FACTORING COMMON MONOMIAL 8

2. WIDE

3. THINK

4. TEAR

5. COMMON
MONOMIAL FACTOR

6. COMMON MONOMIAL FACTOR
•Is a number, a variable, or a
combination of number and
variable which can be found in each
term of a given polynomial.
•GCF – stands for GREATEST
COMMON FACTOR

7. STEPS IN FACTORING POLYNOMIALS
WITH A COMMON MONOMIAL FACTOR
•Find the GCF.
•Divide the polynomial by the GCF. The
quotient is the other factor.
•Express the polynomial as the product of
the quotient and the GCF.

8. EXAMPLE # 1:
16𝒂 𝟐
+ 𝟒𝒂
•STEP 1 : GCF =
4 : 1, 2, 4
16 : 1, 2, 4
4a

9. •STEP 2: Divide by the GCF.
(16𝒂 𝟐 + 𝟒𝒂) ÷ 4a
= (4a + 1)
•STEP 3: Express.
= 4a (4a + 1)

10. EXAMPLE # 2:
63 + 45b
•STEP 1 : GCF =
63 : 1, 3, 7, 9
45 : 1, 3, 5, 9
9

11. •STEP 2: Divide by the GCF.
(63 + 45b) ÷ 9
= (7 + 5b)
•STEP 3: Express.
= 9 (7 + 5b)

13. ASSIGNMENT!
1. 2𝒂 𝟐
b – 2𝒂 𝟐
=2𝒂 𝟐
=2𝒂 𝟐
( 𝒃 − 𝟏)

14. 2. 9𝒙 𝟐
𝒚 𝟐
– 2𝒙𝒚 𝟐
= 𝒙𝒚 𝟐
= 𝒙𝒚 𝟐
( 9𝒙 – 2)

15. 2. 8𝒙 𝟑
𝒚 𝟐
– 6𝒙 𝟐
𝒚 + 2𝒙𝒚 𝟐
= 𝟐𝒙𝒚
= 𝟐𝒙𝒚 ( 4𝒙 𝟐
𝒚 − 𝟑𝒙 + 𝒚)

16. 3. 15𝒂 𝟐
𝒃 𝟑
– 9𝒂𝒃 𝟐
+ 12𝒂 𝟐
𝒃 𝟐
= 𝟑𝒂 𝒃 𝟐
= 𝟑𝒂 𝒃 𝟐
(5 𝒂𝒃 – 3𝒃 + 6𝒂)

17. FACTORING
MONOMIALS BY
GROUPING

18. 𝑥 3
− 4 𝑥 2
+ 3 𝑥 − 12
( 𝑥 3
−4 𝑥 2
) + (3 𝑥 − 12)
𝑥 2
( 𝑥 − 4) 3( 𝑥 − 4)+
( 𝑥 − 4) (𝑥 2
+3)

19. 2 𝑥 2
− 3 𝑥 + 4 𝑥 𝑦 − 6 𝑦
𝑥 (2 𝑥 − 3) +
(2 𝑥 2
−3 𝑥 ) + (4 𝑥 𝑦 − 6 𝑦)
2 𝑦 (2 𝑥 − 3)
2 𝑥 − 3 ( 𝑥 + 2 𝑦)

20. 6 𝑎 𝑏 − 3 𝑏 2 + 8 𝑎 − 4 𝑏
3 𝑏 ( 2 𝑎 − 𝑏 ) + 4(2 𝑎 − 𝑏 )
2 𝑎 − 𝑏 (3 𝑏 + 4)
(6 𝑎 𝑏 − 3 𝑏 2
) + (8 𝑎 − 4 𝑏 )

21. 2 𝑥 2
+ 𝑥 − 2 𝑥 𝑦 − 𝑦
𝑥 (2 𝑥 + 1) +( − 𝑦 ) ( 2 𝑥 + 1 )
2 𝑥 + 1 ( 𝑥 − 𝑦)
2 𝑥 2
+ 𝑥 + [ ( − 2 𝑥 𝑦 + − 𝑦 ]

22. 4 𝑎 𝑏 2 − 12 𝑎 𝑏 + 𝑏 − 3
4 𝑎 𝑏 ( 𝑏 − 3 ) + 1( 𝑏 − 3)
𝑏 − 3 (4 𝑎 𝑏 + 1)
(4 𝑎 𝑏 2
− 12 𝑎 𝑏 ) + ( 𝑏 − 3)

23. 𝑨 𝑺 𝑺 𝑰 𝑮 𝑵 𝑴 𝑬 𝑵 𝑻!
3 𝑥 2
+ 6 𝑥 + 4 𝑥 + 8
5 𝑚 𝑛 + 2 5 𝑚 + 3 𝑛 3
+ 1 5 𝑛 2
6 𝑏 3
+ 16 𝑏 2
− 15 𝑏 − 40
4 𝑎 𝑢 + 2 4 𝑎 𝑣 − 5 𝑏 𝑢 − 3 0 𝑏 𝑣
7 𝑥 𝑦 − 3 𝑛 − 𝑥 + 21 𝑛 𝑦
𝑇 𝑜 𝑏 𝑒
𝑠 𝑢 𝑏 𝑚 𝑖 𝑡 𝑡 𝑒 𝑑
𝑜 𝑛
𝑆 𝐸 𝑃 𝑇 𝐸 𝑀 𝐵 𝐸 𝑅
1 6 , 2 0 2 0
𝑢 𝑛 𝑡 𝑖 𝑙 5 𝑝 𝑚 .

24. 3 𝑥 ( 𝑥 + 2) + 4( 𝑥 + 2)
𝑥 + 2 (3𝑥 + 4)
(3 𝑥 2
+ 6 𝑥) + (4 𝑥 + 8)
3 𝑥 2
+ 6 𝑥 + 4 𝑥 + 8

25. 5 𝑚 ( 𝑛 + 5 ) + 3 𝑛 2
( 𝑛 + 5)
𝑛 + 5 (5 𝑚 + 3 𝑛 2
)
5 𝑚 𝑛 + 2 5 𝑚 + 3 𝑛 3
+ 1 5 𝑛 2
( 5 𝑚 𝑛 + 2 5 𝑚 ) + ( 3 𝑛 3
+ 1 5 𝑛 2
)

26. 2 𝑏 2
( 3 𝑏 + 8 ) + ( − 5 ) ( 3 𝑏 + 8 )
3 𝑏 + 8 (2 𝑏 2
− 5)
6 𝑏 3
+ 1 6 𝑏 2
+ [ ( − 1 5 𝑏 + ( − 4 0 ) ]
6 𝑏 3
+ 16 𝑏 2
− 15 𝑏 − 40

27. 4 𝑎 ( 𝑢 + 6 𝑣 ) +( − 5 𝑏 ) ( 𝑢 + 6 𝑣 )
𝑢 + 6 𝑣 (4 𝑎 − 5 𝑏)
4 𝑎 𝑢 + 2 4 𝑎 𝑣 − 5 𝑏 𝑢 − 3 0 𝑏 𝑣
4 𝑎 𝑢 + 2 4 𝑎 𝑣 + [ − 5 𝑏 𝑢 + − 3 0 𝑏 𝑣 ]

28. 𝑥 (7 𝑦 − 1) + 3 𝑛 ( 7 𝑦 − 1 )
7 𝑦 − 1 ( 𝑥 + 3 𝑛)
𝟕 𝒙 𝒚 − 𝟑 𝒏 − 𝒙 + 𝟐 𝟏 𝒏 𝒚
𝟕 𝒙 𝒚 − 𝒙 + 𝟐 𝟏 𝒏 𝒚 − 𝟑 𝒏
( 𝟕 𝒙 𝒚 − 𝒙 ) + ( 𝟐 𝟏 𝒏 𝒚 − 𝟑 𝒏 )

29. ZERO
PRODUCT RULE

30. THEN A = 0 OR B
= 0
(OR BOTH A=0 AND
B=0)

31. E X A M P L E : SOLVE (X−5)(X−3) = 0
THE "ZERO PRODUCT PROPERT Y" SAYS:
IF (X−5)(X−3) = 0
THEN (X−5) = 0 OR (X−3) = 0
NOW WE JUST SOLVE EACH OF THOSE:
FOR (X−5) = 0 WE GET X = 5
FOR (X−3) = 0 WE GET X = 3
A ND THE ROOTS OF THE E QUATION A RE :
5 and 3

32. 𝑥2
− 12𝑥 = 0
𝑥 = 𝟏𝟐
𝑥(𝑥 − 12) = 0
𝑥 = 0 𝑜 𝑟 𝑥 − 12 = 0
AND THE ROOTS OF THE EQUATION ARE:
0 and 12

33. 2𝑥3
− 8𝑥2
= 0
𝑥 = 𝟒
2 𝑥 2
= 0 𝑜 𝑟 𝑥 − 4 = 0
2𝑥 2
(𝑥 − 4) = 0
AND THE ROOTS OF THE EQUATION ARE:
0 and 4

34. 2 𝑥 2
+ 3 𝑥 − 10 𝑥 − 15 = 0
( 𝟐 𝒙 𝟐
+ 𝟑 𝒙 ) + [ − 𝟏 𝟎 𝒙 + − 𝟏 𝟓 ] = 𝟎
𝒙 ( 𝟐 𝒙 + 𝟑 ) + − 𝟓 𝟐 𝒙 + 𝟑 = 𝟎
𝟐 𝒙 + 𝟑 = 0 𝑜 𝑟 𝑥 − 5 = 0
( 𝟐 𝒙 + 𝟑) 𝒙 − 𝟓 = 𝟎

35. 𝟐 𝒙 + 𝟑 = 0 𝑜 𝑟 𝑥 − 5 = 0
𝟐 𝒙 = −3
2 𝑥
2
= −
3
2
𝒙 = −
3
2
𝒙 = 5
𝒙 − 5 = 0
AND THE ROOTS OF
THE EQUATION ARE:
−
3
2
and 5

36. QUIZ
IN YOUR
NOTEBOOK
ANSWER PAGE
5
# 8 - 13
TO BE
SUBMITTED
TODAY
(SEPTEMBER
22, 2020)
UNTIL 3PM
ONLY.

37. ASSIGNMENT!
TO BE
SUBMITTED ON
THURSDAY
(SEPTEMBER 24,
2020) UNTIL
5PM.
IN YOUR
NOTEBOOK,
ANSWER
PAGE 7-8
# 1-10
# 11-16