Download X + - mrsbybee

8-1 Factors and Greatest Common Factors
5 Minute Warm-Up
Directions: Simplify each problem. Write the
answer in standard form.
1. (4x + 7)(– 2x)
2. -4x2(3x2 + 2x – 6)
3. (x + 6)(x + 9)
4. (-4y + 5)(-7 – 3y)
5. (-8x3 + x – 9x2 + 2) + (8x2 – 2x + 4)
6. (6x2 – x + 3) – (-2x + x2 – 7)
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Warm Up
Tell whether the second number is a factor
of the first number
1. 50, 6
no
2. 105, 7
yes
3. List the factors of 28. ±1, ±2, ±4, ±7,
±14, ±28
Tell whether each number is prime or
composite. If the number is composite, write
it as the product of two numbers.
4. 11 prime
Holt Algebra 1
5. 98 composite; 49  2
8-1 Factors and Greatest Common Factors
Objectives
Write the prime factorization of
numbers.
Find the GCF of monomials.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
The whole numbers that are multiplied to find a
product are called factors of that product. A
number is divisible by its factors.
You can use the factors of a number to write the
number as a product. The number 12 can be
factored several ways.
Factorizations of 12

Holt Algebra 1






8-1 Factors and Greatest Common Factors
The order of factors does not change the product,
but there is only one example below that cannot
be factored further. The circled factorization is
the prime factorization because all the factors
are prime numbers. The prime factors can be
written in any order, and except for changes in
the order, there is only one way to write the
prime factorization of a number.
Factorizations of 12

Holt Algebra 1






8-1 Factors and Greatest Common Factors
Remember!
A prime number has exactly two factors, itself
and 1. The number 1 is not prime because it only
has one factor.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Example 1: Writing Prime Factorizations
Write the prime factorization of 98.
Method 1 Factor tree
Method 2 Ladder diagram
Choose any two factors
Choose a prime factor of 98
of 98 to begin. Keep finding
to begin. Keep dividing by
factors until each branch
prime factors until the
ends in a prime factor.
quotient is 1.
98
2 98
7 49
2  49
7 7

7
7
1
98 = 2  7  7
98 = 2  7  7
The prime factorization of 98 is 2  7  7 or 2  72.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Check It Out! Example 1
Write the prime factorization of each number.
a. 40
40
2  20
2  10
2  5
40 = 23  5
The prime factorization
of 40 is 2  2  2  5 or
23  5.
Holt Algebra 1
b. 33
11 33
3
33 = 3  11
The prime factorization
of 33 is 3  11.
8-1 Factors and Greatest Common Factors
Factors that are shared by two or more whole
numbers are called common factors. The greatest
of these common factors is called the greatest
common factor, or GCF.
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 32: 1, 2, 4, 8, 16, 32
Common factors: 1, 2, 4
The greatest of the common factors is 4.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Example 2A: Finding the GCF of Numbers
Find the GCF of each pair of numbers.
100 and 60
Method 1 List the factors.
factors of 100: 1, 2, 4,
5, 10, 20, 25, 50, 100
List all the factors.
factors of 60: 1, 2, 3, 4, 5,
6, 10, 12, 15, 20, 30, 60
Circle the GCF.
The GCF of 100 and 60 is 20.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Example 2B: Finding the GCF of Numbers
Find the GCF of each pair of numbers.
26 and 52
Method 2 Prime factorization.
26 =
2  13
52 = 2  2  13
2  13 = 26
Write the prime
factorization of each
number.
Align the common
factors.
The GCF of 26 and 52 is 26.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
You can also find the GCF of monomials that
include variables. To find the GCF of monomials,
write the prime factorization of each coefficient
and write all powers of variables as products.
Then find the product of the common factors.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Example 3A: Finding the GCF of Monomials
Find the GCF of each pair of monomials.
15x3 and 9x2
15x3 = 3  5  x  x  x
9x2 = 3  3  x  x
3
Write the prime factorization of
each coefficient and write
powers as products.
Align the common factors.
x  x = 3x2 Find the product of the common
factors.
The GCF of 3x3 and 6x2 is 3x2.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Example 3B: Finding the GCF of Monomials
Find the GCF of each pair of monomials.
8x2 and 7y3
The GCF 8x2 and 7y is 1.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Helpful Hint
If two terms contain the same variable raised to
different powers, the GCF will contain that
variable raised to the lower power.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Check It Out! Example 3a
Find the GCF of each pair of monomials.
18g2 and 27g3
18g2 = 2  3  3 
27g3 =
gg
Write the prime factorization
of each coefficient and
write powers as products.
3  3  3  g  g  g Align the common factors.
33
gg
Find the product of the
common factors.
The GCF of 18g2 and 27g3 is 9g2.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Check It Out! Example 3b
Find the GCF of each pair of monomials.
Write the prime
factorization of
each coefficient
and write powers
as products.
16a6 and 9b
16a6 = 2  2  2  2  a  a  a  a  a  a
9b =
The GCF of 16a6 and 7b is 1.
Holt Algebra 1
33b
Align the common
factors.
There are no common factors
other than 1.
8-1 Factors and Greatest Common Factors
Check It Out! Example 4
Adrianne is shopping for a CD storage unit.
She has 36 CDs by pop music artists and 48
CDs by country music artists. She wants to put
the same number of CDs on each shelf without
putting pop music and country music CDs on
the same shelf. If Adrianne puts the greatest
possible number of CDs on each shelf, how
many shelves does her storage unit need?
The 36 pop and 48 country CDs must be divided into
groups of equal size. The number of CDs in each row
must be a common factor of 36 and 48.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
Check It Out! Example 4 Continued
Find the common
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 factors of 36
and 48.
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
The GCF of 36 and 48 is 12.
The greatest possible number of CDs on each shelf
is 12. Find the number of shelves of each type of
CDs when Adrianne puts the greatest number of
CDs on each shelf.
Holt Algebra 1
8-1 Factors and Greatest Common Factors
36 pop CDs
12 CDs per shelf
= 3 shelves
48 country CDs
12 CDs per shelf
= 4 shelves
When the greatest possible number of CD types
are on each shelf, there are 7 shelves in total.
Holt Algebra 1
8-2 Factoring by GCF
5 Minute Warm-Up
Write the prime factorization of each number.
1. 50
2. 84
Find the GCF of each pair monomial.
3. 18 and 75
4. 20 and 36
5. 12x and 28x3
6. 27x2 and 45x3y2
7. Cindi is planting a rectangular flower bed with 40
orange flower and 28 yellow flowers. She wants to
plant them so that each row will have the same
number of plants but of only one color. How many
rows will Cindi need if she puts the greatest possible
number of plants in each row?
Holt Algebra 1
8-2 Factoring by GCF
Objective
Factor polynomials by using the
greatest common factor.
Holt Algebra 1
8-2 Factoring by GCF
Recall that the Distributive Property states that
ab + ac =a(b + c). The Distributive Property
allows you to “factor” out the GCF of the terms in
a polynomial to write a factored form of the
polynomial.
A polynomial is in its factored form when it is
written as a product of monomials and polynomials
that cannot be factored further. The polynomial
2(3x – 4x) is not fully factored because the terms
in the parentheses have a common factor of x.
Holt Algebra 1
8-2 Factoring by GCF
Example 1A: Factoring by Using the GCF
Factor each polynomial. Check your answer.
2x2 – 4
2x2 = 2 
xx
4=22
Find the GCF.
2
2x2 – (2  2)
The GCF of 2x2 and 4 is 2.
Write terms as products using the
GCF as a factor.
Use the Distributive Property to factor
out the GCF.
Multiply to check your answer.
The product is the original
polynomial.
2(x2 – 2)
Check 2(x2 – 2)
2x2 – 4
Holt Algebra 1
8-2 Factoring by GCF
Writing Math
Aligning common factors can help you find the
greatest common factor of two or more terms.
Holt Algebra 1
8-2 Factoring by GCF
Example 1B: Factoring by Using the GCF
Factor each polynomial. Check your answer.
8x3 – 4x2 – 16x
8x3 = 2  2  2 
x  x  x Find the GCF.
4x2 = 2  2 
xx
16x = 2  2  2  2  x
The GCF of 8x3, 4x2, and 16x is
4x.
22
x = 4x Write terms as products using
the GCF as a factor.
2x2(4x) – x(4x) – 4(4x)
Use the Distributive Property to
4x(2x2 – x – 4)
factor out the GCF.
Check 4x(2x2 – x – 4)
Multiply to check your answer.
The product is the original
8x3 – 4x2 – 16x 
polynomials.
Holt Algebra 1
8-2 Factoring by GCF
Example 1C: Factoring by Using the GCF
Factor each polynomial. Check your answer.
–14x – 12x2
– 1(14x + 12x2)
14x = 2 
7x
12x2 = 2  2  3 
xx
2
–1[7(2x) + 6x(2x)]
–1[2x(7 + 6x)]
–2x(7 + 6x)
Holt Algebra 1
Both coefficients are
negative. Factor out –1.
Find the GCF.
2
The
GCF
of
14x
and
12x
x = 2x
is 2x.
Write each term as a product
using the GCF.
Use the Distributive Property
to factor out the GCF.
8-2 Factoring by GCF
Caution!
When you factor out –1 as the first step, be sure
to include it in all the other steps as well.
Holt Algebra 1
8-2 Factoring by GCF
Example 1D: Factoring by Using the GCF
Factor each polynomial. Check your answer.
3x3 + 2x2 – 10
3x3 = 3
2x2 =
10 =
 x  x  x Find the GCF.
2
xx
25
3x3 + 2x2 – 10
There are no common
factors other than 1.
The polynomial cannot be factored further.
Holt Algebra 1
8-2 Factoring by GCF
Check It Out! Example 1a
Factor each polynomial. Check your answer.
5b + 9b3
5b = 5 
b
9b = 3  3  b  b  b
b
5(b) + 9b2(b)
b(5 + 9b2)
Check b(5 + 9b2)
5b + 9b3 
Holt Algebra 1
Find the GCF.
The GCF of 5b and 9b3 is b.
Write terms as products using
the GCF as a factor.
Use the Distributive Property to
factor out the GCF.
Multiply to check your answer.
The product is the original
polynomial.
8-2 Factoring by GCF
Check It Out! Example 1b
Factor each polynomial. Check your answer.
9d2 – 82
9d2 = 3  3  d  d
82 =
9d2 – 82
Find the GCF.
222222
There are no common
factors other than 1.
The polynomial cannot be factored further.
Holt Algebra 1
8-2 Factoring by GCF
Check It Out! Example 1c
Factor each polynomial. Check your answer.
–18y3 – 7y2
– 1(18y3 + 7y2)
Both coefficients are negative.
Factor out –1.
18y3 = 2  3  3  y  y  y
Find the GCF.
7y2 = 7 
yy
y  y = y2 The GCF of 18y3 and 7y2 is y2.
–1[18y(y2) + 7(y2)]
–1[y2(18y + 7)]
–y2(18y + 7)
Holt Algebra 1
Write each term as a product
using the GCF.
Use the Distributive Property
to factor out the GCF..
8-2 Factoring by GCF
Check It Out! Example 1d
Factor each polynomial. Check your answer.
8x4 + 4x3 – 2x2
8x4 = 2  2  2  x  x  x  x
4x3 = 2  2  x  x  x
Find the GCF.
2x2 = 2 
xx
2
x  x = 2x2 The GCF of 8x4, 4x3 and –2x2 is 2x2.
4x2(2x2) + 2x(2x2) –1(2x2) Write terms as products using the
2x2(4x2 + 2x – 1)
Check 2x2(4x2 + 2x – 1)
8x4 + 4x3 – 2x2
Holt Algebra 1
GCF as a factor.
Use the Distributive Property to factor
out the GCF.
Multiply to check your answer.
The product is the original polynomial.
8-2 Factoring by GCF
To write expressions for the length and width of a
rectangle with area expressed by a polynomial,
you need to write the polynomial as a product.
You can write a polynomial as a product by
factoring it.
Holt Algebra 1
8-2 Factoring by GCF
Example 2: Application
The area of a court for the game squash is
9x2 + 6x m2. Factor this polynomial to find
possible expressions for the dimensions of
the squash court.
A = 9x2 + 6x
= 3x(3x) + 2(3x)
= 3x(3x + 2)
The GCF of 9x2 and 6x is 3x.
Write each term as a product
using the GCF as a factor.
Use the Distributive Property to
factor out the GCF.
Possible expressions for the dimensions of the
squash court are 3x m and (3x + 2) m.
Holt Algebra 1
8-2 Factoring by GCF
Sometimes the GCF of terms is a binomial. This
GCF is called a common binomial factor. You
factor out a common binomial factor the same
way you factor out a monomial factor.
Holt Algebra 1
8-2 Factoring by GCF
Example 3: Factoring Out a Common Binomial Factor
Factor each expression.
A. 5(x + 2) + 3x(x + 2)
5(x + 2) + 3x(x + 2)
(x + 2)(5 + 3x)
The terms have a common
binomial factor of (x + 2).
Factor out (x + 2).
B. –2b(b2 + 1)+ (b2 + 1)
–2b(b2 + 1) + (b2 + 1) The terms have a common
binomial factor of (b2 + 1).
–2b(b2 + 1) + 1(b2 + 1) (b2 + 1) = 1(b2 + 1)
(b2 + 1)(–2b + 1)
Holt Algebra 1
Factor out (b2 + 1).
8-2 Factoring by GCF
Check It Out! Example 3
Factor each expression.
a. 4s(s + 6) – 5(s + 6)
4s(s + 6) – 5(s + 6)
(4s – 5)(s + 6)
The terms have a common
binomial factor of (s + 6).
Factor out (s + 6).
b. 7x(2x + 3) + (2x + 3)
7x(2x + 3) + (2x + 3)
The terms have a common
binomial factor of (2x + 3).
7x(2x + 3) + 1(2x + 3) (2x + 1) = 1(2x + 1)
(2x + 3)(7x + 1)
Holt Algebra 1
Factor out (2x + 3).
8-2 Factoring by GCF
You may be able to factor a polynomial by
grouping. When a polynomial has four terms,
you can make two groups and factor out the
GCF from each group.
Holt Algebra 1
8-2 Factoring by GCF
Example 4A: Factoring by Grouping
Factor each polynomial by grouping.
Check your answer.
6h4 – 4h3 + 12h – 8
(6h4 – 4h3) + (12h – 8) Group terms that have a common
number or variable as a factor.
2h3(3h – 2) + 4(3h – 2) Factor out the GCF of each
group.
2h3(3h – 2) + 4(3h – 2) (3h – 2) is another common
factor.
(3h – 2)(2h3 + 4)
Holt Algebra 1
Factor out (3h – 2).
8-2 Factoring by GCF
Example 4A Continued
Factor each polynomial by grouping.
Check your answer.
Check (3h – 2)(2h3 + 4)
Multiply to check your
solution.
3h(2h3) + 3h(4) – 2(2h3) – 2(4)
6h4 + 12h – 4h3 – 8
6h4 – 4h3 + 12h – 8
Holt Algebra 1
The product is the original
polynomial.
8-2 Factoring by GCF
Example 4B: Factoring by Grouping
Factor each polynomial by grouping.
Check your answer.
5y4 – 15y3 + y2 – 3y
(5y4 – 15y3) + (y2 – 3y)
Group terms.
5y3(y – 3) + y(y – 3)
Factor out the GCF of
each group.
5y3(y – 3) + y(y – 3)
(y – 3) is a common factor.
(y – 3)(5y3 + y)
Factor out (y – 3).
Holt Algebra 1
8-2 Factoring by GCF
Check It Out! Example 4a
Factor each polynomial by grouping.
Check your answer.
6b3 + 8b2 + 9b + 12
(6b3 + 8b2) + (9b + 12)
Group terms.
2b2(3b + 4) + 3(3b + 4)
Factor out the GCF of
each group.
(3b + 4) is a common
factor.
2b2(3b + 4) + 3(3b + 4)
(3b + 4)(2b2 + 3)
Holt Algebra 1
Factor out (3b + 4).
8-2 Factoring by GCF
Check It Out! Example 4b
Factor each polynomial by grouping.
Check your answer.
4r3 + 24r + r2 + 6
(4r3 + 24r) + (r2 + 6)
Group terms.
4r(r2 + 6) + 1(r2 + 6)
Factor out the GCF of
each group.
(r2 + 6) is a common
factor.
4r(r2 + 6) + 1(r2 + 6)
(r2 + 6)(4r + 1)
Holt Algebra 1
Factor out (r2 + 6).
8-2 Factoring by GCF
Helpful Hint
If two quantities are opposites, their sum is 0.
(5 – x) + (x – 5)
5–x+x–5
–x+x+5–5
0+0
0
Holt Algebra 1
8-2 Factoring by GCF
Recognizing opposite binomials can help you factor
polynomials. The binomials (5 – x) and (x – 5) are
opposites. Notice (5 – x) can be written as –1(x – 5).
–1(x – 5) = (–1)(x) + (–1)(–5)
= –x + 5
Simplify.
=5–x
Commutative Property
of Addition.
So, (5 – x) = –1(x – 5)
Holt Algebra 1
Distributive Property.
8-2 Factoring by GCF
Example 5: Factoring with Opposites
Factor 2x3 – 12x2 + 18 – 3x
2x3 – 12x2 + 18 – 3x
(2x3 – 12x2) + (18 – 3x)
2x2(x – 6) + 3(6 – x)
2x2(x – 6) + 3(–1)(x – 6)
2x2(x – 6) – 3(x – 6)
(x – 6)(2x2 – 3)
Holt Algebra 1
Group terms.
Factor out the GCF of
each group.
Write (6 – x) as –1(x – 6).
Simplify. (x – 6) is a
common factor.
Factor out (x – 6).
8-2 Factoring by GCF
Check It Out! Example 5a
Factor each polynomial. Check your answer.
15x2 – 10x3 + 8x – 12
(15x2 – 10x3) + (8x – 12)
5x2(3 – 2x) + 4(2x – 3)
Group terms.
Factor out the GCF of
each group.
5x2(3 – 2x) + 4(–1)(3 – 2x) Write (2x – 3) as –1(3 – 2x).
5x2(3 – 2x) – 4(3 – 2x)
(3 – 2x)(5x2 – 4)
Holt Algebra 1
Simplify. (3 – 2x) is a
common factor.
Factor out (3 – 2x).
8-3 Factoring x2 + bx + c
5 Minute Warm-Up
Factor each polynomial.
1. 16x + 20x3
2. 4m4 – 12m2 + 8m
3. 7k(k – 3) + 4(k – 3)
4. 3y(2y + 3) – 5(2y + 3)
5. 2x3 + x2 – 6x – 3
6. 7p4 – 2p3 + 63p – 18
Holt Algebra 1
8-3 Factoring x2 + bx + c
Objective
Factor quadratic trinomials of the form
x2 + bx + c.
Holt Algebra 1
8-3 Factoring x2 + bx + c
In Chapter 7, you learned how to multiply two
binomials using the Distributive Property or the
FOIL method. In this lesson, you will learn how to
factor a trinomial into two binominals.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Notice that when you multiply (x + 2)(x + 5), the
constant term in the trinomial is the product of
the constants in the binomials.
(x + 2)(x + 5) = x2 + 7x + 10
You can use this fact to factor a trinomial into its
binomial factors. Look for two numbers that are
factors of the constant term in the trinomial.
Write two binomials with those numbers, and
then multiply to see if you are correct.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 1A: Factoring Trinomials by Guess and
Check
Factor x2 + 15x + 36 by guess and check.
(
+
)(
+
) Write two sets of parentheses.
(x + )(x + ) The first term is x2, so the variable
terms have a coefficient of 1.
The constant term in the trinomial is 36.
(x + 1)(x + 36) = x2 + 37x + 36  Try factors of 36 for
the constant
2
(x + 2)(x + 18) = x + 20x + 36 
terms in the
binomials.
(x + 3)(x + 12) = x2 + 15x + 36
The factors of x2 + 15x + 36 are (x + 3)(x + 12).
x2 + 15x + 36 = (x + 3)(x + 12)
Holt Algebra 1
8-3 Factoring x2 + bx + c
Remember!
When you multiply two binomials, multiply:
First terms
Outer terms
Inner terms
Last terms
Holt Algebra 1
8-3 Factoring x2 + bx + c
Check It Out! Example 1a
Factor each trinomial by guess and check.
x2 + 10x + 24
(
+
)(
+
)
Write two sets of parentheses.
(x + )(x + ) The first term is x2, so the variable
terms have a coefficient of 1.
The constant term in the trinomial is 24.
(x + 1)(x + 24) = x2 + 25x + 24 Try factors of 24 for
(x + 2)(x + 12) = x2 + 14x + 24  the constant
terms in the
2

(x + 3)(x + 8) = x + 11x + 24
binomials.
(x + 4)(x + 6) = x2 + 10x + 24 
The factors of x2 + 10x + 24 are (x + 4)(x + 6).
x2 + 10x + 24 = (x + 4)(x + 6)
Holt Algebra 1
8-3 Factoring x2 + bx + c
The guess and check method is usually not the
most efficient method of factoring a trinomial. Look
at the product of (x + 3) and (x + 4).
x2
12
(x + 3)(x +4) = x2 + 7x + 12
3x
4x
The coefficient of the middle term is the sum of 3
and 4. The third term is the product of 3 and 4.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Holt Algebra 1
8-3 Factoring x2 + bx + c
When c is positive, its factors have the same sign.
The sign of b tells you whether the factors are
positive or negative. When b is positive, the
factors are positive and when b is negative, the
factors are negative.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 2A: Factoring x2 + bx + c When c is Positive
Factor each trinomial. Check your answer.
x2 + 6x + 5
(x +
)(x +
)
b = 6 and c = 5; look for factors of 5
whose sum is 6.
Factors of 5 Sum
1 and 5
6
The factors needed are 1 and 5.
(x + 1)(x + 5)
Check (x + 1)(x + 5) = x2 + x + 5x + 5
= x2 + 6x + 5
Holt Algebra 1
Use the FOIL method.
The product is the
original polynomial.
8-3 Factoring x2 + bx + c
Example 2C: Factoring x2 + bx + c When c is Positive
Factor each trinomial. Check your answer.
x2 – 8x + 15
(x +
)(x +
)
b = –8 and c = 15; look for factors of
15 whose sum is –8.
Factors of –15 Sum
–1 and –15 –16 
–3 and –5 –8  The factors needed are –3 and –5 .
(x – 3)(x – 5)
Check (x – 3)(x – 5 ) = x2 – 3x – 5x + 15 Use the FOIL method.
= x2 – 8x + 15 
The product is the
original polynomial.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Check It Out! Example 2b
Factor each trinomial. Check your answer.
x2 – 5x + 6
(x +
)(x+
)
b = –5 and c = 6; look for factors of
6 whose sum is –5.
Factors of 6 Sum
–1 and –6 –7
–2 and –3 –5  The factors needed are –2 and –3.
(x – 2)(x – 3)
Check (x – 2)(x – 3) = x2 –2x – 3x + 6
= x2 – 5x + 6 
Holt Algebra 1
Use the FOIL method.
The product is the
original polynomial.
8-3 Factoring x2 + bx + c
Check It Out! Example 2d
Factor each trinomial. Check your answer.
x2 – 13x + 40
(x +
)(x+
Factors of 40
–2 and –20
–4 and –10
–5 and –8
)
b = –13 and c = 40; look for factors
of 40 whose sum is –13.
Sum
–22 The factors needed are –5 and –8.
–14
–13
(x – 5)(x – 8)
Check (x – 5)(x – 8) = x2 – 5x – 8x + 40 Use the FOIL method.
= x2 – 13x + 40
The product is the
original polynomial.
Holt Algebra 1
8-3 Factoring x2 + bx + c
When c is negative, its factors have opposite
signs. The sign of b tells you which factor is
positive and which is negative. The factor
with the greater absolute value has the same
sign as b.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 3A: Factoring x2 + bx + c When c is Negative
Factor each trinomial.
x2 + x – 20
(x +
)(x +
)
Factors of –20 Sum
–1 and 20
19 
–2 and 10
8
–4 and 5
1
(x – 4)(x + 5)
Holt Algebra 1
b = 1 and c = –20; look for
factors of –20 whose sum is
1. The factor with the greater
absolute value is positive.
The factors needed are +5 and
–4.
8-3 Factoring x2 + bx + c
Example 3B: Factoring x2 + bx + c When c is Negative
Factor each trinomial.
x2 – 3x – 18
(x +
)(x +
Factors of –18
1 and –18
2 and – 9
3 and – 6
)
Sum
–17
– 7
– 3
(x – 6)(x + 3)
Holt Algebra 1
b = –3 and c = –18; look for
factors of –18 whose sum is
–3. The factor with the
greater absolute value is
negative.
The factors needed are 3 and
–6.
8-3 Factoring x2 + bx + c
A polynomial and the factored form of the
polynomial are equivalent expressions. When
you evaluate these two expressions for the
same value of the variable, the results are the
same.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Example 4A: Evaluating Polynomials
Factor y2 + 10y + 21. Show that the original
polynomial and the factored form have the
same value for y = 0, 1, 2, 3, and 4.
y2 + 10y + 21
(y + )(y + )
Factors of 21 Sum
1 and 21
7
3 and 7
10
(y + 3)(y + 7)
Holt Algebra 1
b = 10 and c = 21; look for factors
of 21 whose sum is 10.
The factors needed are 3 and 7.
8-3 Factoring x2 + bx + c
Example 4A Continued
Evaluate the original polynomial and the
factored form for y = 0, 1, 2, 3, and 4.
y2 + 10y + 21
y
(y + 7)(y + 3)
y
0
(0 + 7)(0 + 3) = 21
0
02 + 10(0) + 21 = 21
1
(1 + 7)(1 + 3) = 32
1
12 + 10(1) + 21 = 32
2
(2 + 7)(2 + 3) = 45
2
22 + 10(2) + 21 = 45
3
(3 + 7)(3 + 3) = 60
3
32 + 10(3) + 21 = 60
4
(4 + 7)(4 + 3) = 77
4
42 + 10(4) + 21 = 77
The original polynomial and the factored form
have the same value for the given values of n.
Holt Algebra 1
8-3 Factoring x2 + bx + c
Lesson Quiz: Part I
Factor each trinomial.
1. x2 – 11x + 30 (x – 5)(x – 6)
2. x2 + 10x + 9
(x + 1)(x + 9)
3. x2 – 6x – 27
(x – 9)(x + 3)
4. x2 + 14x – 32
(x + 16)(x – 2)
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Warm Up
Find each product.
1. (x – 2)(2x + 7)
2x2 + 3x – 14
2. (3y + 4)(2y + 9) 6y2 + 35y + 36
3. (3n – 5)(n – 7) 3n2 – 26n + 35
Factor each trinomial.
4. x2 +4x – 32 (x – 4)(x + 8)
5. z2 + 15z + 36 (z + 3)(z + 12)
6. h2 – 17h + 72 (h – 8)(h – 9)
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Objective
Factor quadratic trinomials of the form
ax2 + bx + c.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
In the previous lesson you factored
trinomials of the form x2 + bx + c.
Now you will factor trinomials of the
form ax2 + bx + c, where a ≠ 0.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
When you multiply (3x + 2)(2x + 5), the coefficient
of the x2-term is the product of the coefficients of
the x-terms. Also, the constant term in the trinomial
is the product of the constants in the binomials.
(3x + 2)(2x + 5) = 6x2 + 19x + 10
Holt Algebra 1
8-4 Factoring ax2 + bx + c
To factor a trinomial like ax2 + bx + c into its
binomial factors, write two sets of parentheses
( x + )( x + ).
Write two numbers that are factors of a next to the
x’s and two numbers that are factors of c in the
other blanks. Multiply the binomials to see if you are
correct.
(3x + 2)(2x + 5) = 6x2 + 19x + 10
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 1: Factoring ax2 + bx + c by Guess and Check
Factor 6x2 + 11x + 4 by guess and check.
(
+
)(
+
)
Write two sets of parentheses.
2
( x + )( x + ) The first term is 6x , so at least
one variable term has a
coefficient other than 1.
The coefficient of the x2 term is 6. The constant term in
the trinomial is 4.
(2x + 4)(3x + 1) = 6x2 + 14x + 4
Try factors of 6 for the
2

(1x + 4)(6x + 1) = 6x + 25x + 4
coefficients and
(1x + 2)(6x + 2) = 6x2 + 14x + 4
factors of 4 for the
(1x + 1)(6x + 4) = 6x2 + 10x + 4
constant terms.
(3x + 4)(2x + 1) = 6x2 + 11x + 4
Holt Algebra 1
8-4 Factoring ax2 + bx + c
So, to factor a2 + bx + c, check the factors of a and
the factors of c in the binomials. The sum of the
products of the outer and inner terms should be b.
Product = c
Product = a
(
X+
)(
x+
) = ax2 + bx + c
Sum of outer and inner products = b
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Since you need to check all the factors of a and the
factors of c, it may be helpful to make a table. Then
check the products of the outer and inner terms to
see if the sum is b. You can multiply the binomials
to check your answer.
Product = c
Product = a
(
X+
)(
x+
) = ax2 + bx + c
Sum of outer and inner products = b
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 2A: Factoring ax2 + bx + c When c is Positive
Factor each trinomial. Check your answer.
2x2 + 17x + 21
(
x+
)(
x+
a = 2 and c = 21,
) Outer + Inner = 17.
Factors of 2 Factors of 21
1 and 21
1 and 2
21 and 1
1 and 2
3 and 7
1 and 2
7 and 3
1 and 2
Outer + Inner
1(21) + 2(1) = 23
1(1) + 2(21) = 43
1(7) + 2(3) = 13
1(3) + 2(7) = 17 
Use the Foil method.
(x + 7)(2x + 3)
Check (x + 7)(2x + 3) = 2x2 + 3x + 14x + 21
= 2x2 + 17x + 21 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Remember!
When b is negative and c is positive, the factors
of c are both negative.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 2B: Factoring ax2 + bx + c When c is Positive
Factor each trinomial. Check your answer.
3x2 – 16x + 16
(
x+
)(
x+
)
a = 3 and c = 16,
Outer + Inner = –16 .
Factors of 3 Factors of 16 Outer + Inner
1 and 3
–1 and –16 1(–16) + 3(–1) = –19
1 and 3
– 2 and – 8 1( – 8) + 3(–2) = –14
– 4 and – 4 1( – 4) + 3(– 4)= –16
1 and 3



(x – 4)(3x – 4)
Use the Foil method.
Check (x – 4)(3x – 4) = 3x2 – 4x – 12x + 16
= 3x2 – 16x + 16 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 2b
Factor each trinomial. Check your answer.
9x2 – 15x + 4
( x + )( x +
)
a = 9 and c = 4,
Outer + Inner = –15.
Factors of 9 Factors of 4 Outer + Inner
3 and 3
–1 and – 4 3(–4) + 3(–1) = –15
3 and 3
– 2 and – 2 3(–2) + 3(–2) = –12 
– 4 and – 1 3(–1) + 3(– 4)= –15
3 and 3

(3x – 4)(3x – 1)
Use the Foil method.
Check (3x – 4)(3x – 1) = 9x2 – 3x – 12x + 4
= 9x2 – 15x + 4
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Check It Out! Example 2c
Factor each trinomial. Check your answer.
3x2 + 13x + 12
(
x+
)(
x+
)
Factors of 3 Factors of 12
1 and 3
1 and 12
2 and 6
1 and 3
3 and 4
1 and 3
(x + 3)(3x + 4)
a = 3 and c = 12,
Outer + Inner = 13.
Outer + Inner
1(12) + 3(1) = 15
1(6) + 3(2) = 12 
1(4) + 3(3) = 13 
Use the Foil method.
Check (x + 3)(3x + 4) = 3x2 + 4x + 9x + 12
= 3x2 + 13x + 12 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
When c is negative, one factor of c will be
positive and the other factor will be negative.
Only some of the factors are shown in the
examples, but you may need to check all of the
possibilities.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 3A: Factoring ax2 + bx + c When c is Negative
Factor each trinomial. Check your answer.
3n2 + 11n – 4
(
y+
)(
y+
)
a = 3 and c = – 4,
Outer + Inner = 11 .
Factors of 3 Factors of 4 Outer + Inner
1 and 3
–1 and 4 1(4) + 3(–1) = 1
1(2) + 3(–2) = – 4 
1 and 3
–2 and 2
–4 and 1
1(1) + 3(–4) = –11 
1 and 3
4 and –1
1(–1) + 3(4) = 11
1 and 3

(n + 4)(3n – 1)
Use the Foil method.
Check (n + 4)(3n – 1) = 3n2 – n + 12n – 4
= 3n2 + 11n – 4
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 3B: Factoring ax2 + bx + c When c is Negative
Factor each trinomial. Check your answer.
2x2 + 9x – 18
(
x+
)(
x+
)
a = 2 and c = –18,
Outer + Inner = 9 .
Factors of 2 Factors of – 18 Outer + Inner
1(– 1) + 2(18) = 35
1 and 2
18 and –1
1(– 2) + 2(9) = 16 
1 and 2
9 and –2
6 and –3
1(– 3) + 2(6) = 9
1 and 2

(x + 6)(2x – 3)
Use the Foil method.
Check (x + 6)(2x – 3) = 2x2 – 3x + 12x – 18
= 2x2 + 9x – 18 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 3C: Factoring ax2 + bx + c When c is Negative
Factor each trinomial. Check your answer.
4x2 – 15x – 4
(
x+
)(
x+
Factors of 4 Factors of – 4
1 and 4
–1 and 4
1 and 4
–2 and 2
–4 and 1
1 and 4
(x – 4)(4x + 1)
)
a = 4 and c = –4,
Outer + Inner = –15.
Outer + Inner
1(4) – 1(4) = 0 
1(2) – 2(4) = –6 
1(1) – 4(4) = –15 
Use the Foil method.
Check (x – 4)(4x + 1) = 4x2 + x – 16x – 4
= 4x2 – 15x – 4 
Holt Algebra 1
8-4 Factoring ax2 + bx + c
When the leading coefficient is negative,
factor out –1 from each term before using
other factoring methods.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Caution
When you factor out –1 in an early step, you
must carry it through the rest of the steps.
Holt Algebra 1
8-4 Factoring ax2 + bx + c
Example 4A: Factoring ax2 + bx + c When a is
Negative
Factor –2x2 – 5x – 3.
–1(2x2 + 5x + 3)
–1(
x+
)(
x+
)
Factors of 2 Factors of 3
Outer + Inner
1 and 2
3 and 1
1(1) + 3(2) = 7 
1 and 2
1 and 3
1(3) + 1(2) = 5
(x + 1)(2x + 3)
–1(x + 1)(2x + 3)
Holt Algebra 1
Factor out –1.
a = 2 and c = 3;
Outer + Inner = 5
8-4 Factoring ax2 + bx + c
Check It Out! Example 4a
Factor each trinomial. Check your answer.
–6x2 – 17x – 12
–1(6x2 + 17x + 12)
–1(
x+
)(
x+
)
Factors of 6 Factors of 12
Outer + Inner

17 
2 and 3
4 and 3
2(3) + 3(4) = 18
2 and 3
3 and 4
2(4) + 3(3) =
(2x + 3)(3x + 4)
–1(2x + 3)(3x + 4)
Holt Algebra 1
Factor out –1.
a = 6 and c = 12;
Outer + Inner = 17
8-4 Factoring ax2 + bx + c
Check It Out! Example 4b
Factor each trinomial. Check your answer.
–3x2 – 17x – 10
–1(3x2 + 17x + 10)
–1(
x+
)(
x+
)
Factors of 3 Factors of 10
Outer + Inner

17 
1 and 3
2 and 5
1(5) + 3(2) = 11
1 and 3
5 and 2
1(2) + 3(5) =
(3x + 2)(x + 5)
–1(3x + 2)(x + 5)
Holt Algebra 1
Factor out –1.
a = 3 and c = 10;
Outer + Inner = 17)
8-5 Factoring Special Products
5 Minute Warm-Up
Factor each trinomial.
1. 5x2 + 17x + 6 (5x + 2)(x + 3)
2. 2x2 + 5x – 12 (2x– 3)(x + 4)
3. 6x2 – 23x + 7 (3x – 1)(2x – 7)
(–x + 4)(4x +
4. –4x2 + 11x + 20
5)
5. –2x2 + 7x – 3 (–2x + 1)(x – 3)
6. 8x2 + 27x + 9 (8x + 3)(x + 3)
Holt Algebra 1
8-5 Factoring Special Products
5 Minute Warm-Up
Factor each expression.
1. 40p2 – 10p + 30
2. 5g5 – 10g3 - 15g
3. 2x(x-4) + 9(x-4)
4. 10m3 + 15m2 – 2m - 3
5. x2 + 14x - 120
6. 6x2 – 19x + 15
Holt Algebra 1
8-5 Factoring Special Products
Objectives
Factor perfect-square trinomials.
Factor the difference of two squares.
Holt Algebra 1
8-5 Factoring Special Products
A trinomial is a perfect square if:
• The first and last terms are perfect squares.
• The middle term is two times one factor
from the first term and one factor from
the last term.
9x2
3x
Holt Algebra 1
•
+
12x
+
4
3x 2(3x • 2) 2 • 2
8-5 Factoring Special Products
Holt Algebra 1
8-5 Factoring Special Products
Example 1A: Recognizing and Factoring PerfectSquare Trinomials
Determine whether each trinomial is a perfect
square. If so, factor. If not explain.
9x2 – 15x + 64
9x2 – 15x + 64
3x

3x
2(3x

8) 8  8
2(3x  8) ≠ –15x.
9x2 – 15x + 64 is not a perfect-square trinomial
because –15x ≠ 2(3x  8).
Holt Algebra 1
8-5 Factoring Special Products
Example 1B: Recognizing and Factoring PerfectSquare Trinomials
Determine whether each trinomial is a perfect
square. If so, factor. If not explain.
81x2 + 90x + 25
81x2 + 90x + 25
9x
Holt Algebra 1
●9x
2(9x
● 5)
5 ●5
The trinomial is a perfect
square. Factor.
8-5 Factoring Special Products
Example 1B Continued
Determine whether each trinomial is a perfect
square. If so, factor. If not explain.
Method 2 Use the rule.
81x2 + 90x + 25
a = 9x, b = 5
(9x)2 + 2(9x)(5) + 52
Write the trinomial
as a2 + 2ab + b2.
(9x + 5)2
Write the trinomial
as (a + b)2.
Holt Algebra 1
8-5 Factoring Special Products
Example 1C: Recognizing and Factoring PerfectSquare Trinomials
Determine whether each trinomial is a perfect
square. If so, factor. If not explain.
36x2 – 10x + 14
36x2 – 10x + 14
The trinomial is not a
perfect-square
because 14 is not a
perfect square.
36x2 – 10x + 14 is not a perfect-square trinomial.
Holt Algebra 1
8-5 Factoring Special Products
Example 2: Problem-Solving Application
A rectangular piece of cloth must be cut to
make a tablecloth. The area needed is
(16x2 – 24x + 9) in2. The dimensions of
the cloth are of the form cx – d, where c
and d are whole numbers. Find an
expression for the perimeter of the cloth.
Find the perimeter when x = 11 inches.
Holt Algebra 1
8-5 Factoring Special Products
Example 2 Continued
2
Make a Plan
The formula for the area of a square is
area = (side)2.
Factor 16x2 – 24x + 9 to find the side
length of the tablecloth. Write a formula for
the perimeter of the park, and evaluate the
expression for x = 11.
Holt Algebra 1
8-5 Factoring Special Products
Example 2 Continued
3
Solve
16x2 – 24x + 9
(4x)2
– 2(4x)(3) +
(4x – 3)2
a = 4x, b = 3
32
Write the trinomial as
a2 – 2ab + b2.
Write the trinomial as (a + b)2.
16x2 – 24x + 9 = (4x – 3)(4x – 3)
The side length of the tablecloth is (4x – 3) in.
and (4x – 3) in.
Holt Algebra 1
8-5 Factoring Special Products
Example 2 Continued
Write a formula for the perimeter of the
tablecloth.
= 4(4x – 3)
Write the formula for the
perimeter of a square.
Substitute the side length for s.
= 16x – 12
Distribute 4.
P = 4s
An expression for the perimeter of the
tablecloth in inches is 16x – 12.
Holt Algebra 1
8-5 Factoring Special Products
Example 2 Continued
Evaluate the expression when x = 11.
P = 16x – 12
= 16(11) – 12
Substitute 11 for x.
= 164
When x = 11 in. the perimeter of the
tablecloth is 164 in.
Holt Algebra 1
8-5 Factoring Special Products
In Chapter 7 you learned that the difference of two
squares has the form a2 – b2. The difference of two
squares can be written as the product (a + b)(a – b).
You can use this pattern to factor some polynomials.
A polynomial is a difference of two squares if:
•There are two terms, one subtracted from the
other.
• Both terms are perfect squares.
4x2 – 9
2x
Holt Algebra 1

2x
3

3
8-5 Factoring Special Products
Holt Algebra 1
8-5 Factoring Special Products
Reading Math
Recognize a difference of two squares: the
coefficients of variable terms are perfect squares,
powers on variable terms are even, and
constants are perfect squares.
Holt Algebra 1
8-5 Factoring Special Products
Example 3A: Recognizing and Factoring the
Difference of Two Squares
Determine whether each binomial is a difference
of two squares. If so, factor. If not, explain.
3p2 – 9q4
3p2 – 9q4
3q2  3q2
3p2 is not a perfect square.
3p2 – 9q4 is not the difference of two squares
because 3p2 is not a perfect square.
Holt Algebra 1
8-5 Factoring Special Products
Example 3B: Recognizing and Factoring the
Difference of Two Squares
Determine whether each binomial is a difference
of two squares. If so, factor. If not, explain.
100x2 – 4y2
100x2 – 4y2
10x  10x
2y

2y
(10x)2 – (2y)2
(10x + 2y)(10x – 2y)
The polynomial is a difference
of two squares.
a = 10x, b = 2y
Write the polynomial as
(a + b)(a – b).
100x2 – 4y2 = (10x + 2y)(10x – 2y)
Holt Algebra 1
8-5 Factoring Special Products
Example 3C: Recognizing and Factoring the
Difference of Two Squares
Determine whether each binomial is a difference
of two squares. If so, factor. If not, explain.
x4 – 25y6
x4 – 25y6
x2

x2
5y3  5y3
(x2)2 – (5y3)2
(x2 + 5y3)(x2 – 5y3)
The polynomial is a difference
of two squares.
a = x2, b = 5y3
Write the polynomial as
(a + b)(a – b).
x4 – 25y6 = (x2 + 5y3)(x2 – 5y3)
Holt Algebra 1
8-5 Factoring Special Products
Check It Out! Example 3a
Determine whether each binomial is a difference
of two squares. If so, factor. If not, explain.
1 – 4x2
1 – 4x2
1

1
2x

2x
(1)2 – (2x)2
(1 + 2x)(1 – 2x)
1 – 4x2 = (1 + 2x)(1 – 2x)
Holt Algebra 1
The polynomial is a difference
of two squares.
a = 1, b = 2x
Write the polynomial as
(a + b)(a – b).
8-5 Factoring Special Products
Check It Out! Example 3b
Determine whether each binomial is a difference
of two squares. If so, factor. If not, explain.
p8 – 49q6
p8 – 49q6
p4 ●
p4
7q3 ● 7q3
(p4)2 – (7q3)2
(p4 + 7q3)(p4 – 7q3)
The polynomial is a difference
of two squares.
a = p4, b = 7q3
Write the polynomial as
(a + b)(a – b).
p8 – 49q6 = (p4 + 7q3)(p4 – 7q3)
Holt Algebra 1
8-5 Factoring Special Products
Check It Out! Example 3c
Determine whether each binomial is a difference
of two squares. If so, factor. If not, explain.
16x2 – 4y5
16x2 – 4y5
4x  4x
4y5 is not a perfect square.
16x2 – 4y5 is not the difference of two squares
because 4y5 is not a perfect square.
Holt Algebra 1
8-6 Choosing a Factoring Method
5 Minute Warm-up
Determine whether each trinomial is a perfect
square. If so factor. If not, explain.
1. 64x2 – 40x + 25 Not a perfect-square trinomial
because –40x ≠ 2(8x  5).
2. 121x2 – 44x + 4
(11x – 2)2
3. 49x2 + 140x + 100 (7x2 + 10)2
4. A square fence will be built around a garden with
an area of (49x2 + 56x + 16) ft2. The dimensions
of the garden are cx + d, where c and d are
whole numbers. Find an expression for the
perimeter when x = 5. P = 28x + 16; 156 ft
Holt Algebra 1
8-6 Choosing a Factoring Method
Objectives
Choose an appropriate method for
factoring a polynomial.
Combine methods for factoring a
polynomial.
Holt Algebra 1
8-6 Choosing a Factoring Method
Recall that a polynomial is in its fully factored
form when it is written as a product that cannot
be factored further.
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 1: Determining Whether a Polynomial is
Completely Factored
Tell whether each polynomial is completely
factored. If not factor it.
A. 3x2(6x – 4)
3x2(6x – 4)
6x – 4 can be further factored.
6x2(3x – 2)
Factor out 2, the GCF of 6x and – 4.
6x2(3x – 2) is completely factored.
B. (x2 + 1)(x – 5)
(x2 + 1)(x – 5)
Neither x2 +1 nor x – 5 can be
factored further.
(x2 + 1)(x – 5) is completely factored.
Holt Algebra 1
8-6 Choosing a Factoring Method
Caution
x2 + 4 is a sum of squares, and cannot be
factored.
Holt Algebra 1
8-6 Choosing a Factoring Method
Check It Out! Example 1
Tell whether the polynomial is completely
factored. If not, factor it.
A. 5x2(x – 1)
5x2(x – 1)
Neither 5x2 nor x – 1 can be
factored further.
5x2(x – 1) is completely factored.
B. (4x + 4)(x + 1)
(4x + 4)(x + 1)
4x + 4 can be further factored.
4(x + 1)(x + 1)
Factor out 4, the GCF of 4x and 4.
4(x + 1)2 is completely factored.
Holt Algebra 1
8-6 Choosing a Factoring Method
To factor a polynomial completely, you may need
to use more than one factoring method. Use the
steps below to factor a polynomial completely.
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 2A: Factoring by GCF and Recognizing
Patterns
Factor 10x2 + 48x + 32 completely. Check
your answer.
10x2 + 48x + 32
2(5x2 + 24x + 16)
Factor out the GCF.
2(5x + 4)(x + 4)
Factor remaining trinomial.
Check 2(5x + 4)(x + 4) = 2(5x2 + 20x + 4x + 16)
= 10x2 + 40x + 8x + 32
= 10x2 + 48x + 32
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 2B: Factoring by GCF and Recognizing
Patterns
Factor 8x6y2 – 18x2y2 completely. Check your
answer.
8x6y2 – 18x2y2
2x2y2(4x4 – 9)
Factor out the GCF. 4x4 – 9 is a
perfect-square trinomial of
the form a2 – b2.
2x2y2(2x2 – 3)(2x2 + 3) a = 2x, b = 3
Check 2x2y2(2x2 – 3)(2x2 + 3) = 2x2y2(4x4 – 9)
= 8x6y2 – 18x2y2 
Holt Algebra 1
8-6 Choosing a Factoring Method
Check It Out! Example 2a
Factor each polynomial completely. Check
your answer.
4x3 + 16x2 + 16x
4x3 + 16x2 + 16x
4x(x2 + 4x + 4)
4x(x + 2)2
Factor out the GCF. x2 + 4x + 4 is a
perfect-square trinomial of the
form a2 + 2ab + b2.
a = x, b = 2
Check 4x(x + 2)2 = 4x(x2 + 2x + 2x + 4)
= 4x(x2 + 4x + 4)
= 4x3 + 16x2 + 16x 
Holt Algebra 1
8-6 Choosing a Factoring Method
Check It Out! Example 2b
Factor each polynomial completely. Check
your answer.
2x2y – 2y3
2y(x2 – y2)
Factor out the GCF. 2y(x2 – y2) is a
perfect-square trinomial of the
form a2 – b2.
2y(x + y)(x – y)
a = x, b = y
2x2y – 2y3
Check 2y(x + y)(x – y) = 2y(x2 + xy – xy – y2)
= 2x2y +2xy2 – 2xy2 – 2y3
= 2x2y – 2y3
Holt Algebra 1
8-6 Choosing a Factoring Method
If none of the factoring methods work, the polynomial
is said to be unfactorable.
Helpful Hint
For a polynomial of the form ax2 + bx + c, if
there are no numbers whose sum is b and whose
product is ac, then the polynomial is
unfactorable.
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 3A: Factoring by Multiple Methods
Factor each polynomial completely.
9x2 + 3x – 2
9x2 + 3x – 2
( x + )( x +
)
Factors of 9 Factors of 2
1 and –2
1 and 9
1 and –2
3 and 3
–1 and 2
3 and 3
(3x – 1)(3x + 2)
Holt Algebra 1
The GCF is 1 and there is no
pattern.
a = 9 and c = –2;
Outer + Inner = 3
Outer + Inner
1(–2) + 1(9) = 7
3(–2) + 1(3) = –3
3(2) + 3(–1) = 3 

8-6 Choosing a Factoring Method
Example 3B: Factoring by Multiple Methods
Factor each polynomial completely.
12b3 + 48b2 + 48b The GCF is 12b; (b2 + 4b + 4)
is a perfect-square
12b(b2 + 4b + 4)
trinomial in the form of
(x + )(x + )
a2 + 2ab + b2.
Factors of 4 Sum
1 and 4
5
2 and 2
4 a = 2 and c = 2
12b(b + 2)(b + 2)
12b(b + 2)2
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 3C: Factoring by Multiple Methods
Factor each polynomial completely.
4y2 + 12y – 72
4(y2 + 3y – 18)
(y +
)(y +
)
Factor out the GCF. There is no
pattern. b = 3 and c = –18;
look for factors of –18 whose
sum is 3.
Factors of –18 Sum
–1 and 18
17 
–2 and 9
7
–3 and 6
3  The factors needed are –3
and 6
4(y – 3)(y + 6)
Holt Algebra 1
8-6 Choosing a Factoring Method
Example 3D: Factoring by Multiple Methods.
Factor each polynomial completely.
(x4 – x2)
Holt Algebra 1
x2(x2 – 1)
Factor out the GCF.
x2(x + 1)(x – 1)
x2 – 1 is a difference of two
squares.
8-6 Choosing a Factoring Method
Check It Out! Example 3b
Factor each polynomial completely.
2p5 + 10p4 – 12p3
2p3(p2 + 5p – 6)
(p +
)(p +
)
Factor out the GCF. There is no
pattern. b = 5 and c = –6;
look for factors of –6 whose
sum is 5.
Factors of – 6 Sum
– 1 and 6
5 The factors needed are –1
and 6
2p3(p + 6)(p – 1)
Holt Algebra 1
8-6 Choosing a Factoring Method
Check It Out! Example 3c
Factor each polynomial completely.
9q6 + 30q5 + 24q4
3q4(3q2
(
q+
Factor out the GCF. There is no
pattern.
+ 10q + 8)
)(
q+
)
Factors of 3 Factors of 8
1 and 8
3 and 1
2 and 4
3 and 1
4 and 2
3 and 1
3q4(3q + 4)(q + 2)
Holt Algebra 1
a = 3 and c = 8;
Outer + Inner = 10
Outer + Inner
3(8) + 1(1) = 25
3(4) + 1(2) = 14
3(2) + 1(4) = 10



8-6 Choosing a Factoring Method
Holt Algebra 1
8-6 Choosing a Factoring Method
Lesson Quiz
Tell whether the polynomial is completely
factored. If not, factor it.
1. (x + 3)(5x + 10)
2. 3x2(x2 + 9)
no; 5(x+ 3)(x + 2)
completely factored
Factor each polynomial completely. Check
your answer.
3. x3 + 4x2 + 3x + 12 4. 4x2 + 16x – 48
4(x + 6)(x – 2)
(x + 4)(x2 + 3)
5. 18x2 – 3x – 3
3(3x + 1)(2x – 1)
6. 18x2 – 50y2
2(3x + 5y)(3x – 5y)
7. 5x – 20x3 + 7 – 28x2 (1 + 2x)(1 – 2x)(5x + 7)
Holt Algebra 1