§ 5.4 Factoring Trinomials Factoring Trinomials In section 5.3, we factored certain polynomials having four terms using the method of grouping. Now, we will use.
Download ReportTranscript § 5.4 Factoring Trinomials Factoring Trinomials In section 5.3, we factored certain polynomials having four terms using the method of grouping. Now, we will use.
§
5.4
Factoring Trinomials
Factoring Trinomials
In section 5.3, we factored certain polynomials having
four terms
using the method of
grouping
.
Now, we will use trial and error and a problem solving process to factor
trinomials
.
The polynomials we will begin with will have leading coefficients of one. We will begin by trying to factor these trinomials into a product of two binomials. Polynomials that cannot be factored over a given number set are said to be
prime
.
Blitzer,
Intermediate Algebra
, 5e – Slide #2 Section 5.4
Factoring Trinomials
A Strategy for Factoring
T
1) Enter
x
as the first term of each factor.
x
2
bx
c
2) List pairs of factors of the constant c.
3) Try
various combinations
of these factors. Select the combination in which the sum of the Outside and Inside products is equal to
bx
.
x
x
x
2
bx
c
I O Sum of O + I
4) Check your work by multiplying the factors using the FOIL method. You should obtain the original trinomial.
If none of the possible combinations yield an Outside product
and
an Inside product who sum is equal to
bx
, the trinomial cannot be factored
using integers
and is called
prime
over the set of integers.
Blitzer,
Intermediate Algebra
, 5e – Slide #3 Section 5.4
Factoring Trinomials
EXAMPLE Factor:
x
2 11
x
24
.
SOLUTION
1) Enter x as the first term of each factor.
x
2 11
x
24 To find the second term of each factor, we must find two integers whose product is 24 and whose sum is 11.
2) List pairs of factors of the constant, 24.
Some Factors of 24
24, 1 -24, -1 12, 2 -12, -2 8, 3 -8, -3 6, 4 -6, -4 Blitzer,
Intermediate Algebra
, 5e – Slide #4 Section 5.4
Factoring Trinomials
CONTINUED
3) Try various combinations of these factors.
x
2 11
x
24 The correct the Outside and Inside products is equal to 11
x
. Here is a list of the possible factorizations.
Possible Factorizations of Sum of Outside and Inside
x
2 11
x
24
Products (Should Equal 11x)
(
x
+ 24)(
x
+ 1) (
x
- 24)(
x
- 1) (
x
+ 12)(
x
+ 2) (
x
- 12)(
x
- 2) (
x
+ 8)(
x
+ 3) (
x
- 8)(
x
- 3) (
x
+ 6)(
x
+ 4) (
x
- 6)(
x
- 4)
x
+ 24
x
= 25
x -x
- 24
x
= -25
x
2
x
+ 12
x
= 14
x -
2
x
- 12
x
= -14
x
3
x
+ 8
x
= 11
x
-3
x
- 8
x
= -11
x
4
x
+ 6
x
= 10
x
-4
x
- 6
x
= 10
x
This is the required middle term.
Blitzer,
Intermediate Algebra
, 5e – Slide #5 Section 5.4
Factoring Trinomials
CONTINUED Thus,
x
2 11
x
24
x
8
x
3 . Check this result by multiplying the right side using the FOIL method. You should obtain the original trinomial. Because of the commutative property, the factorization can also be expressed as
x
2 11
x
24
x
3
x
8
. Blitzer,
Intermediate Algebra
, 5e – Slide #6 Section 5.4
Factoring Trinomials
EXAMPLE Factor: SOLUTION
a
2 5
a
14
.
1) Enter a as the first term of each factor.
a
2 5
a
14 To find the second term of each factor, we must find two integers whose product is -14 and whose sum is 5.
2) & 3) List pairs of factors of the constant, -14, and try various combinations of these factors.
Because the desired sum, 5, is positive, the positive factor of -14 must be farther from 0 than the negative factor is. Thus, we will only list pairs of factors of -14 in which the positive factor has the larger absolute value.
Blitzer,
Intermediate Algebra
, 5e – Slide #7 Section 5.4
Factoring Trinomials
CONTINUED
Some Factors of -14 Sum of Factors
14, -1 13 7, -2 5 This is the desired sum.
Thus,
a
2 5
a
14
a
7
a
2 . Blitzer,
Intermediate Algebra
, 5e – Slide #8 Section 5.4
Factoring Trinomials
EXAMPLE Factor: 4
y
3 12
y
2 72
y.
SOLUTION The GCF of the three terms of the polynomial is 4
y
. Therefore, we begin by factoring out 4
y
. Then we factor the remaining trinomial.
4
y
3 12
y
2 4 4 72
y y y y y
2 3
y y
18 Factor out the GCF
y
2 Begin factoring . Find two integers whose product is -18 and whose 4
y
y
3
y
6
Thus, 4
y
3 12
y
2 72
y
sum is 3.
The integers are -3 and 6.
4
y
y
3
y
6
. Blitzer,
Intermediate Algebra
, 5e – Slide #9 Section 5.4
Factoring Trinomials
A Strategy for Factoring
T
Assume, for the moment, that there is no greatest common factor.
1) Find two
F
irst terms whose product is
x
x
ax
2
a bx
2 :
c
. 2) Find two
L
ast terms whose product is
c
:
x
x
ax
2
bx
c
. 3) By trial and error, perform steps 1 and 2 until the sum of the
O
utside product and
I
nside product is
bx
:
x
x
ax
2
bx
c
. I O Sum of O + I If no such combinations exist, the polynomial is prime.
Blitzer,
Intermediate Algebra
, 5e – Slide #10 Section 5.4
Factoring Trinomials
EXAMPLE Factor: 6
x
2 19
x
15
.
SOLUTION
1) Find two First terms whose product is
6
x
2
.
There is more than one choice for our two First terms. Those choices are cataloged below.
6
x
2 19
x
15 6
x
2 19
x
15 6 2
x x
x
3
x
2) Find two Last terms whose product is 15.
There is more than one choice for our two Last terms. Those choices are cataloged below.
6
x
2 19
x
15 6
x
2 19
x
15 1 5 15 3 Blitzer,
Intermediate Algebra
, 5e – Slide #11 Section 5.4
Factoring Trinomials
CONTINUED
3) Try various combinations of these factors.
The correct
x
2
x
Outside and Inside products is equal to 19
x
. Here is a list of
some
of the possible factorizations.
Possible Factorizations of
6
x
2 19
x
15 (6
x
+ 1)(
x
+ 15) (
x
+ 1)(6
x
+ 15) (3
x
+ 3)(2
x
+ 5) (2
x
+ 3)(3
x
+ 5) (6
x
+ 3)(
x
+ 5) (
x
+ 3)(6
x
+ 5)
Sum of Outside & Inside Products (Should Equal 19x)
90
x
+
x
= 91
x
15
x
+ 6
x
= 21
x
15
x
+ 6
x
= 21
x
10
x
+ 9
x
= 19
x
30
x
+ 3
x
= 33
x
5
x
+ 18
x
= 23
x
This is the required middle term.
Blitzer,
Intermediate Algebra
, 5e – Slide #12 Section 5.4
Factoring Trinomials
CONTINUED
x
2 Therefore, the factorization of is: (2
x
+ 3)(3
x
+ 5) .
Determine which possible factorizations were
not
represented in the chart on the preceding page.
Blitzer,
Intermediate Algebra
, 5e – Slide #13 Section 5.4
Factoring Trinomials
EXAMPLE Factor: 10
y
5 17
y
4 3
y
3
.
SOLUTION
y
3 The GCF of the three terms of the polynomial is . We begin by factoring out .
3 10
y
5 17
y
4 3
y
3
y
3 10
y
2 17
y
3
1) Find two First terms whose product is
10
y
2 .
10
y
2 17
y
3 10
y
10
y
2 17
y
3 5
y
y
2
y
Blitzer,
Intermediate Algebra
, 5e – Slide #14 Section 5.4
Factoring Trinomials
CONTINUED
2) Find two Last terms whose product is 3.
The only possible factorization is (-1)(-3) since the sum of the Outside and Inside products must be -17
y
, having a
negative
coefficient.
3) Try various combinations of these factors.
The correct
y
2 factorization of is the one in which the sum of the Outside and Inside products is equal to -17
y
. A list of the possible factorizations can be found on the next page.
Blitzer,
Intermediate Algebra
, 5e – Slide #15 Section 5.4
Factoring Trinomials
CONTINUED
Possible Factorizations of
10
y
2 17
y
3 (10
y
- 1)(
y
- 3) (
y
- 1)(10
y
- 3) (2
y
- 1)(5
y
- 3) (5
y
- 1)(2
y
- 3)
Sum of Outside & Inside Products (Should Equal -17y)
-30
y
-
y
= -31
y
-3
y
- 10
y
= -13
y
-6
y
- 5
y
= -11
y
-15
y
- 2
y
= -17
y
This is the required middle term.
2 The factorization of is (5
y
- 1)(2
y
- 3). Now we include the GCF in the complete factorization of the given polynomial. Thus, 10
y
5 17
y
4 3
y
3
y
3 10
y
2 17
y
3
y
3
5
y
1
2
y
3
. Blitzer,
Intermediate Algebra
, 5e – Slide #16 Section 5.4
Factoring Trinomials
EXAMPLE Factor: 12
x
2 10
xy
8
y
2
.
SOLUTION
1) Find two First terms whose product is
12
x
2 .
12
x
2 10
xy
8
y
2 12
x
2 10
xy
8
y
2 12
x
2 10
xy
8
y
2
12
6
x
4
x x
?
?
y
x
?
y
2
x y
3
x
?
y
?
y
?
y
The question marks inside the parentheses indicate that we are looking for the coefficients of
y
in each factor.
Blitzer,
Intermediate Algebra
, 5e – Slide #17 Section 5.4
Factoring Trinomials
CONTINUED
2) Find two Last terms whose product is -8.
factorizations are as follows.
The possible 12
x
2 10
xy
8
y
2 12
x
2 10
xy
8
y
2 12
x
2 10
xy
8
y
2 12
x
2 10
xy
8
y
2
?
x
?
x
?
?
x x
8
y
?
x
8
y
?
x
4 4
y y
?
?
x x
y
y
2 2
y y
The question marks inside the parentheses indicate that we are looking for the coefficients of
x
in each factor.
Blitzer,
Intermediate Algebra
, 5e – Slide #18 Section 5.4
Factoring Trinomials
CONTINUED
3) Try various combinations of these factors.
The correct 12
x
2 10
xy
8
y
2 the Outside and Inside products is equal to 10
xy
. A list of the possible factorizations can be found below and on the next page.
Possible Factorizations of
12
x
2 10
xy
8
y
2 (12
x
+ 8
y
)(
x
-
y
) (
x
+ 8
y
)(12
x
-
y
) (12
x
+ 4
y
)(
x
- 2
y
) (
x
+ 4
y
)(12
x
- 2
y
) (6
x
+ 8
y
)(2
x
-
y
) (2
x
+ 8
y
)(6
x
-
y
)
Sum of Outside & Inside Products (Should Equal 10xy)
-12
xy
+ 8
xy
= -4
xy
-
xy
+ 96
xy
= 95
xy
-24
xy
+ 4
xy
= -20
xy
-2
xy
+ 48
xy
= 46
xy
-6
xy
+ 16
xy
= 10
xy
-2
xy
+ 48
xy
= 46
xy
Blitzer,
Intermediate Algebra
, 5e – Slide #19 Section 5.4
Factoring Trinomials
CONTINUED
Possible Factorizations of
12
x
2 10
xy
8
y
2 (6
x
+ 4
y
)(2
x
- 2
y
) (2
x
+ 4
y
)(6
x
- 2
y
) (4
x
+ 8
y
)(3
x
-
y
) (3
x
+ 8
y
)(4
x
-
y
) (4
x
+ 4
y
)(3
x
- 2
y
) (3
x
+ 4
y
)(4
x
- 2
y
)
Sum of Outside & Inside Products (Should Equal 10xy)
-12
xy
+ 8
xy
= -4
xy
-4
xy
+ 24
xy
= 20
xy
-4
xy
+ 24
xy
= 20
xy
-3
xy
+ 32
xy
= 29
xy
-8
xy
+ 12
xy
= 4
xy
-6
xy
+ 16
xy
= 10
xy
This is the required middle term.
Only half of the possible factorizations were checked. For example, (12
x
– 8
y
)(
x
+
y
) was not checked. It was not necessary since the corresponding factorization (12
x
+ 8
y
)(
x
-
y
) was checked and the absolute value of the coefficient, -4, was not 10.
Blitzer,
Intermediate Algebra
, 5e – Slide #20 Section 5.4
Factoring Trinomials
CONTINUED Thus, 12
x
2 10
xy
8
y
2
3
x
4
y
4
x
2
y
. NOTE: There was another factorization that resulted in the desired middle term, 10
xy
. That factorization would have worked just as well as the one selected above. That factorization was (6
x
+ 8
y
)(2
x
-
y
) = (2)(3x +4y)(2x – y). The answer we got above is: (3x + 4y)(4x – 2y) = (3x+4y)(2x – y)(2). As you can see, these two answers are equivalent.
This illustrates why the greatest common should always be factored out first. Doing this simplifies the factoring process for you.
Blitzer,
Intermediate Algebra
, 5e – Slide #21 Section 5.4
Factoring Trinomials
EXAMPLE Factor: 2
x
6 13
x
3 15
.
SOLUTION
x
3 Notice that the exponent on is half that of the exponent on
x
6 .
We will let
t
2 Thus, let 2 13
t
15
x
2
t
2
t
2 2
x
3 13
t
3
t
3
x
3 5
equal the variable to the power that is half of 6. 15 3 .
5 Therefore, 2
x
6 13
x
This is the given polynomial, with written as
2 .
t
terms of
t
3 .
.
Factor the trinomial. 3
x
3 Now substitute for
t
. 15 2
x
3 3
x
3 5 .
x
6 Blitzer,
Intermediate Algebra
, 5e – Slide #22 Section 5.4
Factoring Trinomials
2
T
1) Multiply the leading coefficient,
a
, and the constant,
c
.
2) Find the factors of
ac
whose sum is
b
.
3) Rewrite the middle term,
bx
, as a sum or difference using the factors from step 2.
4) Factor by grouping.
Blitzer,
Intermediate Algebra
, 5e – Slide #23 Section 5.4
Factoring Trinomials
EXAMPLE Factor by grouping:
a
2
a
12
.
SOLUTION The trinomial is of the form
ax
2
bx
c
.
a
2
a
12
a
= 1
b
= 1
c
= -12
1) Multiply the leading coefficient, a, and the constant, c.
Using
a
= 1 and
c
= -12.
ac
1
12 Blitzer,
Intermediate Algebra
, 5e – Slide #24 Section 5.4
Factoring Trinomials
CONTINUED
2) Find the factors of ac whose sum is b.
We want the factors of -12 whose sum is
b
, or 1. The factors of -12 whose sum is 1 are 4 and -3.
3) Rewrite the middle term, a, as a sum or difference using the factors from step 2: 4 and -3.
a
2
a
12
a
2 4
a
3
a
12 Blitzer,
Intermediate Algebra
, 5e – Slide #25 Section 5.4
Factoring Trinomials
CONTINUED
4) Factor by grouping.
a
2 4
a
3
a
12
a
a
4
a
4
a
4
a
3
Group terms Factor from each group Factor out
a
+ 4, the common binomial factor Thus,
a
2
a
12
a
4
a
3
. Blitzer,
Intermediate Algebra
, 5e – Slide #26 Section 5.4