TRINOMIALS
See also :
Basic Algebra: Factoring Trinomials
Do you remember the product of two binomials, F OI L, and the fact
that the result is usually in a trinomial? As examples, consider:
Product Binomials F
OI L
(x
+ 2)(
x
+ 5) =
x2
+ 7x
+ 10
(x
+ 2)(
x
5) =
x2
3x
10
(x
2)(
x
+ 5) =
x2
+ 3x
10
(x
2)(
x
5) =
x2
7x
+
10
In each of these examples, you were given a product of two
binomials, and with F
OI
L, in each case you
obtained a trinomial with
x2.
Now the problem will be to work these problems in reverse. What if you
were given a trinomial, such as
x2+7
x+10
and asked to factor itthat is, to express it as a product.
This is the product of the two binomials. When you factor
the trinomial:
x2
+ 7
x
+ 10
you expect the product of binomials: (
)( ).
When factoring a trinomial, instead of thinking
F
OI L,
you need to change the order and think F
L
OI.
In other words, you need to find the correct
F
(first times first) combination, then skip to the
L
(last times last). Finally, check to make sure the
OI (outer
times outer, inner times inner) middle term is correct.
F
OI
L
x2
+
7
x
+ 10 Given
trinomial to factor;
(
)( )
Product of two binomials;
(x
)(
x
)
F term is
x2,
which is
x
times
x;
(x
)(x
)
L term is +10.
Find two numbers whose product is +10. Probably 2 times 5, or it
could be 1 times 10. Try 2 times 5. Since the last sign is
positive, it will be positive times positive, or negative times
negative. Also, the middle terms (O and I) must be added
together.
(x
+ 2)(
x
+ 5)
OI term is 7x.
This means the outer times outer and the inner times inner
terms must add up to 7x.
[NOTICE: The
order does not matter! If you wrote (x+5)(
x+2),
this is exactly equivalent to (x
+ 2)(
x
+ 5).
WHY??]
The following is a helpful summary of trinomial (F L OI) factoring.
RULES: 1. When the sign of the LAST term
is positive, the signs are the SAME.
Find middle term by ADDING the O and I terms.
2.
When the sign of the LAST term is negative, the signs are
OPPOSITE.
Find middle term by SUBTRACTING the O and I terms.
|
It is also worth noting that not all trinomials can be factored. For
examples, the trinomials
x2
+
x
+ 2,
x2
+ 2x
+ 6,
and
x2
4x
6
cannot be factored. These are called prime trinomials.
EXERCISES. Factor each of the following trinomials.
1.
x2
+ 3
x
+ 2
2.
x2
+ 9
x
+ 14
3.
x2
+ 7
x
+ 10
4.
x2
12
x
+ 35
5.
x2
13
x
+ 22
6.
x2
9
x
+ 18
7.
x2
+ 3
x
40
8.
x2
+ 18
x
40
9.
x2
+ 39
x
40
10.
x2
15
x
16
11.
x2
6
x
16
12.
x2
x
72
13.
x2
7
x
18
14.
x2
+ 3
x
18
15.
x2
19
x
+ 60
16.
x2
6
x
27
17.
x2
+ 2
x
24
18.
x2
5
x
24
19.
x2
32
x
+ 60
20.
x2
+ 11
x
60
21.
x2
+ 61
x
+ 60
22.
x2
23
x
+ 60
23.
x2
17
x
60
24.
x2
59
x
60
25.
x2
x
56
26.
x2
x
72
27.
x2
+ 17
x
+ 72
28.
x2
+ 17
x
+ 42
29.
x2
19
x
42
30.
x2
23
x
+ 42
Frequently, it is necessary to FACTOR
THE COMMON
FACTOR
FIRST
(FCFF).
When there is a common factor in the problem,
always remember to FCFF!
(NOTE: These exercises require TWO steps!)
EXERCISES: Factor the trinomials completely. Be sure to
FCFF!!
31. 3x2
+ 6
x
9
32. 5
x2
+ 15
x
20
33. 8
x2
+ 8
x
48
3(x2 + 2x
3)
5(
)
___( )
3( )(
)
5( )(
)
___( )( )
34. 6
x2
+ 6
x
12
35. 12
x2
+ 36
x
120
36. 10
x2
+ 30
x
100
___(
)
___(
)( )
37.
x3
4
x2
5
x
38.
x3
+ 5
x2
+ 6
x
39. 2x3
14
x2
+ 20x
40.
5 x
3 + 5 x 2 10 x
41.
7 x 3
+ 49 x 2 + 42 x
42.
8 x 3
40 x 2 + 32 x
43.
x
4 + x
3 20 x 2
44.
x
4 + 2 x
3 35 x 2
45.
15 x 4 45 x 3 60 x 2
46.
20 x 4
20 x 3 120 x 2
47.
30 x 4 + 90 x 3 300 x 2
48.
18 x 4
+ 54 x3 + 36 x2
See also :
Basic Algebra : Advanced Trinomial Factoring
In the factoring of trinomials of the previous exercises, it may have been
assumed that either the coefficient of
x2
is 1 or that the coefficient can be factored as a common factor of the
entire trinomial. However, consider the example
5x2
+ 6x
+ 1.
Remember that this should be factored by F
L
OI, where the
F term
must be 5x2,
L
must be 1,
and the OI
term must add up to
6x,
as follows:
5
x2
+
6
x
+ 1
( )(
)
Try the following examples (solutions are given below):
3
x
2
+ 4
x
+ 1
8
x
2
+ 9
x
+ 1
8
x
2
+ 6
x
+ 1
( )(
)
( )(
)
( )(
)
Solutions: (3x
+ 1)(x
+ 1)
(8
x
+ 1)(
x
+ 1)
(4
x
+ 1)(2
x
+ 1)
Of course, with larger numbers, with many more combinations of numbers
this can become a very lengthy process of trial and error. There are some
systematic methods of factoring these trinomials, one of which will be
presented at the end of the "Factoring by Grouping" section. In problems
that are not too difficult, the trial and error method will be fairly
simple and more than adequate for now.
Consider the examples
5x2
+ 8x
+
3
and 5x2
+ 16x
+
3.
Again, these are factored by F
L
OI, where the
F term
must be 5x2,
L
must be 3,
and the OI
term must add up to
8x
and
16x
respectively, as follows:
5x2
+ 8
x
+ 3
5x2
+ 16
x
+
3
( )(
) ( )(
)
Try the following examples (again solutions are given below):
3x2
+ 10
x
+ 7
3x2
+ 22
x
+
7
3x2
+ 4
x
7
( )(
)
( )(
)
( )(
)
Solutions: (3
x
+ 7)(
x
+ 1)
(3
x
+ 1)(
x
+ 7)
(3
x
+ 7)(
x
1)
REMEMBER: 1.
When the L term is positive,
add
the O and I terms.
2. When the L term is negative,
subtract
the O and I terms.
|
EXERCISES:
1. 3
x
2
+ 4
x
+ 1
2. 4
x
2
+ 5
x
+ 1
3. 7
x
2
8
x
+ 1
4. 3
x
2
+ 2
x
1
5. 4
x
2
+ 3
x
1
6. 7
x
2
+ 6
x
1
7. 10
x
2
7
x
+ 1
8. 10
x
2
9
x
1
9. 10
x
2
+ 3
x
1
10. 3
x
2
+ 8
x
+ 5
11. 3x2
+ 16x
+ 5
12. 3
x
2
+ 2
x
5
13. 3
x
2
14
x
5
14. 3
x
2
34
x
+ 11
15. 3
x
2
+ 8
x
11
16. 5
x
2
+ 3
x
8
17. 5
x
2
+ 39
x
8
18. 5
x
2
+ 6
x
8
19. 6
x
2
19
x
+ 8
20. 6
x
2
13
x
8
21. 6
x
2
+ 13
x
8
22. 6
x
2
+ 19
x
+ 10
23. 6
x
2
+ 11
x
10
24. 6
x
2
+ 17
x
+ 10
See also :
Basic Algebra: Factoring the Difference of Squares
DIFFERENCE OF SQUARES; PERFECT SQUARE TRINOMIALS
EXAMPLES:
GENERALIZATIONS:
x2
25 =
(x
5)(
x
+ 5)
x
2
y2
=
(x
y)(
x
+
y)
x2
+ 10
x
+ 25 =
(x
+ 5)(
x
+ 5)
x2
+ 2
x y
+
y2
=
(x
+
y)(
x
+
y)
= (x
+ 5)2
= (x
+
y)2
x2
10
x
+ 25 =
(x
5)(
x
5)
x2
2
x y
+
y2
=
(x
y)(
x
y)
= (x
5)2
= (x
y)2
1.
x
2
64
2.
x
2
100
3.
x2
81
(x
____)(
x
+____)
4.
x2
169
5.
x2
a2
6.
x2
b2
7. 4
x
2
9
8. 64
x
2
121
9. 16
x
2
49
(2
x
___)(2
x
+___)
10. 81
x2
25
y2
11. 49
x2
36
y2
12. 25
x2
144a2
Don't forget that the first step in any factoring problem is to factor
the common factor first. This means that each of the following
exercises will require two steps. This is known as the "factoring
twostep."
13. 9
x2
9
14. 3
x2
12
15. 5
x2
45
9(
)
3(
)
___( )
9(
)(
)
3(
)(
)
___( )(
)
16. 4x2
64
17. 4x2
100
18. 8x2
72
19. 3x3
75x
20. 5x3
80x
21. 2x3
50x
22. 64
x4
4
x2y2
23. 12
y4
12
x2
y2
24. 79
y4
79
x2
y2
Factor each of the following perfect square trinomials:
25.
x
2
+ 4
x
+ 4
26.
x
2
+ 14
x
+ 49
27.
x
2
+ 20
x
+ 100
=(
)( )
=
(
)2
28.
x
2
12
x
+ 36
29.
x
2
18
x
+ 81
30.
x
2
24
x
+ 144
Remember to factor the common factor first:
31. 5
x
2
20
x
+ 20
32. 2
x
2
20
x
+ 50
33. 3
x
2
+ 6
x
+ 3
34. 3x2
60
x
+ 300
35. 6x2
36x
+ 54
36. 4x2
+ 32x
+ 64
37.
x
3
+ 4x
2
+ 4x
38.
x
3
12x2
+ 36
x
39. 9x3
18x2
+ 9x
40. 6x
4
36x
3
+ 54x
2
41. 12y
4
48y
3
+ 48y
2
42. 2y
4
28y3
+ 98y
2
In the next exercises, remember that sum of squares, such as
x2
+ 9 or
x2
+ 4,
does not factor by this method.
43.
x
4
16
44. x
4
1
=(
x2
____) (x2
+ ____)
=(
x
___)(
x
+___)(
x
2
+ ___)
45.
x
4
81
46. x
4
y
4
47. 81
x
4
16
y
4
48. 16
x
4
81
49.
x
4
+ 10
x
2
+ 9
50.
x
4
+ 13
x
2
+ 36
51.
x
4
13
x
2
+ 36
52.
x
4
29
x
2
+ 100
53.
y
4
+ 5
y
2
36
54.
y
4
18
y
2
+ 81
55. 25
x
4
64
x
2
56. 9
x
4
81
x
2
57. 9
x
4
36
x
2Y2
58. 9
x
3
54
x
2
+ 81
x
59. 8
x
3
+ 80
x
2
+ 200
x
60. 8
y
4
+ 16
y
3
+ 8
y
2
See also :
Intermediate Algebra: Factoring Difference and Sum of Cubes in
Living Color
SUM AND
DIFFERENCE OF CUBES
In this section, formulas and procedures will enable you to factor
expressions in the form
x3
y3
and also
x3
+
y3.
Recall from previous sections that
x2
y2
= (x
y)(
x
+
y)
and that
x2
+
y2
cannot be factored. Begin with the multiplication problems:
(X Y)(X2
+ XY + Y2)
=
X3
+ X2Y
+ XY2
X2Y
XY2
Y3
= X3
Y3
(X + Y)( X2
XY + Y2
) =
X3
X2Y
+ XY2
+
X2Y
XY2
+ Y3
= X3
+ Y3
.
This derives the formulas, known as the
sum and difference of cubes formulas.
SUM AND DIFFERENCE OF CUBES FORMULAS
x3
y3
= (x
y)(
x2
+
x y
+
y2)
x3
+
y3
= (x
+
y)(
x2
x y
+
y2)
|
Translated into words, this means the sum or difference of two cubes can
be factored into the product of a binomial times a trinomial. Begin by
taking the cube root of the perfect cubes.
In the difference formula, the binomial is the first minus the
second. Then use this binomial to build the trinomial that follows:
take the square of the first plus the product of the first and
second plus the square of the second.
These formulas are really easy to remember. Notice that the x3
y3
formula begins with (x
y).
The
x3
+
y3
formula begins with (x
+
y).
Next notice that the trinomial factor in both formulas is the same except
for one sign. This trinomial factor in each formula involves the square
of the first, the product of the two, and the square of the second.
The first sign in the trinomial is the opposite of the sign of the
binomial, and the last sign is always positive. Finally, you will
never be able to factor the resulting trinomial (by ordinary trinomial
methods), so you need not even try (at this level).
Before getting into the exercises, be sure to be familiar with the perfect
cubes: 13
= 1; 23
= 8; 33
= 27; 43
= 64; and 53
= 125. Be able to recite these from memory: 1, 8, 27, 64, 125. Practice
them as you drive down the highway!
In each of the following, factor completely:
1.
x3
8
2.
x3
125
=
x3
23
[x=first; 2=second]
= ( )3
( )3
= (
)(
x
2
+ 2
x
+ 22)
=
(
)( + +
)
= (
)(
)
= (
)(
)
3.
x3
64
4.
x3
27
= ( )3
( )3
= ( )3
( )3
= ( )(
)
= (
)(
)
5.
x3
+ 8
6.
x3
+ 64
=
x3
+ 23
[x=first;
2=second]
= ( )3
+ ( )3
= ( +
)(
x2
+
)
= ( + )(
+ )
= (
)(
)
= (
)(
)
7.
x3
+ 125
8.
x3
+ 27
= ( )3
+ ( )3
= ( )3
+ ( )3
= (
)(
)
= ( )(
)
9. 8
x3
125
10. 27
x3
8
y3
= (2
x)3
53
[2
x
=first; 5=second]
= ( )3
( )3
= ( )[(2
x)2
+ (2
x)(5)
+ 52]
= ( )[ + +
]
= ( )(
)
= ( )(
)
11. 64
x3
+ 125
12. 27
x3
+ 8
y3
= ( )3
+ ( )3
= ( )3
+ ( )3
= ( )[
]
= ( )[
]
= ( )(
)
= ( )(
)
13. 8
x3
27
y3
14. 125
y3
8
x3
15. 8
x3
+ 1
16. 125
y3
1
In the next exercises, dont forget the common factor first.
17. 16
x4
54
x
18. 3
x3
24
y3
= 2x
(8
x3
27)
= 2x
[( 2
x
)3
( 3 )3
]
= 2x
( ____
____ )( ____+_____+____ )
19. 5
x4
+ 40
x
20. 10
x5
y
+ 80
x2
y4
21. 3
x5
y5
81
x2
y2
22. 16
x2
y2
+ 250
x2
y5
23.
x6
y9
24.
x6
+
y9
= (
x2
)3
(
y3
)3
= ( )3
+ ( )3
= ( )[(
)2
+ ( )( )+(
)2]
= (
) [
]
= ___________________________
= __________________________
25. 8
x6
+ 125
y6
26. 8
x6
125
y6
= ( )3
+ ( )3
=
=
Remember, if there is a common factor, factor this first.
27. 5
x7
+ 40
x y9
28. 5
x7
40
x y9
= 5
x(
)
= 5
x[(
)3
+ ( )3
]
=
=
29. 16
x4
2x
y6
30. 25
x5
+ 200
x8
y9
31.
x6
+ 2x3
+
1
32.
x6
9x3
+ 8
= (x3
)(
x3
)
=
=
33.
x6
7x3
8
34.
x6
64
= (x3
)(x3
+ )
=
See also :
Basic Algebra: Factoring by Grouping
Intermediate Algebra: Factoring by Grouping in Living Color
FACTORING
BY GROUPING
This lesson introduces the technique of factoring by grouping. It builds
on the ideas that were presented in the section on factoring the common
factor. There are many different ways to group terms, some of
which will be successful in the factoring process, others will not. The
method of grouping then is one of trial and error, with a few insights
that may be helpful. There is no substitute for practice and experience.
EXAMPLE 1: Factor X3
+ 2X2
+ 8X + 16
SOLUTION:
There are no common factors to all four terms. It is not a trinomial, and
nothing discussed so far works to factor this. So, try grouping the first
two terms together, and the last two terms together, and factor out the
common factor within each grouping as follows:
(X3
+ 2X2)
+ (8X + 16)
= X2
(X + 2)+ 8(X
+ 2)
Notice that there is a common factor of (X+2) that can be factored out:
= (X + 2)( X2
+ 8)
EXAMPLE 2: Factor XY
4Y
+ 3X 12
SOLUTION:
Again, there are no common factors, and this is not a trinomial. Group
the first two and the last two terms together, and factor out the common
factor from each grouping:
(XY
4Y)
+ (3X 12) = Y(X 4) + 3(X 4) Now, take out the common factor,
which is (X 4):
= (X 4)(Y + 3)
EXAMPLE 3: Factor XY
4Y
3X + 12
SOLUTION:
Group the first two and the last two terms together, and factor out the
common factor from each grouping:
(XY
4Y)
+ (3X + 12)
= Y(X 4) + 3(X + 4)
This time there is no common factor. Try again, this time
factoring a 3 from the last grouping. This works!
XY
4Y
3X + 12 =
Y(X 4) 3(X 4)
= (X 4)(Y 3)
EXAMPLE 4: Factor XY
4Y
+ 3X + 12
SOLUTION: Group the first two and the last two terms:
XY
4Y
+ 3X + 12 = Y(X 4) + 3(X + 4)
At this point, it is important to realize that no common factor resulted.
Do not try to factor out something that is not common to both groupings.
In fact, there is no way to group this problem to get a common factor.
This one cannot be factored by grouping. In fact, it cant be factored by
any method. Remember, not all problems can be factored.
Remember that the entire process of grouping is one of trial and error,
and, as you will see later, there are different types of grouping.
EXERCISES:
Factor each of the following by grouping:
1.
XY + 7X + 4Y + 28
2. 2XY + 5X + 10Y + 25
= X(
) + 4( )
=
= (
)(
)
=
3. X3
+ 3X2
+ 9X + 27
4. X3
3X2
+ 9X 27
=
=
=
=
5. XY 5X 2Y + 10
6. X2Y
+ XY2
5X 5Y
= X( ) 2(
)
= XY( ) 5(
)
=
=
7. X3
X2
9X + 9
8. X3
5X2
4X + 20
= X2
( ) 9(
)
= X2
( ) 4(
)
= ( )
( )
=
= (
)( )(
)
=
9. X3
+ 7X2
X 7
10. X3
5X2
+ 25X 125
= X2
( ) 1(
)
=
=
=
=
Does X2
+ 25 factor?
11. X3
+ 5X2
25X 125
12. X3
5X2
25X + 125
13. X3 4X2 + 9X 36
14. X3 + 4X2 9X 36
15. X3 4X2 9X + 36
16. X3 9X2 4X + 36
