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5-Minute Check 1

A B C D. Factor 70 x 2 y. 2 ● 5 ● 7 ● x ● x ● y. 5-Minute Check 1. A B C D. What is the GCF of 4 a 2 b , 8 ab 2 , and –4 b 3 ?. 4 b. 5-Minute Check 4. A B C D.

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5-Minute Check 1

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  1. A B C D Factor 70x2y. 2 ● 5 ● 7 ● x ● x ● y 5-Minute Check 1

  2. A B C D What is the GCF of 4a2b, 8ab2, and –4b3? 4b 5-Minute Check 4

  3. A B C D David is separating 72 students into at least three teams. What size teams will divide the students into equal groups of no fewer than 15 students each? 3 teams of 24 students 5-Minute Check 5

  4. A B C D What is the greatest monomial that can be factored out of 12r4n3 – 24r2n2t + 18r3n4t2? 6r2n2 5-Minute Check 6

  5. Use the Distributive Property to factor polynomials. • Solve quadratic equations of the form ax2 + bx = 0. Then/Now

  6. Remember doing this? 5x(x – 4) Vocabulary

  7. Use the Distributive Property A. Use the Distributive Property to factor 15x + 25x2. First, find the GCF of 15x + 25x2. 15x = 3 ● 5 ● x Factor each monomial. Circle the common prime factors. 25x2 = 5 ● 5 ● x ● x GCF: 5 ● x or 5x Write each term as the product of the GCF and its remaining factors. Then use the Distributive Property to factor out the GCF. Example 1

  8. Use the Distributive Property = 5x(3) + 5x(5x) Simplify remaining factors. 15x + 25x2 = 5x(3) + 5x(5 ● x) Rewrite each term using the GCF. = 5x(3 + 5x) Distributive Property Answer: The completely factored form of 15x + 25x2 is 5x(3 + 5x). Example 1

  9. 12xy = 2 ● 2 ● 3 ● x ● y Factor each monomial. 24xy2 = 2 ● 2 ● 2 ● 3 ● x ● y ● y 30x2y4 = 2 ● 3 ● 5 ● x ● x● y ● y ● y ● y Use the Distributive Property B. Use the Distributive Property to factor 12xy + 24xy2 – 30x2y4. Circle the common prime factors. GCF: 2 ● 3 ● x ● y or 6xy Example 1

  10. Use the Distributive Property 12xy + 24xy2 – 30x2y4 = 6xy(2) + 6xy(4y) + 6xy(–5xy3) Rewrite each term using the GCF. = 6xy(2 + 4y – 5xy3) Distributive Property Answer: The factored form of 12xy + 24xy2 – 30x2y4 is 6xy(2 + 4y – 5xy3). Example 1

  11. A B C D A. Use the Distributive Property to factor the polynomial 3x2y + 12xy2. 3xy(x + 4y) Example 1

  12. A B C D B. Use the Distributive Property to factor the polynomial 3ab2 + 15a2b2 + 27ab3. 3ab2(1 + 5a + 9b) Example 1

  13. Remember doing this? (5x + 3)(x – 4) Vocabulary

  14. Concept

  15. Factor by Grouping Factor 2xy + 7x – 2y – 7. 2xy + 7x – 2y – 7 = (2xy – 2y) + (7x – 7) Group terms with common factors. = 2y(x – 1) + 7(x – 1) Factor the GCF from each grouping. = (x – 1)(2y + 7) Distributive Property Answer: (x – 1)(2y + 7) Example 2

  16. A B C D Factor 4xy + 3y – 20x – 15. (y – 5)(4x + 3) Example 2

  17. Factor by Grouping with Additive Inverses Factor 15a – 3ab + 4b – 20. 15a – 3ab + 4b – 20 = (15a – 3ab) + (4b – 20) Group terms with common factors. = 3a(5 – b) + 4(b – 5) Factor GCF from each grouping. = 3a(–1)(b – 5) + 4(b – 5) 5 – b = –1(b – 5) = –3a(b – 5) + 4(b – 5) 3a(–1) = –3a = (b – 5)(–3a + 4) Distributive Property Answer: (b – 5)(–3a + 4) Example 3

  18. A B C D Factor –2xy – 10x + 3y + 15. (y + 5)(–2x + 3) Example 3

  19. -6mp + 4m + 18p - 12 (-2m + 6)(3p - 2)

  20. Concept

  21. Solve Equations A. Solve (x – 2)(4x – 1) = 0. Check the solution. If (x – 2)(4x – 1) = 0, then according to the Zero Product Property, either x – 2 = 0 or 4x – 1 = 0. (x – 2)(4x – 1) = 0 Original equation x – 2 = 0 or 4x – 1 = 0 Zero Product Property. x = 2 4x = 1 Solve each equation. Example 4

  22. Check Substitute 2 and for x in the original equation. ? ? (2 – 2)(4 ● 2 – 1) = 0 ? ? (0)(7) = 0 Solve Equations (x – 2)(4x – 1) = 0 (x – 2)(4x – 1) = 0 0 = 0 0 = 0   Example 4

  23. y = 0 Solve Equations B. Solve 4y = 12y2. Check the solution. Write the equation so that it is of the form ab = 0. 4y = 12y2 Original equation 4y – 12y2 = 0 Subtract 12y2 from each side. 4y(1 – 3y) = 0 Factor the GCF of 4y and 12y2, which is 4y. 4y = 0 or 1 – 3y = 0 Zero Product Property –3y = –1 Solve each equation. Example 4

  24. Answer: The roots are 0 and . Check by substituting 0 and for y in the original equation. 1 1 __ __ 3 3 Solve Equations Example 4

  25. A B C D A. Solve (s – 3)(3s + 6) = 0. Then check the solution. {3, –2} Example 4

  26. A B C D B. Solve 5x – 40x2 = 0. Then check the solution. Example 4

  27. Use Factoring FOOTBALL A football is kicked into the air. The height of the football can be modeled by the equation h = 16x2 + 16x, where h is the height reached by the ball after x seconds. Find the values of x when h = 0. h = 16x2 + 16 Original equation 0= 16x2 + 16h = 0 0 = 16x(–x + 1) Factor by using the GCF. 16x = 0 or – x + 1 = 0 Zero Product Property x = 0 x = 1 Solve each equation. Answer: 0 seconds, 1 second Example 5

  28. A B C D Juanita is jumping on a trampoline in her back yard. Juanita’s jump can be modeled by the equation h = –14t2 + 21t, where h is the height of the jump in feet at t seconds. Find the values of t when h = 0. t = 0 or 1.5 Example 5

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