group cycloid factoring

SECOND QUARTER’s

PERFORMANCE TASK:

FACTORINGICL: Showcase of Solution and Final

Answer

Group Name: _CYCLOID____

Second Quarter Grade 8 Mathematics Performance Task Factoring 2

KRYSTAL MIZONA

DUSTINE DANAE V. VILLA

CARLO TORRES

NAME OF STUDENT 4

Krystal Anne V. Mizona


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GCMF

BINOMIAL•DTS•STC•DTC

QUADRATIC TRINOMIALS•PST•QT 1•QT 2•GQT

Factoring by GROUPING / Factoring COMPETELY

Name:GREATEST COMMON MONOMIAL FACTOR (GCMF)• QUESTION 1:• 6y + 4y2

• GCF: 2y

Solution:2y ÷ 6y = 3y2y ÷ 2y2 = y

Answer:2y (3y + y)Second Quarter Grade 8 Mathematics Performance Task Factoring

4

Name:GREATEST COMMON MONOMIAL FACTOR (GCMF)• QUESTION 2:• 9jk4 + 15i + 12yGCF: 3

Solution:3 ÷ 9jk4 = 3k3

3 ÷ 15i = 53 ÷ 12y = 4

Answer: 3 (3k3 + 5 + 4)

Second Quarter Grade 8 Mathematics Performance Task Factoring

5

Name:DIFFERENCE OF TWO SQUARES (DTS)

• QUESTION 1:• x2 - 16k4

Solution:√x2 = x√-16k4 = -4k2 & 4k2

Answer:=(x + 4k2) (x - 4k2)


6

Name:DIFFERENCE OF TWO SQUARES (DTS)

• QUESTION 2:• 16s2 - 576k4

Solution:√16s2 = 4s√-576k4 = -24k2 & 24k2

Answer:=(4s + 24k2) (4s - 24k2)


7

Name:SUM OF TWO CUBES (STC)• QUESTION 1:• 216m9 + 27o6

Solution: Binomial Factor3√216m9 = 6m33√27o6 = 3o2

Answer: (6m3 + 3o2)3

Trinomial Factor:(6m3 + 3o2)2 = (36m6 + 18m3o2 + 9o4)Final Answer: (6m3 + 3o2) (36m6 + 18m3o2 + 9o4)


8

Name:SUM OF TWO CUBES (STC)• QUESTION 2:• 512s15 + 729i6

Solution: Binomial Factor3√512s15 =8s53√-729i6 =9i2

Answer= (8s5 + 9i2)3

Trinomial FactorAnswer: (8s5 + 9i2)2 = (64s10 + 72s5i2 + 81i4)Final Answer: (8s5 + 9i2) (64s10 + 72s5i2 + 81i4)


9

Name:DIFFERENCE OF TWO CUBES (DTC)

• QUESTION 1:• 64n3 - 125m12

Solution: Binomial Factor3√64n3 =4n3√-125m12 =-5m4

Answer: (4n - 5m4)3

Trinomial Factor:(4n - 5m4)2 = (16n2 - 20nm4 + 25m8)Final Answer: (4n - 5m4) (16n2 - 20nm4 + 25m8)


10

Name:DIFFERENCE OF TWO CUBES (DTC)

• QUESTION 2:• 1728y21 - 5832h27

Solution: Binomial Factor:3√1728y21 =12y73√-5832h27 =-18h9

Answer: (12y7 - 18h9)3

Trinomial Factor:(12y7 - 18h9)2 = (144y14 - 216y7h9 + 324h18)Final Answer: (12y7 - 18h9) (144y14 - 216y7h9 + 324h18)


11

Name:PERFECT SQUARE TRINOMIAL (PST)

• QUESTION 1:• x2 + 8xk + 12k2

Solution:√x2

√12k2

Answer:=(x + 4k)2


12

Name:PERFECT SQUARE TRINOMIAL (PST)

• QUESTION 2:• x2 + 24xy2 + 144y4

Solution:√x2 = x√144y4 = 12y2

Answer:=(x + 12y2)


13

Name:Quadratic Trinomial in the form: ax2 + bx + c where a = 1 and c is POSITIVE (QT1)

• QUESTION 1:• x2 + 41 + 210

Solution:AC TEST

Answer:(x + 35) (x + 6)


14

41 6

35210

Name:Quadratic Trinomial in the form: ax2

+ bx + c where a = 1 and c is POSITIVE (QT1)

• QUESTION 2:• j2 + 68 + 867

Solution:

Answer:(j + 17) (j + 51)


15

867

68 51

17

Name:Quadratic Trinomial in the form: ax2 + bx + c where a = 1 and c is NEGATIVE (QT2)• QUESTION 1:•x2 + 20 - 125

Solution:

Answer:(x + 25) (x - 5)


16

-125

20 -5

25

Name:Quadratic Trinomial in the form: ax2 + bx + c where a = 1 and c is NEGATIVE (QT2)• QUESTION 2:• a2 + 27 - 90

Solution:AC TEST

Answer:(a + 30) (a - 3)


17

-90

27 -3

30

Name:GENERAL QUADRATIC TRINOMIAL (GQT)

• QUESTION 1:• 2o2 + 13 + 15

Solution:

Answer:(o2 + 5) (o2 + 3)


18

30 10

313

10

2

3

2

=5 =3

AC TEST

• QUESTION 2:• 2i2 -17 + 15

Solution:

Answer:(i2 - 15) (i2 - 2)


19Name:GENERAL QUADRATIC TRINOMIAL (GQT)

30 -15

-17 -2

AC TEST

Name:FACTORING BY GROUPING

• QUESTION 1:• x2 + 2x + yx + 2y

Solution:GCMF: x (x + 2)GCMF: y (x + 2)

Answer: (x + 2) (x + y)


20

Name:FACTORING COMPLETELY


21• QUESTION 1: n2 + 4n + 2 - v4

SOLUTION: PST: n2 + 4n + 2 = (n + 2)2 DTS: (n + 2)2 - v4

= [(n + 2)2- v2] [(n + 2)2- v2]

FINAL ANSWER:[(n + 2)2- v2] [(n + 2)2- v2]

DUSTINE VILLA


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GCMF

BINOMIAL•DTS•STC•DTC

QUADRATIC TRINOMIALS•PST•QT 1•QT 2•GQT

Factoring by GROUPING / Factoring COMPETELY

DUSTINE DANAE V. VILLAGREATEST COMMON MONOMIAL

FACTOR (GCMF)• QUESTION 1: 10x + 6xGCF: 2x SOLUTION:2x/10x =5 2x/6x =3

FINAL ANSWER:2x (5 + 3)


23

DUSTINE DANAE V. VILLAGREATEST COMMON MONOMIAL

FACTOR (GCMF)• QUESTION 2:• 20c4 - 40c3z2 + 60cz2

GCF: 20c

SOLUTION:20c/20c4 = c3

20c/-40c3z2 = -2c2z2 20c/60c = 3

FINAL ANSWER: 20c (c3 - 2c2z2 + 3z2)


24

DUSTINE DANAE V. VILLADIFFERENCE OF TWO

SQUARES (DTS)• QUESTION 1: 16v4 - 25d2n4

SOLUTION:√16v4 = 4v2 (principal root)√-25d2n4 = 5dn2 and -5dn2 (+/- root)

FINAL ANSWER:(4v2 + 5dn2) (4v2 - 5dn2 )


25

DUSTINE DANAE V. VILLADIFFERENCE OF TWO

SQUARES (DTS)• QUESTION 2: 144x10v2 - 121d6n4

SOLUTION:√144x10v2 = 12x5v (principal root)√-121d6n4 = 11d3n2 (+/- root)

FINAL ANSWER:( 12x5v + 11d3n2 ) (12x5v - 11d3n2 )


26

DUSTINE DANAE V. VILLASUM OF TWO CUBES (STC)

• QUESTION 1: 8x15 + 27n12

SOLUTION: BINOMIAL FACTORS:∛8x15 = 2x5

∛27n12 = 3n4

TRINOMIAL FACTORS:(2x5)2 -(2x5)(3n4) + (3n4)2

FINAL ANSWER:(2x5 + 3n4) (4x10 - 6x5n4 + 9n8)


27

DUSTINE DANAE V. VILLASUM OF TWO CUBES (STC)

• QUESTION 2: 64d21 + 125n12

SOLUTION: BINOMIAL FACTORS:∛64d21 = 4d7

∛125n12 = 5n4

TRINOMIAL FACTORS:(4d7)2 -(4d7)(5n4) + (5n4)2

FINAL ANSWER:(4d7 + 5n4) (16d14 - 20d5n4 + 25n8)


28

DUSTINE DANAE V. VILLADIFFERENCE OF TWO CUBES

(DTC)• QUESTION 1: 8c21 - 27v12

SOLUTION: BINOMIAL FACTORS:∛8c21 = 2c7

∛27v12 = 3v4 = -(3v4)TRINOMIAL FACTORS:(2c7)2 + (2c7)(3v4) + (3v4)2

FINAL ANSWER:(2c7 - 3v4) (4c14 + 6c7v4 + 9v8)


29

DUSTINE DANAE V. VILLADIFFERENCE OF TWO CUBES

(DTC)• QUESTION 2: 64r24 + 216a30

SOLUTION: BINOMIAL FACTORS:∛64r24 = 4r8

∛216a30 = 6a10 = -(6a10 )TRINOMIAL FACTORS:(4r8)2 (4r8)(6a10 ) + (6a10 )2

FINAL ANSWER:(4r8 - 6a10 ) (16r16 + 24r8a10 + 36a10)


30

DUSTINE DANAE V. VILLAPERFECT SQUARE TRINOMIAL

(PST)• QUESTION 1: v2 + 8vx + 16x2

SOLUTION:√v2 = v√16x2 = 4x

FINAL ANSWER:(V + 4x)2


31

DUSTINE DANAE V. VILLAPERFECT SQUARE TRINOMIAL

(PST)• QUESTION 2: 4v4 - 12v2n2 + 9n4

SOLUTION:√4v4 = 2v2

√9n4 = 3n2

FINAL ANSWER:(2v2 - 3n2)2


32

NAME:QUADRATIC TRINOMIAL IN THE

FORM: AX2 + BX + C WHERE A = 1 AND C IS POSITIVE (QT1)

• QUESTION 1: v2 + 11v + 30SOLUTION:AC TEST

√v2 = vFACTORS OF THE THIRD TERM: 5 & 6

FINAL ANSWER:(v+5)(v+6)


33

30 5

11 6


NAME:QUADRATIC TRINOMIAL IN THE

FORM: AX2 + BX + C WHERE A = 1 AND C IS POSITIVE (QT1)

• QUESTION 2: c2 - 6c + 40SOLUTION:AC TEST

√c2= cFACTORS OF THE THIRD TERM: 10 & -4

FINAL ANSWER:(c+10)(c-4)


34

40 10

6 -4


DUSTINE DANAE V. VILLAQUADRATIC TRINOMIAL IN THE FORM:

AX2 + BX + C WHERE A = 1 AND C IS NEGATIVE (QT2)• QUESTION 1: r2 + 6r - 55

SOLUTION:AC TEST

√r2= rFACTORS OF THE THIRD TERM: 11 & -5FINAL ANSWER:(r+11)(r-5)


35

-55 11

6 -5

DUSTINE DANAE V. VILLA:QUADRATIC TRINOMIAL IN THE FORM:

AX2 + BX + C WHERE A = 1 AND C IS NEGATIVE (QT2)• QUESTION 2: j2 - 15j -100

SOLUTION:AC TEST

√r2= jFACTORS OF THE THIRD TERM: -20 & 5FINAL ANSWER:(r -20)(r+5)


36

-100 -20

-15 5

DUSTINE DANAE V. VILLAGENERAL QUADRATIC TRINOMIAL

(GQT)• QUESTION 1: 2x2 + 5x + 7SOLUTION:AC TEST = FRACTIONS: 2/7 & 2/-2 =SIMPLIFIED FORM: 2/7 & 1/-1

FINAL ANSWER:(2x + 7) (x-1) 2/2/15 2/-2 = (simplified) 2/15 1/-115 2/-2 = (simplified) 2/15 1/-1 2/15 2/-2 = (simplified) 2/15 1/-1


37

-14 7

5 -2

• QUESTION 2: FACTOR 4r^2 – 5 -9SOLUTION:AC TEST = FRACTIONS: 4/-9 & 4/-4 =SIMPLIFIED FORM: 4/-9 & 1/-1

FINAL ANSWER:(4r - 9) (x+1) 2/2/15 2/-2 = (simplified) 2/15 1/-115 2/-2 =


38DUSTINE DANAE V. VILLAGENERAL QUADRATIC TRINOMIAL (GQT)

-36 -9

-5 4

DUSTINE DANAE V. VILLA FACTORING BY GROUPING

• QUESTION 1: ca + ja - jy - cySOLUTION: GCMF• (ca - cy) + ( ja - jy)GCMF: c ( a-y) + j ( a -y)GCMF: (a-y) (c+j)

FINAL ANSWER:( a - y )( c + y )


39

DUSTINE DANAE V. VILLAFACTORING COMPLETELY

• QUESTION 1: n2 + 4n + 2 - v4

SOLUTION: PST: n2 + 4n + 2 = (n + 2)2 DTS: (n + 2)2 - v4

= [(n + 2)2- v2] [(n + 2)2- v2]

FINAL ANSWER:[(n + 2)2- v2] [(n + 2)2- v2]


40

REFLECTIONGroup Name ____

1 2

3 4

Krystal Mizona

I IMPROVED AND UNDERSTAND THE TOPICS IN FACTORING, BUT SOMETIMES I EASILY FORGET HOW AM I GOING TO FACTOR OR SOLVE SOME QUESTIONS. HONESTLY SOMETIMES WHEN I CREATE A QUESTION I PUT THE ANSWERS FIRST THAN THE ONE THAT IM GOING TO FACTOR BECAUSE I COULDN’T THINK OF ANY QUESTION.


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DUSTINE DANAE V. VILLAI HAVE LEARNED AND MASTERED THE DIFFERENT FACTORING TECHNIQUES THROUGH THIS PETA AND AT THE SAME TIME I COULD HELP OTHER STUDENTS LIKE ME IF THEY HAVE MISCONCEPTIONS, ETC. I ALSO FULLY UNDERSTAND THAT I COULD HELP OTHER PEOPLE THROUGH THIS WEBSITES OR GOOGLE FORMS AND ELEAP AND I COULD REACH OUT TO THEM WITHOUT ANY COST. I HAVE LEARNED TO HELP WITHOUT WASTING THROUGH THIS PAPERLESS PETA. I HAVE FULLY UNDERSTAND THE TRUE MEANING OF TEAMWORK AND IT IS IMPORTANT TO HELP AND COMMUNICATE ON ANOTHER TO ACHIEVE YOUR GOALS. I ALSO LEARNED THAT NOTHING WOULD BE HARD IF WE ONLY WORK TOGETHER. I LOOK FORWARD TO MORE PETAS LIKE THIS.


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group cycloid factoring

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