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Warm Up

Solve Polynomial Equations in Factored Form. Warm Up. Lesson Presentation. Lesson Quiz. 3.

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Warm Up

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  1. Solve Polynomial Equations in Factored Form Warm Up Lesson Presentation Lesson Quiz

  2. 3. The number (in hundreds) of sunscreen and sun tanning products sold at a pharmacy from 1999–2005 can be modeled by –0.8t2 + 0.3t + 107, where t is the number of years since 1999. About how many products were sold in 2002? 4 ANSWER about 10,070 ANSWER 6 ANSWER Warm-Up 1.Find theGCF of 12 and 28. 2.Find the GCF of 18 and 42.

  3. ANSWER The solutions of the equation are4and –2. Example 1 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.

  4. ? ? (2  4)(2 + 2) = 0 6  0 = 0 0 = 0 (4  4)(4 + 2) = 0 0  6 = 0 0 = 0 ? ? Example 1 CHECK Substitute each solution into the original equation to check.

  5. ANSWER The solutions of the equation are5and 1. Guided Practice 1. Solve the equation (x – 5)(x – 1) = 0.

  6. 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 Factor out the greatest common monomial factor. a.12x + 42y b. 4x4+ 24x3 SOLUTION

  7. 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 SOLUTION

  8. ANSWER 14m+ 35n = 7(2m + 5n) Guided Practice 2. Factor out the greatest common monomial factor from 14m + 35n.

  9. ANSWER The solutions of the equation are 0 and – 4. Example 3 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.

  10. 5 5 2 2 n = ANSWER . The solutions of the equation are 0 and Example 4 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.

  11. 1 2 ANSWER ANSWER ANSWER 0 and – 5 0 and 3 0 and Guided Practice Solve the equation. 3. a2+ 5a = 0 4. 3s2– 9s = 0 5.4x2 = 2x

  12. Example 5 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?

  13. Example 5 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.

  14. ANSWER The armadillo lands on the ground 0.875 second after the armadillo jumps. Example 5 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.

  15. ANSWER The armadillo lands on the ground 0.75 second after the armadillo jumps. 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?

  16. ANSWER 0, – 5, 9 ANSWER 3 2 3 – , 4 ANSWER 2 10 ANSWER 0, 3 Lesson Quiz Solve the equation. 1. (y + 5)(y – 9) = 0 2. (2n + 3)(n – 4) = 0 3. 6x2 = 20x 4. 12x2 = 18x

  17. 5. A dog jumps in the air with an initial velocity of 18 feet per second to catch a flying disc. How long does the dog remain in the air? ANSWER 1.125 sec Lesson Quiz

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