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Do Now

Do Now. Pass out calculators. Work on practice EOC Week # 8. Quick Check:. (x 2 – 3x + 5) + (-2x 2 + 11x + 1) (8y 3 – 7y 2 + y) – (9y 2 – 5y + 7) -3x 2 (x 3 – 3x 2 ) (2r + 11)(r – 6) (m + 3)(-2m 2 + 5m – 1) (5w + 9z) 2. Answers:. –x 2 + 8x +6 8y 3 – 16y 2 +6y – 7

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Do Now

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

  2. Quick Check: • (x2 – 3x + 5) + (-2x2 + 11x + 1) • (8y3 – 7y2 + y) – (9y2 – 5y + 7) • -3x2(x3 – 3x2) • (2r + 11)(r – 6) • (m + 3)(-2m2 + 5m – 1) • (5w + 9z)2

  3. Answers: • –x2 + 8x +6 • 8y3 – 16y2 +6y – 7 • -3x5 + 9x4 • 2r2 – r – 66 • -2m3 – m2 +14m – 3 • 25w2 90wz +81z2

  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. ANSWER The solutions of the equation are4and –2. EXAMPLE 1 Use the zero-product property Solve (x – 4)(x + 2) = 0. (x – 4)(x +2) = 0 Write original equation. x – 4 = 0 or x + 2 = 0 Zero-product property x = 4 or x = – 2 Solve forx.

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

  8. ANSWER The solutions of the equation are5and 1. for Example 1 GUIDED PRACTICE 1. Solve the equation (x – 5)(x – 1) = 0.

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

  10. b. The GCF of 4 and 24 is 4. The GCF of x4 and x3 is x3. So, the greatest common monomial factor of the terms is 4x3. ANSWER 4x4+ 24x3 = 4x3(x + 6) EXAMPLE 2 Find the greatest common monomial factor Factor out the greatest common monomial factor. b. 4x4+ 24x3 SOLUTION

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

  12. ANSWER The solutions of the equation are 0 and – 4. EXAMPLE 3 Solve an equation by factoring Solve2x2+ 8x = 0. 2x2+ 8x = 0 Write original equation. 2x(x + 4) = 0 Factor left side. or x + 4 = 0 2x= 0 Zero-product property or x = 0 x =– 4 Solve for x.

  13. 5 5 2 2 n = ANSWER . The solutions of the equation are 0 and EXAMPLE 4 Solve an equation by factoring Solve 6n2 = 15n. 6n2– 15n = 0 Subtract 15nfrom each side. 3n(2n – 5) =0 Factor left side. or 2n – 5= 0 3n= 0 Zero-product property or n = 0 Solve for n.

  14. 1 1 2 2 ANSWER ANSWER ANSWER 0 and – 5 0 and 3 . 0 and for Examples 3 and 4 GUIDED PRACTICE Solve the equation. 3. a2+ 5a = 0 5.4x2 = 2x. 4. 3s2– 9s = 0

  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 = -16t2 + 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. EXAMPLE 5 Solve a multi-step problem ARMADILLO 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. EXAMPLE 5 Solve a multi-step problem SOLUTION STEP 1 Write a model for the armadillo’s height above the ground. h =–16t2+vt +s Vertical motion model h =–16t2+14t +0 Substitute 14 for vand 0 for s. h =–16t2 + 14t Simplify.

  18. ANSWER The armadillo lands on the ground 0.875 second after the armadillo jumps. 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 =–16t2+ 14t Substitute 0 for h. 0 =2t(–8t+ 7) Factor right side. 2t = 0 –8t + 7 = 0 or Zero-product property or t = 0 t = 0.875 Solve for t.

  19. ANSWER The armadillo lands on the ground 0.75 second after the armadillo jumps. for Example 5 GUIDED PRACTICE 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?

  20. Exit Ticket • 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. 12y2 + 21y

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