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Warm – up #4 Factor. 1. 6

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Presentation on theme: "Warm – up #4 Factor. 1. 6 "β€” Presentation transcript:

1 Warm – up #4 Factor π‘₯ 2 βˆ’3x 2. π‘₯ 2 βˆ’ π‘₯ 4 βˆ’ 3𝑦 4

2 Warm – up #4 Solutions Factor π‘₯ 2 βˆ’3x = 3x(2x – 1) 2. π‘₯ 2 βˆ’25 = ( ) 2 βˆ’ 2 = (x – 5)(x + 5) x 5

3 Warm – up #4 Solutions 3. 48π‘₯ 4 βˆ’ 3𝑦 4 = 3(16 π‘₯ 4 βˆ’ 𝑦 4 ) = ( ) 2 βˆ’ 2 = 3(4 π‘₯ 2 βˆ’ 𝑦 2 )(4 π‘₯ 2 + 𝑦 2 ) = 3(2x – y)(2x + y)(4 π‘₯ 2 + 𝑦 2 ) 4 π‘₯ 2 𝑦 2 2x y

4 Homework Log Fri 11/6 Lesson 4 – 4 Learning Objective:
To factor difference and sum of cubes & by grouping Hw: Factoring WS 2

5 11/6/15 Lesson 4 – 4 Factoring Day 2
Algebra II

6 Learning Objective To factor difference of cubes
To factor sum of cubes To factor by grouping

7 Factoring Difference of Squares π‘₯ 2 βˆ’ 𝑦 2 =(π‘₯βˆ’π‘¦)(π‘₯+𝑦) Sum of Squares π‘₯ 2 + 𝑦 2 = Can’t be factored!

8 Factoring Difference of Cubes π‘₯ 3 βˆ’ 𝑦 3 =(π‘₯βˆ’π‘¦)( π‘₯ 2 +π‘₯𝑦+ 𝑦 2 ) Sum of Cubes π‘₯ 3 + 𝑦 3 =(π‘₯+𝑦)( π‘₯ 2 βˆ’π‘₯𝑦+ 𝑦 2 )

9 Difference & Sum of Cube β€œSong”
Pull out the cube roots Square the first term Change the Sign First times the second term Plus the last term squared

10 Difference of Cubes π‘₯ 3 βˆ’ 𝑦 3 =(π‘₯βˆ’π‘¦)( π‘₯ 2 +π‘₯𝑦+ 𝑦 2 ) 1. 8π‘₯ 3 βˆ’1 = ( ) 3 βˆ’ ( ) 3 = (2x – 1) 2. 27π‘₯ 3 βˆ’8 = (3x – 2) 2x 1 (4 π‘₯ 2 + 2x + 1) 3x 2 (9 π‘₯ 2 + 6x + 4)

11 Sum of Cubes π‘₯ 3 + 𝑦 3 =(π‘₯+𝑦)( π‘₯ 2 βˆ’π‘₯𝑦+ 𝑦 2 ) 3. 64π‘₯ = ( ) 3 + ( ) 3 = (4x + 5) 4. π‘₯ 𝑦 3 = (x+6y) 4x 5 (16 π‘₯ 2 - 20x + 25) x 6y ( π‘₯ 2 - 6xy + 36 𝑦 2 )

12 Sum of Cubes 5. π‘₯ 9 +1 = ( ) 3 + ( ) 3 = ( π‘₯ 3 + 1)( π‘₯ 6 - π‘₯ 3 +1) = (x + 1)( π‘₯ 2 βˆ’π‘₯+1)( π‘₯ 6 - π‘₯ 3 +1) π‘₯ 3 1 x 1

13 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 β€œCombine” like terms

14 Review Factoring Factor 6. 4π‘₯ 2 π‘¦βˆ’6π‘₯ = 2x(2xy – 3) 7. π‘₯ 2 βˆ’π‘₯ = x(x – 1)

15 Factor by Grouping 8. 3π‘₯ 2 +2π‘₯𝑦+10𝑦+15π‘₯ = x(3x + 2y)+5(2y + 3x) =
9. 7π‘₯ 2 π‘¦βˆ’6 π‘₯ 2 βˆ’7𝑦+6 = π‘₯ 2 (7y – 6)- 1(7y - 6) = (7y – 6) ( π‘₯ 2 βˆ’1) = (7y – 6)(x + 1)(x – 1)

16 Factor by Grouping 10. 32ax + 12bx – 48ay – 18by = 2(16ax + 6bx – 24ay - 9by) = 2(2x(8a + 3b) – 3y(8a + 3b)) = 2 (8a + 3b) (2x – 3y)

17 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)

18 Factor 12. π‘₯ 3 βˆ’3 π‘₯ 2 +π‘₯βˆ’3 = π‘₯ 2 (x – 3)+1(x – 3) = (x – 3) ( π‘₯ 2 + 1)
13. 2π‘₯ 3 βˆ’16 π‘₯ 2 βˆ’8π‘₯+64 = 2( π‘₯ 3 βˆ’8 π‘₯ 2 βˆ’4π‘₯+32) = 2( π‘₯ 2 π‘₯βˆ’8 βˆ’4(π‘₯βˆ’8)) = 2 (x – 8) ( π‘₯ 2 βˆ’4) = 2(x – 8)(x + 2)(x – 2)

19 Factor 14. 8π‘₯ 3 βˆ’8 = 8( π‘₯ 3 βˆ’1) = ( ) 3 βˆ’ ( ) 3 = 8(x – 1)( π‘₯ 2 + x + 1) x 1

20 Factor π‘₯ = ( ) 3 + ( ) 3 = (6x + 4)(36 π‘₯ x+16) = 2(3x + 2)4(9 π‘₯ 2 βˆ’6π‘₯+4) = 8(3x + 2)(9 π‘₯ 2 βˆ’6π‘₯+4) OR from 216π‘₯ οƒ  8(27 π‘₯ 3 +8) 6x 4 3x 2

21 Ticket Out the Door Factor 54π‘₯ 3 + 16𝑦 3 54π‘₯ 3 βˆ’ 16𝑦 3
What’s the difference in your answers?

22 Assignment: Factoring WS 2


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