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Warm β up #4 Factor π₯ 2 β3x 2. π₯ 2 β π₯ 4 β 3π¦ 4
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Warm β up #4 Solutions Factor π₯ 2 β3x = 3x(2x β 1) 2. π₯ 2 β25 = ( ) 2 β 2 = (x β 5)(x + 5) x 5
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Warm β up #4 Solutions 3. 48π₯ 4 β 3π¦ 4 = 3(16 π₯ 4 β π¦ 4 ) = ( ) 2 β 2 = 3(4 π₯ 2 β π¦ 2 )(4 π₯ 2 + π¦ 2 ) = 3(2x β y)(2x + y)(4 π₯ 2 + π¦ 2 ) 4 π₯ 2 π¦ 2 2x y
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Homework Log Fri 11/6 Lesson 4 β 4 Learning Objective:
To factor difference and sum of cubes & by grouping Hw: Factoring WS 2
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11/6/15 Lesson 4 β 4 Factoring Day 2
Algebra II
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Learning Objective To factor difference of cubes
To factor sum of cubes To factor by grouping
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Factoring Difference of Squares π₯ 2 β π¦ 2 =(π₯βπ¦)(π₯+π¦) Sum of Squares π₯ 2 + π¦ 2 = Canβt be factored!
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Factoring Difference of Cubes π₯ 3 β π¦ 3 =(π₯βπ¦)( π₯ 2 +π₯π¦+ π¦ 2 ) Sum of Cubes π₯ 3 + π¦ 3 =(π₯+π¦)( π₯ 2 βπ₯π¦+ π¦ 2 )
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Difference & Sum of Cube βSongβ
Pull out the cube roots Square the first term Change the Sign First times the second term Plus the last term squared
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Difference of Cubes π₯ 3 β π¦ 3 =(π₯βπ¦)( π₯ 2 +π₯π¦+ π¦ 2 ) 1. 8π₯ 3 β1 = ( ) 3 β ( ) 3 = (2x β 1) 2. 27π₯ 3 β8 = (3x β 2) 2x 1 (4 π₯ 2 + 2x + 1) 3x 2 (9 π₯ 2 + 6x + 4)
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Sum of Cubes π₯ 3 + π¦ 3 =(π₯+π¦)( π₯ 2 βπ₯π¦+ π¦ 2 ) 3. 64π₯ = ( ) 3 + ( ) 3 = (4x + 5) 4. π₯ π¦ 3 = (x+6y) 4x 5 (16 π₯ 2 - 20x + 25) x 6y ( π₯ 2 - 6xy + 36 π¦ 2 )
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Sum of Cubes 5. π₯ 9 +1 = ( ) 3 + ( ) 3 = ( π₯ 3 + 1)( π₯ 6 - π₯ 3 +1) = (x + 1)( π₯ 2 βπ₯+1)( π₯ 6 - π₯ 3 +1) π₯ 3 1 x 1
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Factor By Grouping Factor By Grouping
Mainly used when there are 4 terms Group the first two terms & the last two terms Factor from each group βCombineβ like terms
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Review Factoring Factor 6. 4π₯ 2 π¦β6π₯ = 2x(2xy β 3) 7. π₯ 2 βπ₯ = x(x β 1)
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Factor by Grouping 8. 3π₯ 2 +2π₯π¦+10π¦+15π₯ = x(3x + 2y)+5(2y + 3x) =
9. 7π₯ 2 π¦β6 π₯ 2 β7π¦+6 = π₯ 2 (7y β 6)- 1(7y - 6) = (7y β 6) ( π₯ 2 β1) = (7y β 6)(x + 1)(x β 1)
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Factor by Grouping 10. 32ax + 12bx β 48ay β 18by = 2(16ax + 6bx β 24ay - 9by) = 2(2x(8a + 3b) β 3y(8a + 3b)) = 2 (8a + 3b) (2x β 3y)
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Factor by Grouping 11. 30ac + 80bd + 40ad + 60bc = 10(3ac + 8bd + 4ad + 6bc) = 10(3ac + 4ad + 8bd + 6bc) = 10(a(3c + 4d) + 2b(4d + 3c)) = 10 (3c + 4d) (a + 2b)
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Factor 12. π₯ 3 β3 π₯ 2 +π₯β3 = π₯ 2 (x β 3)+1(x β 3) = (x β 3) ( π₯ 2 + 1)
13. 2π₯ 3 β16 π₯ 2 β8π₯+64 = 2( π₯ 3 β8 π₯ 2 β4π₯+32) = 2( π₯ 2 π₯β8 β4(π₯β8)) = 2 (x β 8) ( π₯ 2 β4) = 2(x β 8)(x + 2)(x β 2)
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Factor 14. 8π₯ 3 β8 = 8( π₯ 3 β1) = ( ) 3 β ( ) 3 = 8(x β 1)( π₯ 2 + x + 1) x 1
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Factor π₯ = ( ) 3 + ( ) 3 = (6x + 4)(36 π₯ x+16) = 2(3x + 2)4(9 π₯ 2 β6π₯+4) = 8(3x + 2)(9 π₯ 2 β6π₯+4) OR from 216π₯ ο 8(27 π₯ 3 +8) 6x 4 3x 2
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Ticket Out the Door Factor 54π₯ 3 + 16π¦ 3 54π₯ 3 β 16π¦ 3
Whatβs the difference in your answers?
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Assignment: Factoring WS 2
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