Second Quarter Group F Math Peta - Factoring (GCMF, DTS, STC, DTC, PST, QT1, QT2, GQT, FbG, FC)

SECOND QUARTER’s
PERFORMANCE TASK:
FACTORING
ICL: 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:
• QUESTION 2:
5

Name:
DIFFERENCE OF TWO SQUARES (DTS)
• QUESTION 1:
6

Name:
• QUESTION 2:
7

Name:
SUM OF TWO CUBES (STC)
• QUESTION 1:
8

Name:
• QUESTION 2:
9

Name:
DIFFERENCE OF TWO CUBES (DTC)
• QUESTION 1:
10

Name:
• QUESTION 2:
11

Name:
PERFECT SQUARE TRINOMIAL (PST)
• QUESTION 1:
12

Name:
• 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:
where a = 1 and c is NEGATIVE (QT2)
• QUESTION 1:
16

Name:
where a = 1 and c is NEGATIVE (QT2)
• QUESTION 2:
17

Name:
GENERAL QUADRATIC TRINOMIAL (GQT)
• QUESTION 1:
18

• QUESTION 2:
19
Name:

Name:
FACTORING BY GROUPING
• QUESTION 1:
20

Name:
FACTORING COMPLETELY
• QUESTION 1:
21

RAFAEL BIAG
Insert Picture
Here
GCMF
BINOMIAL
•DTS
•STC
•DTC
•PST
•QT 1
•QT 2
•GQT
COMPETELY

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:
FACTOR (GCMF)
• QUESTION : 2n3+6n2+10n
SOLUTION:
< 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 - 9
SOLUTION:
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:
(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:
• QUESTION 1: x3 – 8
SOLUTION:
x3 – 8 = x3 – 23
= (x – 2)(x2 + 2x + 22)
FINAL ANSWER:
= (x – 2)(x2 + 2x + 4)
27

NAME:
• QUESTION 2: 27x3 + 1
SOLUTION:
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 - 125
SOLUTION:
∛nm3 = nm
∛125 = 5
FINAL ANSWER:
(nm-5)(nm2+5nm+25)
29

NAME:
(DTC)
• QUESTION 2: 343 - 729 a15
SOLUTION:
∛343= 7
∛729 a15 = 9a5
FINAL ANSWER:
(7- 9a5) (49+ 63a5 + 81a10)
30

NAME:
PERFECT SQUARE TRINOMIAL
(PST)
• QUESTION 1: x+ 12x + 36
SOLUTION:
>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 + 1
SOLUTION:
< 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 + 21
SOLUTION:
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

NAME:
QUADRATIC TRINOMIAL IN THE
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

NAME:
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 = n
Factors of the 3rd term= 7 and -6
FINAL ANSWER:
(n+7)(n-6)
35

NAME:
FORM: AX2 + BX + C WHERE A = 1
AND C IS NEGATIVE (QT2)• QUESTION : z2 + 2z - 35
Solution: AC Test: -35 | 7
2 | -5
√z2 = z
Factors of the 3rd term= 7 and -5
FINAL ANSWER:
(z7)(z-5)
36

NAME:
GENERAL QUADRATIC TRINOMIAL
(GQT)
• QUESTION 1: 5x2 + 11x + 2
SOLUTION:
<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)
37

• QUESTION 2: 4x2 + 7x – 15
SOLUTION:
< 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)
38
Name:
(GQT)

NAME:
• QUESTION 1: x2−4
SOLUTION:
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:
• QUESTION 1: 5yz2 - 30yz + 40
SOLUTION:
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
BINOMIAL
•DTS
•STC
•DTC
•PST
•QT 1
•QT 2
•GQT
COMPETELY

Name:Daniel Paul Perez
• 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

• QUESTION 2:15p - 3q
GCF:3
Factors of 15:1,3,5,15
Factors of 3:1,3
15p - 3q = 3*5p - 3*q
Answer:3(5p-q)
Task Factoring
43

Name:
• QUESTION 1:81c2 - 144m2
What times itself will give 81c^2?9
What times itself will give 144m^2?12
Answer:
Factored form:(9c - 12m) (9c + 12m)
Task Factoring
44

Name:
• 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)
Task Factoring
45

Name:
• QUESTION 1:x3+27
solution:
(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)
Task Factoring
46

Name:
• 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 get
answer:
(3x+4y)(9x2-12xy+16y2)
Task Factoring
47

• 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)
Task Factoring
48

• QUESTION 2:y3-8
solution:
(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)
Task Factoring
49

• QUESTION 1:a2 + 4a + 4
solution:
√a2=a
√4a=2
√4=2
answer:
(a + 2)(a + 2)
there are all possitive because the middle term is positive
and we going to add both of them then the twice of it.
Task Factoring
50

• QUESTION 2:25x2+30xy+9y2
solution:
√25x2=5x
√30xy=15xy
√9y2=3y
answer:
(5+3)(5+3)
to get the middle we need to add the first and the last and then the twice
of it.
Task Factoring
51

c where a = 1 and c is POSITIVE (QT1)
• QUESTION 1:x2 + 5y + 6.
solution:
x2+5y+6
a = 1, b = 5 and c = 6
answer:
(x+3)(x+2)
Task Factoring
52

Quadratic Trinomial in the form: ax2 +
bx + c where a = 1 and c is POSITIVE
(QT1)
• QUESTION 2:x2+12x+36
solution:
x2+12x+36
a=1,b=12 and c=36
answer:
(x+6)(x+6)
Task Factoring
53

where a = 1 and c is NEGATIVE (QT2)• QUESTION 1:x2+1x-42
solution:
x2+1x-42
a=1,b=1 and c=42
answer:
(x+6)(x-7)
Task Factoring
54

where a = 1 and c is NEGATIVE (QT2)• QUESTION 2:x2-14x-14
solution:
x2-14x-14
a=1,b=14 and c=14
answer:
(x-9)(x-5)
Task Factoring
55

• QUESTION 1:6x^2+22x+20
solution:
6x2 20
^ ^
3x 2x 5 4
Answer:
(3x+5)(2x+4)
Task Factoring
56

• QUESTION 2:6x2+2x-8
Solution:
6x2 - 8
^ ^
6x x -8 1
Answer:
(6x-8)(x+1)
Task Factoring
57

• QUESTION 1:xy+2y+3x+6
solution:
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)
Task Factoring
58

• 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
Task Factoring
59

NAME OF STUDENT 1
Insert Picture
Here
GCMF
BINOMIAL
•DTS
•STC
•DTC
•PST
•QT 1
•QT 2
•GQT
COMPETELY

Name:
FACTOR (GCMF)
• QUESTION 1:
Second Quarter Grade 8 Mathematics
Performance Task Factoring 61

Name:
FACTOR (GCMF)
• QUESTION 2:

Name:
(DTS)
• QUESTION 1:

Name:
(DTS)
• QUESTION 2:

Name:
• QUESTION 1:

Name:
• QUESTION 2:

Name:
(DTC)
• QUESTION 1:

Name:
(DTC)
• QUESTION 2:

Name:
• QUESTION 1:

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

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

Name:
(GQT)
• QUESTION 1:

• QUESTION 2:
Name:

Name:
• QUESTION 1:

REFLECTION
Group Name ____
1 2
3 4

Name of Member 1
1 2
3 4

Name of Member 2
1 2
3 4

Name of Member 3
1 2
3 4

Name of Member 4
1 2
3 4

Second Quarter Group F Math Peta - Factoring (GCMF, DTS, STC, DTC, PST, QT1, QT2, GQT, FbG, FC)

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