIL. Example 1

  13. Answer: Translate Sentences into Equations Multiply each side by 2 so b and c are integers. Example 1

  14. A. ans B. ans C. ans D. ans Example 1

  15. Concept

  16. y = 0 Solve each equation. Answer: Factor GCF A.Solve 9y2 + 3y = 0. 9y2 + 3y = 0 Original equation 3y(3y) + 3y(1) = 0 Factor the GCF. 3y(3y + 1) = 0 Distributive Property 3y = 0 3y + 1 = 0 Zero Product Property Example 2

  17. Factor GCF B.Solve 5a2 – 20a = 0. 5a2 – 20a = 0 Original equation 5a(a) – 5a(4) = 0 Factor the GCF. 5a(5a – 4) = 0 Distributive Property 5a = 0 a – 4 = 0 Zero Product Property a = 0 a = 4 Solve each equation. Answer:0, 4 Example 2

  18. Solve 12x – 4x2 = 0. A. 3, 12 B. 3, –4 C. –3, 0 D. 3, 0 Example 2

  19. Perfect Squares and Differences of Squares A.Solve x2 – 6x + 9 = 0. x2 = (x)2; 9 = (3)2 First and last terms are perfect squares. 6x = 2(x)(3) Middle term equals 2ab. x2 – 6x + 9 is a perfect square trinomial. x2 + 6x + 9 = 0 Original equation (x – 3)2 = 0 Factor using the pattern. x – 3 = 0 Take the square root of each side. x = 3 Add 3 to each side. Answer:3 Example 3

  20. Perfect Squares and Differences of Squares B.Solve y2 = 36. y2 = 32 Original equation y2 – 36 = 0 Subtract 36 from each side. y2 – (6)2 = 0 Write in the form a2 – b2. (y + 6)(y – 6) = 0 Factor the difference of squares. y + 6 = 0 y – 6 = 0 Zero Product Property y = –6 y = 6 Solve each equation. Answer:–6, 6 Example 3

  21. Solve x2 – 16x + 64 = 0. A. 8, –8 B. 8, 0 C. 8 D. –8 Example 3

  22. Factor Trinomials A.Solve x2 – 2x – 15 = 0. ac = –15 a = 1, c = –15 Example 4

  23. Factor Trinomials x2 – 2x – 15 = 0 Original equation x2 + mx + px – 15 = 0 Write the pattern. x2 + 3x – 5x – 15 = 0 m = 3 and p = –5 (x2 + 3x) – (5x + 15) = 0 Group terms with common factors. x(x + 3) – 5(x + 3) = 0 Factor the GCF from each grouping. (x – 5)(x + 3) = 0 Distributive Property x – 5 = 0 x + 3 = 0 Zero Product Property x = 5 x = –3 Solve each equation. Answer:5, –3 Example 4

  24. Factor Trinomials B.Solve 5x2 + 34x + 24 = 0. ac = 120 a = 5, c = 24 Example 4

  25. x = –6 Solve each equation. Factor Trinomials 5x2 + 34x + 24 = 0 Original equation 5x2 + mx + px + 24 = 0 Write the pattern. 5x2 + 4x + 30x + 24 = 0 m = 4 and p = 30 (5x2 + 4x) + (30x + 24) = 0 Group terms with common factors. x(5x + 4) + 6(x + 4) = 0 Factor the GCF from each grouping. (x + 6)(5x + 4) = 0 Distributive Property x + 6 = 0 5x + 4 = 0 Zero Product Property Example 4

  26. Answer: Factor Trinomials Example 4

  27. A. B. C. D. Solve 6x2 – 5x – 4 = 0. Example 4

  28. B.Factor 3s2 – 11s – 4. A. (3s + 1)(s – 4) B. (s + 1)(3s – 4) C. (3s + 4)(s – 1) D. (s – 1)(3s + 4) Example 4

  29. Solve Equations by Factoring ARCHITECTURE The entrance to an office building is an arch in the shape of a parabola whose vertex is the height of the arch. The height of the arch is given by h = 9 – x2, where x is the horizontal distance from the center of the arch. Both h and x are measured in feet. How wide is the arch at ground level? To find the width of the arch at ground level, find the distance between the two zeros. Example 5

  30. Solve Equations by Factoring 9 – x2 = 0 Original expression x2 – 9 = 0 Multiply both sides by –1. (x + 3)(x – 3) = 0 Difference of squares x + 3 = 0 or x – 3 = 0 Zero Product Property x = –3 x = 3 Solve. Answer:The distance between 3 and – 3 is3 – (–3) or 6 feet. Example 5

  31. ? ? 9 – (3)2 = 0 or 9 – (–3)2 = 0 ? ? 9 – 9 = 0 9 – 9 = 0 Solve Equations by Factoring Check 9 – x2 = 0   0 = 0 0 = 0 Example 5

  32. TENNIS During a match, Andre hit a lob right off the court with the ball traveling in the shape of a parabola whose vertex was the height of the shot. The height of the shot is given by h = 49 – x2, where x is the horizontal distance from the center of the shot. Both h and x are measured in feet. How far was the lob hit? A. 7 feet B. 11 feet C. 14 feet D. 25 feet Example 5

  33. End of the Lesson

More Related