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# 9.5

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• End of day 1
• Day 2
• 1. -9, 1 2. s(3s^3+16) 3. 0, 7/2
• 1. 0, 1/2 2. 4xy(5xy-1) 3. 0, 5/2
• ### 9.5

1. 1. Zero-Product Property  If ab 0 , then a 0 or b 0 .
2. 2. Example 1 Use the zero-product property Solve ( x – 4 ) ( x + 2 ) = 0. (x – 4) (x + 2) = 0 x – 4= 0 or x = 4 or x +2= 0 x = –2 Write original equation. Zero-product property Solve for x. ANSWER The solutions of the equation are 4 and – 2.
3. 3. Example 1 Use the zero-product property CHECK Substitute each solution into the original equation to check. ? (4 – 4) (4 + 2) = 0 ? (– 2 – 4 ) (– 2 + 2 ) = 0 ? 0 •6 = 0 ? –6 • 0 = 0 0 = 0 0 = 0
4. 4. Factoring  To solve a polynomial equation using the zero-product property, factor the polynomial, which involves writing it as a product of other polynomials.  First step in factoring is to look for the greatest common monomial factor of the polynomial’s terms.
5. 5. Example 2 Find the greatest common monomial factor Factor out the greatest common monomial factor. a. 12x + 42y b. 4x4 + 24x3 SOLUTION 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)
6. 6. Example 2 Find the greatest common monomial factor 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 )
7. 7. Example 3 Solve an equation by factoring Solve 2x2 = – 8x. 2x2 = – 8x 2x2 + 8x = 0 2x ( x + 4 ) = 0 2x = 0 or x= 0 or x +4 = 0 x = –4 Write original equation. Add 8x to each side. Factor left side. Zero-product property Solve for x. ANSWER The solutions of the equation are 0 and – 4.
8. 8. Roots  A root of polynomial involving x is a value of x for which the corresponding value of the polynomial is 0.
9. 9. Example 4 Find the roots of a polynomial Find the roots of 6x2 – 15x. 6x2 – 15x = 0 Set polynomial equal to 0. 3x ( 2x – 5 ) = 0 Factor the polynomial. 3x = 0 or 2x – 5 = 0 x= 0 5 x = 2 or Zero-product property Solve for x. ANSWER 5 The roots of the polynomial are 0 and . 2
10. 10. 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 projectile, 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 is modeled by h 16t 2 vt s  t is the time (in seconds) the object has been in the air.  v is the initial vertical velocity (in feet per second).  s is the initial height (in feet).
11. 11. Example 5 Solve a multi-step problem SALMON As a salmon swims upstream, it leaps into the air with an initial vertical velocity of 10 feet per second. After how many seconds does the salmon return to the water? SOLUTION STEP 1 Write a model for the salmon’s height above the surface of the water. h = – 16t2 + vt + s Vertical motion model h = – 16t2 + 10t + 0 Substitute 10 for v and 0 for s. h = – 16t2 + 10t Simplify.
12. 12. Example 5 Solve a multi-step problem STEP 2 Substitute 0 for h. When the salmon returns to the water, its height above the surface is 0 feet. Solve for t. 0 = – 16t2 + 10t Substitute 0 for h. 0 = 2t (– 8t + 5 ) Factor right side. 2t = 0 or – 8t + 5 = 0 t = 0 or t = 0.625 Zero-product property Solve for t. ANSWER The salmon returns to the water 0.625 second after leaping.
13. 13. 9.5 Warm-Up (Day 1) Solve the equation 1. ( y 9)( y 1) 0 Factor out the greatest common monomial factor. 2. 3s 4 16s Solve the equation 3. 10n 2 35n 0
14. 14. 9.5 Warm-Up (Day 2) Find the roots of the polynomial. 1. 6h2 3h Factor out the greatest common monomial factor. 2. 20 x 2 y 2 4 xy Solve the equation 3. 12 p 2 30 p