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Warm-Up Any questions about homework?.

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Presentation on theme: "Warm-Up Any questions about homework?."β€” Presentation transcript:

1 Warm-Up Any questions about homework?

2 Agenda: Factoring Sum and Difference of Cubes Factor Relays Homework

3 Factoring Sum and Difference of Cubes
β€œPatterns” π‘Ž 3 + 𝑏 3 =(π‘Ž+𝑏)( π‘Ž 2 βˆ’π‘Žπ‘+ 𝑏 2 ) π‘Ž 3 βˆ’ 𝑏 3 =(π‘Žβˆ’π‘)( π‘Ž 2 +π‘Žπ‘+ 𝑏 2 ) Remember signs: same-different-positive

4 1. π‘₯ A = x b =5 2. π‘₯ 3 βˆ’27 A = x b =3 π‘Ž 3 + 𝑏 3 π‘Ž 3 βˆ’ 𝑏 3 (π‘₯) 3 + (5) 3 (π‘₯) 3 βˆ’ (3) 3 (π‘Ž+𝑏)(π‘Ž 2 βˆ’π‘Žπ‘+ 𝑏 2 ) (π‘Žβˆ’π‘)(π‘Ž 2 +π‘Žπ‘+ 𝑏 2 ) (π‘₯+5)(π‘₯ 2 βˆ’5π‘₯+25) (π‘₯βˆ’3)(π‘₯ 2 +3π‘₯+9)

5 3. 8 π‘₯ A = 2x b =7 4. 64 π‘₯ 4 βˆ’27π‘₯ A = 4x b =3 π‘Ž 3 + 𝑏 3 x(64 π‘₯ 3 βˆ’27) (2π‘₯) 3 + (7) 3 π‘₯(π‘Ž 3 βˆ’ 𝑏 3 ) (π‘Ž+𝑏)(π‘Ž 2 βˆ’π‘Žπ‘+ 𝑏 2 ) π‘₯((4π‘₯) 3 βˆ’ ) (2π‘₯+7)((2π‘₯) 2 βˆ’14π‘₯+49) π‘₯(π‘Žβˆ’π‘)(π‘Ž 2 +π‘Žπ‘+ 𝑏 2 ) (2π‘₯+7)(4π‘₯ 2 βˆ’14π‘₯+49) x (4π‘₯βˆ’3)(16π‘₯ 2 +12π‘₯+9)

6 Other Cubics by Grouping
5. 12π‘₯ 3 βˆ’ 9π‘₯ 2 +4π‘₯ βˆ’3 6. 2π‘₯ 3 + 5π‘₯ 2 +6π‘₯+15 3π‘₯ 2 (4π‘₯ βˆ’3) +1(4π‘₯ βˆ’3) π‘₯ 2 (2π‘₯+5) +3(2π‘₯+5) (3 π‘₯ 2 +1)(4π‘₯ βˆ’3) ( π‘₯ 2 +3)(2π‘₯+5) 7. 10𝑝 3 βˆ’ 20𝑝 2 +6𝑛 βˆ’12 2( 5𝑝 3 βˆ’ 10𝑝 2 +3𝑛 βˆ’6) 5𝑝 2 (π‘βˆ’2) +3(π‘βˆ’2) 2( 5𝑝 2 +3)(π‘βˆ’2)

7 Factor Relays Directions: Get into groups of 4
As a new questions appears you have a new spokesperson for the group Work as a team to get the answer Spokesperson runs up to the board and writes their answer (make sure to keep it covered If your team gets it right, you get one point

8 5 π‘₯ 2 +5π‘₯ βˆ’10 5(π‘₯+2)(π‘₯βˆ’1)

9 2 𝑛 3 βˆ’ 𝑛 2 +2𝑛 βˆ’1 (2π‘₯βˆ’1)( π‘₯ 2 +1)

10 18 π‘₯ 2 βˆ’2 2(3π‘₯βˆ’1)(3π‘₯+1)

11 2 𝑑 2 +12𝑑 βˆ’16 2( 𝑑 2 +6π‘‘βˆ’8)

12 125 π‘₯ 3 βˆ’216 (5π‘₯βˆ’6)(25 π‘₯ 2 +30π‘₯+36)

13 βˆ’ 𝑣 2 +2𝑣 βˆ’1 βˆ’ (π‘£βˆ’1) 2

14 π‘₯ 3 +64 (π‘₯+4)( π‘₯ 2 βˆ’4π‘₯+16)

15 5 𝑝 2 +4𝑝 βˆ’12 (5π‘₯βˆ’6)(π‘₯+2)

16 𝑣 3 βˆ’ 4𝑣 2 +4𝑣 βˆ’16 (π‘₯βˆ’4)( π‘₯ 2 +4)

17 2 π‘Ž 7 βˆ’32 π‘₯ 3 2 π‘₯ 3 (π‘₯+2)(π‘₯βˆ’2)( π‘₯ 2 +4)

18 25 π‘₯ 2 βˆ’10π‘₯+1 (5π‘₯βˆ’1) 2

19 π‘₯ 4 βˆ’10 π‘₯ 2 +16 ( π‘₯ 2 +2)( π‘₯ 2 +8)

20 6 𝑦 𝑦 2 +15π‘₯ 3π‘₯(π‘₯+1)(2π‘₯+5)


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