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Over Lesson 8– 4

Over Lesson 8– 4

Using the Distributive Property Lesson 8 -5

Understand how to use the Distributive Property to factor polynomials and solve equations of the form ax 2 + bx = 0.

Use the Distributive Property A. Use the Distributive Property to factor 15 x + 25 x 2. First, find the GCF of 15 x + 25 x 2. 15 x = 3 ● 5 ● x Factor each monomial. 25 x 2 = 5 ● x ● x Circle the common prime factors. GCF = 5 ● x or 5 x Write each term as the product of the GCF and its remaining factors. Then use the Distributive Property to factor out the GCF.

Use the Distributive Property 15 x + 25 x 2 = 5 x(3) + 5 x(5 ● x) Rewrite each term using the GCF. = 5 x(3 + 5 x) Distributive Property Answer: The completely factored form of 15 x + 25 x 2 is 5 x(3 + 5 x).

Use the Distributive Property B. Use the Distributive Property to factor 12 xy + 24 xy 2 – 30 x 2 y 4. 12 xy = 2 ● 3 ● x ● y 24 xy 2 = 2 ● 2 ● 3 ● x ● y Factor each term. – 30 x 2 y 4 = – 1 ● 2 ● 3 ● 5 ● x ● y ● y Circle common factors. GCF = 2 ● 3 ● x ● y or 6 xy

Use the Distributive Property 12 xy + 24 xy 2 – 30 x 2 y 4 = 6 xy(2) + 6 xy(4 y) + 6 xy(– 5 xy 3) Rewrite each term using the GCF. = 6 xy(2 + 4 y – 5 xy 3) Distributive Property Answer: The factored form of 12 xy + 24 xy 2 – 30 x 2 y 4 is 6 xy(2 + 4 y – 5 xy 3).

A. Use the Distributive Property to factor the polynomial 3 x 2 y + 12 xy 2. B. Use the Distributive Property to factor the polynomial 3 ab 2 + 15 a 2 b 2 + 27 ab 3.

Factor by Grouping Factor 2 xy + 7 x – 2 y – 7 Answer: = (2 xy – 2 y) + (7 x – 7) Group terms with common factors. = 2 y(x – 1) + 7(x – 1) Factor the GCF from each group. = (2 y + 7)(x – 1) Property Distributive (2 y + 7)(x – 1) or (x – 1)(2 y + 7)

Factor 4 xy + 3 y – 20 x – 15.

Factor by Grouping with Additive Inverses Factor 15 a – 3 ab + 4 b – 20 = (15 a – 3 ab) + (4 b – 20) Group terms with common factors. = 3 a(5 – b) + 4(b – 5) Factor the GCF from each group. = 3 a(– 1)(b – 5) + 4(b – 5) 5 – b = – 1(b – 5) = – 3 a(b – 5) + 4(b – 5) 3 a(– 1) = – 3 a = (– 3 a + 4)(b – 5) Distributive Property Answer: (– 3 a + 4)(b – 5) or (3 a – 4)(5 – b)

Factor – 2 xy – 10 x + 3 y + 15.

Solve Equations A. Solve (x – 2)(4 x – 1) = 0. Check the solution.

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

Solve Equations B. Solve 4 y = 12 y 2. Check the solution.

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

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

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

Juanita is jumping on a trampoline in her back yard. Juanita’s jump can be modeled by the equation h = – 14 t 2 + 21 t, where h is the height of the jump in feet at t seconds. Find the values of t when h = 0.

Homework p. 498 #15 -45 odd, 52, 75 -79 odd