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Section 8.5 Day 2 Using the Distributive Property
Algebra 1
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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)
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(2π₯π¦)(__+__+__)=6 π₯ 2 π¦ 7 β8 π₯ 8 π¦ 2 +2π₯π¦
GCF Puzzle Problem 2 Fill in the blanks knowing the final answer. (2π₯π¦)(__+__+__)=6 π₯ 2 π¦ 7 β8 π₯ 8 π¦ 2 +2π₯π¦ 2π₯π¦ 3π₯ π¦ 6 β4 π₯ 7 π¦+1 =6 π₯ 2 π¦ 7 β8 π₯ 8 π¦ 2 +2π₯π¦
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Factoring Binomials: Example 1
1. Find GCF of both terms GCF =9π¦ 2. Pull out the GCF and write the remainder 9π¦(3π¦+2)
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Factoring Binomials: Example 2
GCF=3π₯ π¦ 3 3π₯ π¦ 3 (2π₯π¦β3)
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Factoring Polynomials: Example 3
1. GCF GCF: 2ππ 2. Factor out GCF 2ππ(β2πβ4π+1)
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Factoring Polynomials: Example 4
GCF: π’βπ‘=π’π‘ π’π‘(7π’π‘+21π‘β1)
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Factoring Polynomials: Example 5
GCF: 3 π 2 π‘ 3 π 2 π‘(6ππ‘+2π‘β2)
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Zero Product Property If the product of two factors is 0, then at least one of the factors must be 0. ____ β_____=0 Option 1: # β0=0 Option 2: 0β#=0 Option 3: 0β0=0
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Zero Product Property: Example 6
Solve 2π+6 3πβ15 =0 By ZPP, you can set each factor equal to zero and solve. 2π+6=0 3πβ15=0 2π=β6 3π=15 π=β and π=5 The solutions of an equation are also called roots.
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Zero Product Property: Example 7
Solve 4π₯ 8π₯β6 =0 4π₯=0 8π₯β6=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
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Zero Product Property: Example 8
Solve π 2 =3π Get everything to one side. Need to have "=0" ! π 2 β3π=0 (solve) π πβ3 =0 (factor) π=0 πβ3=0 π=0 and π=3
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Zero Product Property: Example 9
Solve 8 π 2 =40π 1. Solve 8 π 2 β40π=0 2. Factor 8π πβ5 =0 3. ZPP 8π=0 πβ5=0 π= and π=5
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