Integers whose product is 9
1, 9
1,
9
3, 3
3,
3
Determine the sum of the integers.
Integers whose product is 9
1, 9
1,
9
3, 3
3,
3
Sum
10
10
6
6
Select 1, 9 because the product is 9 (last term
of the trinomial) and the sum is 10 (middle
term coefficient of the trinomial).
x
2
10
x
9
(
x
9)(
x
1)
Check:
(
x
9)(
x
1)
x
2
1
x
9
x
9
x
2
10
x
9
✓
■
YOUR TURN
Factor
x
2
9
x
20.

0.4
Factoring Polynomials
43
■
Answer:
(2
t
3)(
t
1)
Factors of
a
a
x
2
b
x
c
(
n
x
n
)(
n
x
n
)
Factors of
c
EXAMPLE 8
Factoring a Trinomial Whose Leading
Coefficient Is Not
1
Factor 5
x
2
9
x
2.
Solution:
S
TEP
1
Start with the fi
r
st
term. Note that 5
x
x
5
x
2
.
(5
x
;
n
)(
x
;
n
)
S
TEP
2
The product of the last
terms should yield
2.
1, 2 or 1,
2
S
TEP
3
Consider all possible factors based on Steps 1 and 2.
(5
x
1)(
x
2)
(5
x
1)(
x
2)
(5
x
2)(
x
1)
(5
x
2)(
x
1)
Since the outer
and inner
products must sum
to 9
x
, the factored form must be:
5
x
2
9
x
2
(5
x
1)(
x
2)
Check:
(5
x
1)(
x
2)
5
x
2
10
x
1
x
2
5
x
2
9
x
2
✓
■
YOUR TURN
Factor 2
t
2
t
3.
When the leading coefficient of the trinomial is not equal to 1, then we consider all
possible factors using the following procedure, which is based on the FOIL method in
reverse.
EXAMPLE 9
Factoring a Trinomial Whose Leading
Coefficient Is Not
1
Factor 15
x
2
x
6.
Solution:
S
TEP
1
Start with the fi
r
st
term.
(5
x
;
n
)(3
x
;
n
) or (15
x
;
n
)(
x
;
n
)
S
TEP
2
The product of the last
terms
should yield
6.
1, 6 or 1,
6 or 2,
3 or
2, 3
Step 1:
Find two
F
irst terms whose product is the first term of the trinomial.
Step 2:
Find two
L
ast terms whose product is the last term of the trinomial.
Step 3:
Consider all possible combinations found in Steps 1 and 2 until the sum of
the
O
uter and
I
nner products are equal to the middle term of the trinomial.
F
ACTORING A TRINOMIAL WHOSE LEADING
COEFFICIENT IS NOT
1

44
CHAPTER 0
Prerequisites and Review
S
TEP
3
Consider all possible factors
(5
x
1)(3
x
6)
(15
x
1)(
x
6)
based on Steps 1 and 2.
(5
x
6)(3
x
1)
(15
x
6)(
x
1)
(5
x
1)(3
x
6)
(15
x
1)(
x
6)
(5
x
6)(3
x
1)
(15
x
6)(
x
1)
(5
x
2)(3
x
3)
(15
x
2)(
x
3)
(5
x
3)(3
x
2)
(15
x
3)(
x
2)
(5
x
2)(3
x
3)
(15
x
2)(
x
3)
(5
x
3)(3
x
2)
(15
x
3)(
x
2)
Since the outer
and inner
products must
sum to
x
, the factored form must be:
15
x
2
x
6
(5
x
3)(3
x
2)
Check:
(5
x
3)(3
x
2)
15
x
2
10
x
9
x
6
15
x
2
x
6
✓
■
YOUR TURN
Factor 6
x
2
x
12.
■
Answer:
(3
x
4)(2
x
3)
EXAMPLE 11
Factoring a Polynomial by Grouping
Factor
x
3
x
2
2
x
2.
Solution:
Group the terms that have a common factor.
(
x
3
x
2
)
(2
x
2)
Factor out the common factor in each pair of parentheses.
x
2
(
x
1)
2(
x
1)
Use the distributive property.
(
x
2
2)(
x
1)
EXAMPLE 10
Identifying Prime (Irreducible) Polynomials
Factor
x
2
x
8.
Solution:
Write the trinomial as a product of
two binomials in general form.
x
2
x
8
(
x
n
)(
x
n
)
Write all of the
integers whose
product is
8.
Factoring by Grouping
Much of our attention in this section has been on factoring trinomials. For polynomials
with more than three terms we first look for a common factor to all terms. If there is no
common factor to all terms of the polynomial, we look for a group of terms that have a
common factor. This strategy is called
factoring by grouping
.

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- Summer '17
- juan alberto
- Fractions, Elementary arithmetic, Greatest common divisor