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Section 8.5 Day 2 Using the Distributive Property

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1 Section 8.5 Day 2 Using the Distributive Property
Algebra 1

2 GCF Puzzle Problem 1 Fill in the blanks knowing the final answer.
(5___)(___π‘₯+6)=45 π‘₯ 2 +30π‘₯ 5π‘₯ 9π‘₯+6 =45 π‘₯ 2 +30π‘₯ Correct Answer: 15π‘₯(3π‘₯+2)

3 (2π‘₯𝑦)(__+__+__)=6 π‘₯ 2 𝑦 7 βˆ’8 π‘₯ 8 𝑦 2 +2π‘₯𝑦
GCF Puzzle Problem 2 Fill in the blanks knowing the final answer. (2π‘₯𝑦)(__+__+__)=6 π‘₯ 2 𝑦 7 βˆ’8 π‘₯ 8 𝑦 2 +2π‘₯𝑦 2π‘₯𝑦 3π‘₯ 𝑦 6 βˆ’4 π‘₯ 7 𝑦+1 =6 π‘₯ 2 𝑦 7 βˆ’8 π‘₯ 8 𝑦 2 +2π‘₯𝑦

4 Factoring Binomials: Example 1
1. Find GCF of both terms GCF =9𝑦 2. Pull out the GCF and write the remainder 9𝑦(3𝑦+2)

5 Factoring Binomials: Example 2
GCF=3π‘₯ 𝑦 3 3π‘₯ 𝑦 3 (2π‘₯π‘¦βˆ’3)

6 Factoring Polynomials: Example 3
1. GCF GCF: 2π‘Žπ‘ 2. Factor out GCF 2π‘Žπ‘(βˆ’2π‘Žβˆ’4𝑏+1)

7 Factoring Polynomials: Example 4
GCF: π‘’βˆ™π‘‘=𝑒𝑑 𝑒𝑑(7𝑒𝑑+21π‘‘βˆ’1)

8 Factoring Polynomials: Example 5
GCF: 3 π‘Ÿ 2 𝑑 3 π‘Ÿ 2 𝑑(6π‘Ÿπ‘‘+2π‘‘βˆ’2)

9 Zero Product Property If the product of two factors is 0, then at least one of the factors must be 0. ____ βˆ™_____=0 Option 1: # βˆ™0=0 Option 2: 0βˆ™#=0 Option 3: 0βˆ™0=0

10 Zero Product Property: Example 6
Solve 2𝑑+6 3π‘‘βˆ’15 =0 By ZPP, you can set each factor equal to zero and solve. 2𝑑+6=0 3π‘‘βˆ’15=0 2𝑑=βˆ’6 3𝑑=15 𝑑=βˆ’ and 𝑑=5 The solutions of an equation are also called roots.

11 Zero Product Property: Example 7
Solve 4π‘₯ 8π‘₯βˆ’6 =0 4π‘₯=0 8π‘₯βˆ’6=0 π‘₯=0 8π‘₯=6 π‘₯=0 and π‘₯= 6 8 = 3 4 π‘₯=0 & π‘₯= 3 4 are the solutions/roots of the equation

12 Zero Product Property: Example 8
Solve 𝑐 2 =3𝑐 Get everything to one side. Need to have "=0" ! 𝑐 2 βˆ’3𝑐=0 (solve) 𝑐 π‘βˆ’3 =0 (factor) 𝑐=0 π‘βˆ’3=0 𝑐=0 and 𝑐=3

13 Zero Product Property: Example 9
Solve 8 𝑏 2 =40𝑏 1. Solve 8 𝑏 2 βˆ’40𝑏=0 2. Factor 8𝑏 π‘βˆ’5 =0 3. ZPP 8𝑏=0 π‘βˆ’5=0 𝑏= and 𝑏=5


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