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Example 1 Quiz on Monday on 8-1 to 8-3 and 8-5 Bell Ringer 1.Use the Distributive Property to factor the polynomial 3ab 2 + 15a 2 b 2 + 27ab 3. 2.15w –

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Presentation on theme: "Example 1 Quiz on Monday on 8-1 to 8-3 and 8-5 Bell Ringer 1.Use the Distributive Property to factor the polynomial 3ab 2 + 15a 2 b 2 + 27ab 3. 2.15w –"— Presentation transcript:

1 Example 1 Quiz on Monday on 8-1 to 8-3 and 8-5 Bell Ringer 1.Use the Distributive Property to factor the polynomial 3ab 2 + 15a 2 b 2 + 27ab 3. 2.15w – 3v 3.Video on Factoring by Grouping

2 Example 2 Factor 4xy + 3y – 20x – 15.

3 Example 3 Factor by Grouping with Additive Inverses Factor 15a – 3ab + 4b – 20.

4 Concept The solution to an equation is called the root. The root is any value that makes the equation true.

5 Example 4 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) = 0Original equation x – 2 = 0 or 4x – 1= 0 Zero Product Property x = 2 4x= 1Solve each equation. Divide.

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

7 Example 4 Solve Equations B. Solve 4y = 12y 2. Check the solution. Write the equation so that it is of the form ab = 0. 4y=12y 2 Original equation 4y – 12y 2 =0Subtract 12y 2 from each side. 4y(1 – 3y)=0Factor the GCF of 4y and 12y 2, which is 4y. 4y = 0 or 1 – 3y=0Zero Product Property y = 0–3y=–1Solve each equation. Divide.

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

9 Example 4 A. Solve (s – 3)(3s + 6) = 0. Then check the solution.

10 Example 4 B. Solve 5x – 40x 2 = 0. Then check the solution.

11 Example 5 Use Factoring FOOTBALL A football is kicked into the air. The height of the football can be modeled by the equation h = –16x 2 + 48x, where h is the height reached by the ball after x seconds. Find the values of x when h = 0. h= –16x 2 + 48x Original equation 0= –16x 2 + 48x h = 0 0= 16x(–x + 3)Factor by using the GCF. 16x= 0 or –x + 3= 0Zero Product Property x= 0 x= 3Solve each equation. Answer: 0 seconds, 3 seconds

12 Example 5 A.0 or 1.5 seconds B.0 or 7 seconds C.0 or 2.66 seconds D.0 or 1.25 seconds Juanita is jumping on a trampoline in her back yard. Juanita’s jump can be modeled by the equation h = –14t 2 + 21t, where h is the height of the jump in feet at t seconds. Find the values of t when h = 0.


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