1. Do Now Pass out calculators. Work on practice EOC Week # 8.

2. Quick Check: 1.(x 2 – 3x + 5) + (-2x 2 + 11x + 1) 2.(8y 3 – 7y 2 + y) – (9y 2 – 5y + 7) 3. -3x 2 (x 3 – 3x 2 ) 4.(2r + 11)(r – 6) 5.(m + 3)(-2m 2 + 5m – 1) 6. (5w + 9z) 2

3. Answers: 1.–x 2 + 8x +6 2.8y 3 – 16y 2 +6y – 7 3.-3x 5 + 9x 4 4.2r 2 – r – 66 5. -2m 3 – m 2 +14m – 3 6. 25w 2 90wz +81z 2

4. Objective: To use the zero product property and factor using the greatest common factor.

5. Zero – Product Property: The zero-product property is used to solve an equation when one side is zero and the other side is two polynomials being multiplied. The solutions of an equations like are called roots.

6. Use the zero-product property EXAMPLE 1 Solve (x – 4)(x + 2) = 0. (x – 4)(x + 2) = 0 Write original equation. x – 4 = 0 x = 4 Zero-product property Solve for x. ANSWER The solutions of the equation are 4 and –2. or x + 2 = 0 x = – 2

7. Use the zero-product property EXAMPLE 1 CHECK Substitute each solution into the original equation to check. (4  4)(4 + 2) = 0 0  6 = 0 0 = 0 ? ? (  2  4)(  2 + 2) = 0  6  0 = 0 0 = 0 ? ?

8. GUIDED PRACTICE for Example 1 1. Solve the equation (x – 5)(x – 1) = 0. ANSWER The solutions of the equation are 5 and 1.

9. SOLUTION EXAMPLE 2 Find the greatest common monomial factor Factor out the greatest common monomial factor. a. 12x + 42y a. The GCF of 12 and 42 is 6. The variables x and y have no common factor. So, the greatest common monomial factor of the terms is 6. ANSWER 12x + 42y = 6(2x + 7y)

10. EXAMPLE 2 Find the greatest common monomial factor b.The GCF of 4 and 24 is 4. The GCF of x 4 and x 3 is x 3. So, the greatest common monomial factor of the terms is 4x 3. ANSWER 4x 4 + 24x 3 = 4x 3 (x + 6) SOLUTION Factor out the greatest common monomial factor. b. 4x 4 + 24x 3

11. GUIDED PRACTICE for Example 2 2. Factor out the greatest common monomial factor from 14m + 35n. ANSWER 14m + 35n = 7(2m + 5n)

12. EXAMPLE 3 Solve an equation by factoring Solve 2x 2 + 8x = 0. 2x 2 + 8x = 0 2x(x + 4) = 0 2x = 0 x = 0 or x + 4 = 0 or x = – 4 ANSWER The solutions of the equation are 0 and – 4. Solve for x. Zero-product property Factor left side. Write original equation.

13. EXAMPLE 4 Solve an equation by factoring Solve 6n 2 = 15n. 6n 2 – 15n = 0 3n(2n – 5) = 0 3n = 0 n = 0 2n – 5 = 0 n = 5 2 or Solve for n. Zero-product property Factor left side. Subtract 15n from each side. ANSWER The solutions of the equation are 0 and 5 2.

14. GUIDED PRACTICE for Examples 3 and 4 Solve the equation. ANSWER 0 and – 5 3. a 2 + 5a = 0 4. 3s 2 – 9s = 0 ANSWER 0 and 3 5. 4x 2 = 2x. ANSWER 0 and 1 2. 1 2

15. Vertical Motion: A projectile is an object that is propelled into the air but has no power to keep itself in the air. A thrown ball is a projective, but an airplane is not. The height of a projectile can be described by the vertical motion model. The height h (in feet) of a projectile can be modeled by: h = -16t 2 + vt + x t = time (in seconds) the object has been in the air v = initial velocity (in feet per second) s = the initial height (in feet)

16. ARMADILLO EXAMPLE 5 Solve a multi-step problem A startled armadillo jumps straight into the air with an initial vertical velocity of 14 feet per second. After how many seconds does it land on the ground ?

17. SOLUTION EXAMPLE 5 Solve a multi-step problem STEP 1 Write a model for the armadillo’s height above the ground. h = –16t 2 + vt + s h = –16t 2 + 14t + 0 h = –16t 2 + 14t Vertical motion model Substitute 14 for v and 0 for s. Simplify.

18. EXAMPLE 5 Solve a multi-step problem STEP 2 Substitute 0 for h. When the armadillo lands, its height above the ground is 0 feet. Solve for t. 0 = –16t 2 + 14t 0 = 2t(–8t + 7) 2t = 0 t = 0 –8t + 7 = 0 t = 0.875 or Solve for t. Zero-product property Factor right side. Substitute 0 for h. ANSWER The armadillo lands on the ground 0.875 second after the armadillo jumps.

19. GUIDED PRACTICE for Example 5 6. WHAT IF ? In Example 5, suppose the initial vertical velocity is 12 feet per second.After how many seconds does armadillo land on the ground ? ANSWER The armadillo lands on the ground 0.75 second after the armadillo jumps.

20. Exit Ticket 1.Solve (x + 3)(x – 5) = 0 Why does this type of problem have two solutions? 2. Factor out the greatest common monomial factor. a. 8x +12 y b. 12y 2 + 21y