Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
258
Greatest Common Factors
Largest factor that divides into everything evenly
12.1 What does the fundamental theorem of arithmetic guarantee?
According to the fundamental theorem of arithmetic, all natural numbers
greater than 1 can be expressed as a unique product of prime numbers—no two
natural numbers have the same set of prime factors.
12.2 Generate the prime factorization of 21.
Only two natural numbers have a product of 21: 3 and 7. Therefore, the prime
factorization of 21 is 3 · 7.
Note: Problems 12.3–12.4 demonstrate that a number has the same prime factorization
regardless of the initial pair of products chosen.
12.3 Generate the prime factorization of 20, given 20 = 4 · 5.
To generate a prime factorization, begin by expressing the number to be
factored as a product of two natural numbers. This problem directs you to
begin with the product 4 · 5.
20 = 4 · 5
Your goal is to express 20 as a product of prime numbers. Whereas 5 is a prime
number, 4 is not, so you should express 4 as a product of natural numbers,
much like 20 was expressed as the product 4 · 5 one step earlier.
20 = (2 · 2) · 5
Because 2 and 5 are prime numbers, 2 · 2 · 5 is the prime factorization of 20.
Complete the problem by expressing 2 · 2 in exponential notation (2 · 2 = 2
2
).
20 = 2
2
· 5
Note: Problems 12.3–12.4 demonstrate that a number has the same prime factorization
regardless of the initial pair of natural numbers chosen.
12.4 Generate the prime factorization of 20, given 20 = 2 · 10.
Of the given factors, 2 is prime but 10 is not. Note that 10 is divisible by
5 (10 ÷ 5 = 2). Express 10 as a product of natural numbers.
Natural
numbers are
positive numbers
with no fractions or
decimals: 1, 2, 3, 4,
5, 6, and so on.
Prime numbers
aren’t divisible
by any numbers
except themselves
and 1. (And 1 divides
into everything, so
that’s no big deal.) 2,
3, 5, 7, and 11 are prime
numbers but 4 is not
(because it’s divisible by
2) and 100 isn’t prime
(because it’s divisible
by 2, 4, 5, 10, and a
bunch of other
numbers as
well).
In Problem
12.2, those numbers
happened to be prime,
but they don’t have to
be in the rst step. In
fact, they usually won’t
be. You’ll just factor
the ones that aren’t
until they’re all
prime.
4 is not
prime because
you can divide it
by 2: 4 ÷ 2 = 2. That
means 2
˙
2 = 4. Write
(2
˙
2) where 4 was in
the factorization
of 20.
Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
259
Factors 2 and 5 are prime, so the prime factorization of 20 is 2 · 2 · 5. As
directed in Problem 12.3, write the repeated factor using exponential notation.
20 = 2
2
· 5
This prime factorization exactly matches the prime factorization generated by
Problem 12.3. The initial natural numbers chosen to factor 20 (whether 4 · 5 or
2 · 10) had no effect on the final answer.
Note: Problems 12.5–12.6 demonstrate that a number has the same prime factorization
regardless of the initial pair of natural numbers chosen.
12.5 Generate the prime factorization of 48 given 6 · 8 = 48.
Neither of the initial factors (neither 6 nor 8) are prime, so both should be
expressed as the product of natural numbers. There is only one way to factor
each into a pair of natural numbers: 2 · 3 = 6 and 2 · 4 = 8.
Of the three factors (2, 3, and 4), only 4 is not prime. Express it as a product:
2 · 2 = 4.
Both of the remaining factors (2 and 3) are prime. Use exponential notation to
indicate that the factor 2 appears four times.
48 = 2
4
· 3
Note: Problems 12.5–12.6 demonstrate that a number has the same prime factorization
regardless of the initial pair of natural numbers chosen.
12.6 Generate the prime factorization of 48 given 4 · 12 = 48.
Neither 4 nor 12 is prime. Express each as a product of natural numbers. Note
that 12 can be expressed either as 2 · 6 or 3 · 4, but both will result in the same
prime factorization.
The number
2 appears twice in
the factorization, so
write it once but raise
it to the second
power.
Well, you could
write 6 as 3
˙
2,
but that’s not really
different. It’s just the
same numbers in a
different order.
Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
260
One composite factor remains. Express it as a product of natural numbers.
This prime factorization exactly matches the factorization generated by
Problem 12.5, despite the fact that the problems began with different natural
number factors.
Note: Problems 12.7–12.9 illustrate the steps used to calculate the greatest common factor of
64 and 112.
12.7 Generate the prime factorization of 64.
Because 64 is an even number, it is divisible by 2: 64 ÷ 2 = 32. Therefore,
64 = 2 · 32.
64 = 2 · 32
The number 32 is composite, so express it as a product.
64 = 2 · (4 · 8)
Neither 4 nor 8 is prime, so factor each: 4 = 2 · 2 and 8 = 2 · 4.
Only one composite factor remains: 4.
Note: Problems 12.7–12.9 illustrate the steps used to calculate the greatest common factor of
64 and 112.
12.8 Generate the prime factorization of 112.
Notice that 112 is an even number, so it is divisible by 2: 112 ÷ 2 = 56.
112 = 2 · 56
Factor the composite number 56.
112 = 2 · (4 · 14)
“Composite”
is the opposite
of prime. The
composite factor
in 2
˙
2
˙
3
˙
4
is 4.
Here, “factor”
means “write 56 as
two natural numbers
multiplied together.” You
could use 8
˙
7 instead
of the product used
by the book, 4
˙
14.
Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
261
Factor the newly introduced composite numbers, 4 and 14.
Note: Problems 12.7–12.9 illustrate the steps used to calculate the greatest common factor of
64 and 112.
12.9 Identify the greatest common factor of 64 and 112.
According to Problems 12.7 and 12.8, 64 = 2
6
and 112 = 2
4
· 7. To construct the
greatest common factor identify the common factors shared by 64 and 112. The
number 112 has two different prime factors, 2 and 7, but 64 has only a single
prime factor, 2. Thus the only factor 64 and 112 have in common is 2.
In the factorization of 64, 2 is raised to the sixth power, and in the factorization
of 112, 2 is raised to the fourth power. The greatest common factor is 2 raised
to the lower of those powers: 2
4
. Therefore, the greatest common factor of 64
and 112 is 2
4
= 16.
Note: Problems 12.10–12.11 illustrate the steps used to calculate the greatest common factor
of 36 and 90.
12.10 Generate the prime factorizations of 36 and 90.
Write each number as a product of natural numbers and factor all of the
resulting composite numbers.
The greatest common
factor has to divide into both
numbers evenly, so choosing the lower
power means that all of the 2s will cancel
out if you divide both numbers by 2
4
:
Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
262
Note: Problems 12.10–12.11 illustrate the steps used to calculate the greatest common factor
of 36 and 90.
12.11 Identify the greatest common factor of 36 and 90.
According to Problem 12.10, 36 and 90 have two factors in common, 2 and 3.
The prime factorization of 36 contains 2
2
and 3
2
; the prime factorization of 90
contains 2
1
and 3
2
. Choose the lower power of each to include in the greatest
common factor (GCF).
GCF of 36 and 90: 2
1
· 3
2
= 2 · 9 = 18
12.12 Identify the greatest common factor of 9x
2
and 15x
3
.
Generate the prime factorizations of coefficients 9 and 15. As you factor the
numbers, do not alter the variable expressions.
9x
2
and 15x
3
have two factors in common: 3 and x. The greatest common factor
(GCF) will include the lower power of both factors, 3
1
and x
2
.
GCF of 9x
2
and 15x
3
: 3
1
· x
2
= 3x
2
.
Note: Problems 12.13–12.14 present the steps used to factor the expression 18xy
4
+ 27x
3
y
2
.
12.13 Identify the greatest common factor of 18xy
4
and 27x
3
y
2
.
Generate the prime factorizations of coefficients 18 and 27.
18xy
4
and 27x
3
y
2
have three factors in common: 3, x, and y. Choose the lower
power of each from the factorizations to calculate the greatest common factor
(GCF).
GCF of 18xy
4
and 27x
3
y
2
: 3
2
· x
1
· y
2
= 9xy
2
Note: Problems 12.13–12.14 present the steps used to factor the expression 18xy
4
+ 27x
3
y
2
.
12.14 Factor the greatest common factor out of the expression and verify your
answer.
According to Problem 12.13, the greatest common factor of 18xy
4
and 27x
3
y
2
is
9xy
2
. Divide each term of the expression by 9xy
2
.
Both
factorizations
contain 3
2
, so
technically there is
no “lower power” of 3;
they’re both equal.
When this happens,
use the matching
power: 3
2
.
Instead of 3
2
Instead of x
3
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