Transcript Beginning & Intermediate Algebra, 4ed
Chapter 6
Factoring Polynomials
§ 6.1
The Greatest Common Factor and Factoring by Grouping
Factors
Factors
(either numbers or polynomials) When an integer is written as a product of integers, each of the integers in the product is a
factor
of the original number.
When a polynomial is written as a product of polynomials, each of the polynomials in the product is a
factor
of the original polynomial.
Factoring
– writing a polynomial as a product of polynomials
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Greatest Common Factor
Greatest common factor
– largest quantity that is a factor of all the integers or polynomials involved.
Finding the GCF of a List of Integers or Terms
1) Write each number as a product of prime numbers.
2) Identify the common prime factors.
3) The product of all common prime factors found in step 2 is the greatest common factor. If there are no common prime factors, the greatest common factor is 1.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Greatest Common Factor Example:
Find the GCF of each list of numbers.
1) 12 and 8 12 =
2
·
2
· 3 8 =
2
·
2
· 2 So the GCF is
2
·
2
= 4.
2) 7 and 20 7 = 1 · 7 20 = 2 · 2 · 5 There are no common prime factors so the GCF is 1.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Greatest Common Factor Example:
Find the GCF of each list of numbers.
1) 6, 8 and 46 6 =
2
· 3 8 = 46 =
2 2
· 2 · 2 · 23 So the GCF is 2.
2) 144, 256 and 300 144 = 256 =
2 2
· ·
2 2
· 2 · 3 · 3 · 2 · 2 · 2 · 2 · 2 · 2 300 =
2
·
2
· 3 · 5 · 5 So the GCF is
2
·
2
= 4.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Greatest Common Factor Example:
Find the GCF of each list of terms.
1)
x
3 and
x
7
x
3 =
x
·
x
·
x
x
7 =
x
·
x
·
x
·
x
·
x
·
x
·
x
So the GCF is
x
·
x
·
x
=
x
3 2) 6
x
5 and 4
x
3 6
x
5 =
2
· 3 ·
x
·
x
·
x
·
x
·
x
4
x
3 =
2
· 2 ·
x
·
x
·
x
So the GCF is
2
·
x
·
x
·
x
= 2
x
3
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Greatest Common Factor Example:
Find the GCF of the following list of terms.
a
3
b
2 ,
a
2
b
5 and
a
4
b
7
a a
3 2
b b
2 5 = =
a a
· ·
a a
·
a
·
b
· ·
b b
a
4
b
7 =
a
·
a
So the GCF is
a
·
a
·
a
·
a
·
b
· ·
b
b
·
b
· ·
b
·
b b
= ·
b
·
a
2
b b
2 ·
b
·
b
·
b
·
b
Notice that the GCF of terms containing variables will use the smallest exponent found amongst the individual terms for each variable.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Factoring Polynomials
The first step in factoring a polynomial is to find the GCF of all its terms. Then we write the polynomial as a product by
factoring out
the GCF from all the terms. The remaining factors in each term will form a polynomial.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Factoring out the GCF Example:
Factor out the GCF in each of the following polynomials.
1) 6
x
3 – 9
x
2 + 12
x
=
3
·
x
· 2 ·
x
2 –
3
·
x
· 3 ·
x
+
3
·
x
· 4 =
3x
(2
x
2 – 3
x
+ 4) 2) 14
x
3
y
+ 7
x
2
y
– 7
xy =
7
·
x
·
y
· 2 ·
x
2 +
7
·
x
·
y
· x –
7
·
x
·
y
· 1 =
7xy
(2
x
2 +
x
– 1)
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Factoring out the GCF Example:
Factor out the GCF in each of the following polynomials.
1) 6(
x
+ 2) –
y
(
x
+ 2) = 6 ·
(x + 2)
–
y
·
(x + 2)
=
(x + 2)
(6 –
y
) 2)
xy
(
y
+ 1) – (
y
+ 1) =
xy
·
(y + 1)
– 1 ·
(y + 1)
=
(y + 1)
(
xy
– 1)
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Factoring
Remember that factoring out the GCF from the terms of a polynomial should always be the first step in factoring a polynomial. This will usually be followed by additional steps in the process.
Example:
Factor 90 + 15
y
2 – 18
x
– 3
xy
2 .
90 + 15
y
2 – 18
x
– 3
xy
2 = 3(30 + 5
y
2 – 6
x
–
xy
2 ) = 3(
5
· 6 +
5
·
y
2 – 6 ·
x
–
x
·
y
2 ) = 3(
5 (6 + y 2 )
–
x
(6 + y 2 )
) = 3
(6 + y 2 )
(
5
–
x
)
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Factoring by Grouping
Factoring polynomials often involves additional techniques after initially factoring out the GCF.
One technique is
factoring by grouping
.
Example:
Factor
xy
+
y
+ 2
x
+ 2 by grouping.
Notice that, although 1 is the GCF for all four terms of the polynomial, the first 2 terms have a GCF of
y
and the last 2 terms have a GCF of 2.
xy
+
y
+ 2
x
+ 2 =
x
·
y
+ 1 ·
y
+
2
·
x
+
2
· 1 =
y
(x + 1)
+
2 (x + 1)
=
(x + 1)
(
y
+
2
)
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Factoring by Grouping
To Factor a Four-Term Polynomial by Grouping
1) Group the terms in two groups of two terms so that each group has a common factor.
2) Factor out the GFC from each group.
3) If there is now a common binomial factor in the groups, factor it out.
4) If not, rearrange the terms and try these steps again.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Factoring by Grouping Example:
Factor each of the following polynomials by grouping.
1)
x
3 + 4
x
+
x
2 + 4 =
x
·
x
2 +
x
· 4 +
1
·
x
2 +
1
· 4 =
x
(x 2 + 4)
+
1 (x 2 + 4)
=
(x 2 + 4)
(
x
+
1
) 2) 2
x
3 –
x
2 – 10
x
+ 5 =
x
2
· 2x –
x
2
· 1 –
5
· 2
x
–
5
· (– 1) =
x
2 (2x – 1)
–
5 (2x – 1)
=
(2x – 1)
(
x
2
–
5
)
Martin-Gay, Beginning and Intermediate Algebra, 4ed
Factoring by Grouping Example:
Factor 2
x
– 9
y
+ 18 –
xy
by grouping.
Neither pair has a common factor (other than 1).
So, rearrange the order of the factors.
2
x
+ 18 – 9
y
–
xy
=
2
·
x
+
2
· 9 – 9 ·
y
–
x
·
y
=
2
(
x
+ 9) –
y
(9 +
x
) =
2 (x + 9)
–
y
(x + 9)
= (make sure the factors are identical)
(x + 9)
(
2
–
y
)
Martin-Gay, Beginning and Intermediate Algebra, 4ed