Factoring the Sum and
Difference of Two cubes a 3 + b 3 a 3 – b 3

1
1
Count
1 • How long is the edge?
• How many squares in the face?
• How many blocks?

Count
Edge Face Blocks
1
2
1
4
1
8

Count
Edge Face Blocks
1
2
3
1
4
9
1
8
27

Count
Edge Face Blocks n n 2 n 3
1 1 1
5
4
2
3
4
9
16
25
8
27
64
125

2
3 n
1
4
5
6
7
8
9
10
Memorize the First 10 Perfect Cubes
4
9 n 2
1
16
25
36
49
64
81
100 n 3
1
8
27
64
125
216
343
512
729
1000

Recall the Difference of Two
Squares Formula a 2 – b 2
=(a + b)(a – b) x 2 – 9 =(x + 3)(x – 3)
• There are similar formulas for the sum and difference of two cubes.

Multiply a Binomial by a Trinomial
( x
2  xy
( x
 y ) x
3  2 x y
 2 x y
 y
2
) The Sum of Cubes x
3  y
3
 xy
2
 xy
2  y
3

( x

)(
2  xy
 y
2
) x
3  y
3

Difference of Cubes
( x
2  xy
 y
2
) x
3  2 x y
 2 x y
( x
 y ) x
3  y
3
 xy
2
 xy
2  y
3

( x

)(
2  xy
 y
2
) x
3  y
3

Compare the Formulas
The Sum of Cubes x
3  y
3 
( x

)(
2  xy
 y
2
)
The Difference of Cubes x
3  y
3

( x

)(
2  xy
 y
2
)
They are just alike except for where they are different.

Using the Difference of Cubes x
3  y
3

( x

)(
2  xy
 y
2
) x 3 - 8
Recall 2 3 = 8
= (x - 2)(x 2 + 2x + 4)

Using the Sum of Cubes x
3  y
3 
( x

)(
2  xy
 y
2
) y 3 + 27
Recall 3 3 = 27
= (y + 3)(y 2 – 3y + 9)

Factor Out the Common Factor
3xa + 2x + 21a + 14 =
3 x a + 2 x + 3 (7) a + 2 (7) = x (3a + 2) + 7 (3a + 2) = (3a + 2) (x +7)
This is called factoring by grouping.

What is factoring by grouping?
Factoring a common monomial from pairs of terms, then looking for a common binomial factor is called factor by grouping.
When do I use factoring by grouping?
*when the problem consists of 4 terms
How will my answer look?
*it will be the product of two binomials

Factor the expression
5 x
2
( x
  x

2)
5 x
2
( x

2)

3( x

2)
Pull the common factor out of each term.
( x

2) (5 x
2 
3)
Notice there are two terms
Notice what each term has in common.
x

2
Notice what is left in each term after factoring out the common factor.

Try this example:
7 ( 5) 3( y

5)
( y

5)(7 y

3)

m
3 
7 m
2 
2 m

14
Form two binomials with a + sign
( m
3 
7 m
2   m

14) between them.
m
2
( m

7 )

2 ( m

7)
( m

7) ( m
2 
2 )

Try this example:
9 x
3 
9 x
2 
7 x

7
(9 x
3 
9 x
2
) ( 7 7)
9 x
2
( x
  x

1)
( x

1)(9 x
2 
7)

6x 2 – 3x – 4x + 2 by grouping
6x 2 – 3x – 4x + 2
= (6x 2 – 3x) + (– 4x + 2)
= 3x(2x – 1) + -2(2x - 1)
= (2x – 1)(3x – 2)

WB pp 89 and 90
Book p. 78 #1-27 0dd, p. 79 #1-27 odd

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