second quarter group f math peta - factoring (gcmf, dts, stc, dtc, pst, qt1, qt2, gqt, fbg, fc)

SECOND QUARTER’s

PERFORMANCE TASK:

FACTORINGICL: Showcase of Solution and Final

Answer

Group Name: _____Factoring_____

Second Quarter Grade 8 Mathematics Performance Task Factoring 2

Justin Bacon

Rafael Biag

Daniel Paul Perez

Justin Bacon


Insert Picture Here

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:

Second Quarter Grade 8 Mathematics Performance Task Factoring

4

Name:GREATEST COMMON MONOMIAL FACTOR (GCMF)• QUESTION 2:


5

Name:DIFFERENCE OF TWO SQUARES (DTS)

• QUESTION 1:


6


• QUESTION 2:


7

Name:SUM OF TWO CUBES (STC)

• QUESTION 1:


8


• QUESTION 2:


9

Name:DIFFERENCE OF TWO CUBES (DTC)

• QUESTION 1:


10

Name:DIFFERENCE OF TWO CUBES (DTC)

• QUESTION 2:


11

Name:PERFECT SQUARE TRINOMIAL (PST)

• QUESTION 1:


12


• QUESTION 2:


13

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

• QUESTION 1:


14

Name:Quadratic Trinomial in the form: ax2

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

• QUESTION 2:


15

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


16

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


17

Name:GENERAL QUADRATIC TRINOMIAL (GQT)

• QUESTION 1:


18

• QUESTION 2:


19Name:GENERAL QUADRATIC TRINOMIAL (GQT)

Name:FACTORING BY GROUPING

• QUESTION 1:


20

Name:FACTORING COMPLETELY

• QUESTION 1:


21

RAFAEL BIAG


Insert Picture Here

GCMF




NAME:GREATEST COMMON MONOMIAL

FACTOR (GCMF)• QUESTION 1:SOLUTION:we see that the number 3 will divide into all three terms. However, we can't find anything else that will divide into all three terms. So 3 must be the GREATEST common factor.

FINAL ANSWER:


23

NAME:GREATEST COMMON MONOMIAL

FACTOR (GCMF)• QUESTION : 2n3+6n2+10nSOLUTION:< We see that there is a power of "n" in each term. The smallest exponent in all three terms is the understood 1 on the last term, so "n" will divide evenly into each term along with the 2 found above. That makes the GREATEST common factor for this example "2n".

FINAL ANSWER:2n(n2+3n+5)


24

NAME:DIFFERENCE OF TWO

SQUARES (DTS)• QUESTION 1: x2 - 9SOLUTION:What times itself will give x2 ? The answer is x.What times itself will give 9 ? The answer is 3.

FINAL ANSWER:(x + 3) (x - 3) or (x - 3) (x + 3)


25

NAME:DIFFERENCE OF TWO

SQUARES (DTS)• QUESTION 2: 4y2 - 36y6 SOLUTION:What times itself will give 1? The answer is 1.What times itself will give 9y4 ? The answer is 3y2

FINAL ANSWER:4y2 (1 + 3y2) (1 - 3y2) or 4y2 (1 - 3y2) (1 + 3y2)


26

NAME:SUM OF TWO CUBES (STC)

• QUESTION 1: x3 – 8SOLUTION:x3 – 8 = x3 – 23 = (x – 2)(x2 + 2x + 22)

FINAL ANSWER: = (x – 2)(x2 + 2x + 4)


27

NAME:SUM OF TWO CUBES (STC)

• QUESTION 2: 27x3 + 1SOLUTION:Remember that 1 can be regarded as having been raised to any power you like, so this is really (3x)3 + 13.

27x3 + 1 = (3x)3 + 13 = (3x + 1)((3x)2 – (3x)(1) + 12) FINAL ANSWER:= (3x + 1)(9x2 – 3x + 1)


28

NAME:DIFFERENCE OF TWO CUBES

(DTC)• QUESTION 1: nm3 - 125SOLUTION:

∛nm3 = nm ∛125 = 5

FINAL ANSWER: (nm-5)(nm2+5nm+25)


29

NAME:DIFFERENCE OF TWO CUBES

(DTC)• QUESTION 2: 343 - 729 a15SOLUTION:

∛343= 7∛729 a15 = 9a5

FINAL ANSWER:

(7- 9a5) (49+ 63a5 + 81a10)


30

NAME:PERFECT SQUARE TRINOMIAL

(PST)• QUESTION 1: x+ 12x + 36SOLUTION:>Both x2 and 36 are perfect squares, and 12x is twice the product of x and 6.>Since all signs are positive, the pattern is (a + b)2 = a2 + 2ab + b2.Let a = x and b = 6.

FINAL ANSWER:= (x + 6)2 or (x + 6)(x + 6)


31

NAME:PERFECT SQUARE TRINOMIAL

(PST)• QUESTION 2: 9a2 - 6a + 1SOLUTION:< Both 9a2 and 1 are perfect squares, and 6a is twice the product of 3a and 1.<Since the middle term is negative, the pattern is (a - b)2 = a2 - 2ab + b2.Let a = 3a and b = 1.

FINAL ANSWER:

(3a - 1)2 or (3a - 1)(3a - 1)


32

NAME:QUADRATIC TRINOMIAL IN THE

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

• QUESTION 1: x2 + 10x + 21SOLUTION:21, we will factor 21 in as many pairs as possible. The pairs are: {1, 21} and {3, 7}. Now we must find a pair that has a sum equal to 'b,' which is 10. The pair {1, 21} has a sum of 22, but this does not match our value for 'b.' The pair {3, 7} has a sum of 10.AC TEST: 21 3 | 7 = 10 1| 21 = 22 FINAL ANSWER: (x + 3)(x + 7)


33


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

• QUESTION : y2 + 13y + 42

Solution: AC Test: 42|7 13|6

√y2 = 7

FINAL ANSWER:(y7)(y6)


34


FORM: AX2 + BX + C WHERE A = 1 AND C IS NEGATIVE (QT2)• QUESTION 1: n2 + 13n - 42

Solution: AC Test: -42 | 7 13 | -6

√n2 = nFactors of the 3rd term= 7 and -6

FINAL ANSWER: (n+7)(n-6)


35


FORM: AX2 + BX + C WHERE A = 1 AND C IS NEGATIVE (QT2)• QUESTION : z2 + 2z - 35

Solution: AC Test: -35 | 7 2 | -5

√z2 = zFactors of the 3rd term= 7 and -5

FINAL ANSWER: (z7)(z-5)


36

NAME:GENERAL QUADRATIC TRINOMIAL

(GQT)• QUESTION 1: 5x2 + 11x + 2SOLUTION:<Find the product ac:<Think of two factors of 10 that ad:<Write the 11x as the sum of 1x and 10x:

<Group the two pairs of terms:

<Remove common factors from each group:<Notice that the two quantities in parentheses are now identical. That means we can factor out a common factor of (5x + 1):

FINAL ANSWER:

(5x + 1)(x + 2)Second Quarter Grade 8 Mathematics Performance Task Factoring

37

• QUESTION 2: 4x2 + 7x – 15SOLUTION:< Find the product ac:

< Think of two factors of -60 that add up to 7:

< Write the 7x as the sum of -5x and 12x:

< Group the two pairs of terms:

< Remove common factors from each group:

< Notice that the two quantities in parentheses are now identical. That means we can factor out a common factor of (4x - 5):

FINAL ANSWER:(4x – 5)(x + 3)


38Name:GENERAL QUADRATIC TRINOMIAL (GQT)

NAME:FACTORING BY GROUPING

• QUESTION 1: x2−4SOLUTION:The polynomial cannot be factored using the grouping method. Try a different method, or if you aren't

sure, choose 'Factor using any method.

The polynomial cannot be factored using the grouping method.

The binomial can be factored using the difference of squares formula, because both terms are perfect

squares.

FINAL ANSWER:

(x+2)(x−2)


39

NAME:FACTORING COMPLETELY

• QUESTION 1: 5yz2 - 30yz + 40SOLUTION:

GCMF: 5 (yz2 - 6yz + 8)QT1: 5 (yz-4) (yz-2)

FINAL ANSWER:5 (yz-4) (yz-2)


40

Daniel Paul Perez


Insert Picture Here

GCMF




Name:Daniel Paul PerezGREATEST COMMON MONOMIAL FACTOR (GCMF)• QUESTION 1:10x4y2+8xy3-12xy5

GCMF:2xy2

Divide each term of the polynomial by the GCMF:5x3y+4y-6y3

Answer:Factored form:2xy2(5x3y=4y-6y3)

Second Quarter Grade 8 Mathematics

Performance Task Factoring

42

Name:Daniel Paul PerezGREATEST COMMON MONOMIAL FACTOR (GCMF)• QUESTION 2:15p - 3qGCF:3Factors of 15:1,3,5,15Factors of 3:1,315p - 3q = 3*5p - 3*qAnswer:3(5p-q)



43


• QUESTION 1:81c2 - 144m2

What times itself will give 81c^2?9What times itself will give 144m^2?12Answer:Factored form:(9c - 12m) (9c + 12m)



44


• QUESTION 2:4a2b2 - 49n6

• What times itself will give 4a2b2?2ab• What times itself will give 49n6?7n3

Answer:Factored form:(2ab - 7n3) (2ab + 7n3)



45

Name:SUM OF TWO CUBES (STC)• QUESTION 1:x3+27solution:(x)3+(3)3

(x+3)[(x)2-(x)(3)+(3)2]The term with variable x is okay but the 27 should be taken care of. Obviously we know that 27 = (3)(3)(3) = 33.Rewrite the original problem as sum of two cubes, and then simplify. Since this is the "sum" case, the binomial factor and trinomial factor will have positive and negative middle signs, respectively.answer:(x+3)(x2-3x+9)



46

Name:SUM OF TWO CUBES (STC)• QUESTION2:27x3+64y3

solution:(3x)3+(4y)3

(3x+4y)[(3x)2-(3x)(4y)+(4y)2]The first step as always is to express each term as cubes. We know that 27 = 33 and 64 = 43. Rewrite the problem as sum of two cubic terms and apply the rule, so we getanswer:(3x+4y)(9x2-12xy+16y2)



47

Name:Daniel Paul PerezDIFFERENCE OF TWO CUBES (DTC)

• QUESTION 1:2:1-216x3y3

solution:(1)3-(6xy)3

(1-6xy)[(1)2+(1)(6xy)+(6xy)2]Now for the number, it is easy to see that that 1 = (1)(1)(1) = 13 while 216 = (6)(6)(6) = 63. This is really a case of difference of two cubes.answer:(1-6xy)(1+6xy+36x2y2)



48

Name:Daniel Paul PerezDIFFERENCE OF TWO CUBES (DTC)

• QUESTION 2:y3-8solution:(y)3-(2)3

(y-2)[(y)2+(y)(2)+(2)2]This is a case of difference of two cubes since the number 8 can be written as a cube of a number, where 8 = (2)(2)(2) = 23.Apply the rule for difference of two cubes, and simplify. Since this is the "difference" case, the binomial factor and trinomial factor will have negative and positive middle signs, respectively.answer:(y-2)(y2+2y+4)



49

Name:Daniel Paul PerezPERFECT SQUARE TRINOMIAL (PST)

• QUESTION 1:a2 + 4a + 4solution: √a2=a √4a=2 √4=2answer: (a + 2)(a + 2)there are all possitive because the middle term is positiveand we going to add both of them then the twice of it.



50

Name:Daniel Paul PerezPERFECT SQUARE TRINOMIAL (PST)• QUESTION 2: 25x2+30xy+9y2

solution:√25x2=5x√30xy=15xy√9y2=3yanswer:(5+3)(5+3)to get the middle we need to add the first and the last and then the twice of it.



51

Name:Daniel Paul PerezQuadratic Trinomial in the form: ax2

+ bx + c where a = 1 and c is POSITIVE (QT1)• QUESTION 1:x2 + 5y + 6.

solution:x2+5y+6a = 1, b = 5 and c = 6answer:(x+3)(x+2)



52

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

• QUESTION 2:x2+12x+36solution:x2+12x+36a=1,b=12 and c=36answer:(x+6)(x+6)



53

Name:Daniel Paul PerezQuadratic Trinomial in the form: ax2 + bx + c where a = 1 and c is NEGATIVE (QT2)• QUESTION 1:x2+1x-42solution:x2+1x-42a=1,b=1 and c=42answer:(x+6)(x-7)



54

Name:Daniel Paul PerezQuadratic Trinomial in the form: ax2 + bx + c where a = 1 and c is NEGATIVE (QT2)• QUESTION 2:x2-14x-14solution:x2-14x-14a=1,b=14 and c=14answer:(x-9)(x-5)



55

Name:Daniel Paul PerezGENERAL QUADRATIC TRINOMIAL (GQT)

• QUESTION 1:6x^2+22x+20solution: 6x2 20 ^ ^ 3x 2x 5 4Answer:(3x+5)(2x+4)



56

• QUESTION 2:6x2+2x-8Solution: 6x2 - 8 ^ ^ 6x x -8 1Answer:(6x-8)(x+1)



57

Name:Daniel Paul PerezGENERAL QUADRATIC TRINOMIAL (GQT)

Name:Daniel Paul PerezFACTORING BY GROUPING• QUESTION 1:xy+2y+3x+6solution:y(x+2)+3(x+2)when you finish the solution just copy them replace the other x+2 then put y+3 so the answer will be (x+2)(y+3)answer:(x+2)(y+3)or(y+3)(x+2)



58

Name:Daniel Paul PerezFACTORING COMPLETELY• QUESTION 1: 4x2 + 16x + 16 Get the GCMF because they have common factors

4(x2 + 4x + 4)

Get the PST because they are perfect squares

(x + 2)2

FA= (x + 2)2



59

NAME OF STUDENT 1


Insert Picture Here

GCMF




Name:GREATEST COMMON MONOMIAL

FACTOR (GCMF)• QUESTION 1:


Name:GREATEST COMMON MONOMIAL

FACTOR (GCMF)• QUESTION 2:


Name:DIFFERENCE OF TWO SQUARES

(DTS)• QUESTION 1:


Name:DIFFERENCE OF TWO SQUARES

(DTS)• QUESTION 2:



• QUESTION 1:



• QUESTION 2:


Name:DIFFERENCE OF TWO CUBES

(DTC)• QUESTION 1:


Name:DIFFERENCE OF TWO CUBES

(DTC)• QUESTION 2:



• QUESTION 1:



• QUESTION 2:


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

(QT1)• QUESTION 1:


Name:Quadratic Trinomial in the form: ax2

+ bx + c where a = 1 and c is POSITIVE (QT1)• QUESTION 2:


Name:Quadratic Trinomial in the form: ax2 + bx +

c where a = 1 and c is NEGATIVE (QT2)• QUESTION 1:


Name:Quadratic Trinomial in the form: ax2 + bx +

c where a = 1 and c is NEGATIVE (QT2)• QUESTION 2:


Name:GENERAL QUADRATIC TRINOMIAL

(GQT)• QUESTION 1:


• QUESTION 2:


Name:GENERAL QUADRATIC TRINOMIAL (GQT)

Name:FACTORING BY GROUPING

• QUESTION 1:


Name:FACTORING COMPLETELY

• QUESTION 1:


REFLECTIONGroup Name ____

1 2

3 4

Name of Member 1


1 23 4

Name of Member 2


1 23 4

Name of Member 3


1 23 4

Name of Member 4


1 23 4

second quarter group f math peta - factoring (gcmf, dts, stc, dtc, pst, qt1, qt2, gqt, fbg, fc)

Education