Example 275.
5
xy
−
8
x
−
10
y
+
16
Split the problem into two groups
5
xy
−
8
x
−
10
y
+
16
GCF on left is
x,
on right we need
a
negative
,
so we use
−
2
x
(5
y
−
8)
−
2(5
y
−
8)
(5
y
−
8)
is the same
!
Factor out this GCF
(5
y
−
8)(
x
−
2)
Our Solution
217

Sometimes when factoring the GCF out of the left or right side there is no GCF
to factor out. In this case we will use either the GCF of 1 or
−
1
. Often this is all
we need to be sure the two binomials match.
Example 276.
12
ab
−
14
a
−
6
b
+7
Split the problem into two groups
12ab
−
14
a
−
6
b
+7
GCF on left is
2
a,
on right
,
no GCF
,
use
−
1
2
a
(6
b
−
7)
−
1(6
b
−
7)
(6
b
−
7)
is the same
!
Factor out this GCF
(6
b
−
7)(2
a
−
1)
Our Solution
Example 277.
6
x
3
−
15
x
2
+2
x
−
5
Split problem into two groups
6
x
3
−
15
x
2
+2
x
−
5
GCF on left is
3
x
2
,
on right
,
no GCF
,
use
1
3
x
2
(2
x
−
5)
+1(2
x
−
5)
(2
x
−
5)
is the same
!
Factor out this GCF
(2
x
−
5)(3
x
2
+1)
Our Solution
Another problem that may come up with grouping is after factoring out the GCF
on the left and right, the binomials don’t match, more than just the signs are dif-
ferent. In this case we may have to adjust the problem slightly. One way to do
this is to change the order of the terms and try again. To do this we will move
the second term to the end of the problem and see if that helps us use grouping.
Example 278.
4
a
2
−
21
b
3
+6
ab
−
14
ab
2
Split the problem into two groups
4
a
2
−
21
b
3
+6
ab
−
14
ab
2
GCF on left is
1
,
on right is
2
ab
1(4
a
2
−
21
b
3
)
+2
ab
(3
−
7
b
)
Binomials don
′
t
match
!
Move second term to end
4
a
2
+6
ab
−
14
ab
2
−
21
b
3
Start over
,
split the problem into two groups
4
a
2
+6
ab
−
14
ab
2
−
21
b
3
GCF on left is
2
a,
on right is
−
7
b
2
2
a
(2
a
+3
b
)
−
7
b
2
(2
a
+3
b
)
(2
a
+3
b
)
is the same
!
Factor out this GCF
(2
a
+3
b
)(2
a
−
7
b
2
)
Our Solution
When rearranging terms the problem can still be out of order. Sometimes after
factoring out the GCF the terms are backwards. There are two ways that this can
happen, one with addition, one with subtraction. If it happens with addition, for
218

example the binomials are
(
a
+
b
)
and
(
b
+
a
)
, we don’t have to do any extra
work. This is because addition is the same in either order (
5+3=3+5=8
).
Example 279.
7+
y
−
3
xy
−
21
x
Split the problem into two groups
7+
y
−
3
xy
−
21
x
GCF on left is
1
,
on the right is
−
3
x
1(7+
y
)
−
3
x
(
y
+7)
y
+7
and
7+
y
are the same
,
use either one
(
y
+7)(1
−
3
x
)
Our Solution
However, if the binomial has subtraction, then we need to be a bit more careful.
For example, if the binomials are
(
a
−
b
)
and
(
b
−
a
)
, we will factor out the oppo-
site of the GCF on one part, usually the second. Notice what happens when we
factor out
−
1
.
Example 280.
(
b
−
a
)
Factor out
−
1
−
1(
−
b
+
a
)
Addition can be in either order
,
switch order
−
1(
a
−
b
)
The order of the subtraction has been switched
!

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- Spring '16
- artbellen
- Algebra, Addition, Subtraction, ........., Elementary arithmetic, Negative and non-negative numbers