SECTION 18.1: FACTORING GCF
FROM A POLYNOMIAL
A greatest common factor(GCF) is the "largest" factor common to a number of
terms in a polynomial. Here is the procedure for finding the GCF.
HOW TO FIND THE GCF OF A GIVEN POLYNOMIAL 
Step 1) Rewrite each term as "factors of primes."
Step 2) Determine the factors, BOTH NUMERICAL AND VARIABLE, that are common to all of the terms in a given
polynomial.
Comment: Often there will be more than one factor common to
all the terms.
Step 3) Multiply all the common factors together. This will produce the GCF.
Step 4) Multiply the GCF found in step 1) by the polynomial that will result in the original polynomial 
Example 1) Factor the GCF from
Step 1)
2x^{2}= (2)(x)(x)
4x = (2)(2)(x)
24 = (8)(3) = (2)(2)(2)(3)
Step 2 and 3)
The GCF is 2:
Why? 2 is the largest factor common to each term
Step 4)
2(x + 2x + 12)
Check:
Check is correct!
Example 2) Factor the GCF from
18x + 27x + 81
Step 1)
18x^{2}= (2)(3)(3)(x)(x)
27x = (3)(3)(3)(x)
81 = (3)(3)(3)(3)
Step 2 and 3) The GCF is 9:
Why? There are PAIRS of 3's
in common WITH EACH TERM and I multiplied the 3's together.
Step 4)
9(2x + 3x + 9)
Check: check is correct!
Example 3) Factor the GCF from
Step 1)
Step 2) The GCF is 9x:
Why? There are pairs of 3's in common WITH EACH
TERM as well as one x in common with each term. I multiplied the pair of 3's
and the x together
Step 3)
9x(2x + 3x + 9)
Check: check is correct!
KEYS TO UNLOCKING THE GCF 
Key 1) The numerical GCF is the LARGEST NUMBER that will divide WITHOUT REMAINDER the numerical parts of EVERY TERM.
Key 2) The variable GCF will be the variable exponential with the lowest exponent; and the variable exponential's base MUST BE COMMON TO ALL THE TERMS.
Key 3) The GCF will be the product of the numerical and variable GCF's. 
Example 1) Factor the GCF from
KEY 1) 30 divides 30, 60 and 90 of EACH TERM without
remainder!
KEY 2) x is the lowest variable exponential and x is COMMON to all
terms.
KEY 3) The GCF is
30x^{2}
answer:
Example 2) Factor the GCF from
KEY 1) 8 divides 16, 8 and 24 of EACH TERM without remainder!
KEY 2) x^{10} is the lowest variable exponential and x is COMMON to all
terms.
KEY 3) The GCF is
8x^{10}
answer:
Example 3) Factor the GCF from
KEY 1) 8 divides 16, 8 and 24 of EACH TERM without remainder!
KEY 2) x^{13 }is the lowest variable exponential, but x is NOT IN COMMON to
all terms. Therefore there is NO variable GCF!
KEY 3) The GCF is 8.
answer:
Example 4) Factor the GCF from
KEY 1) 8 divides 16 and 8 without remainder; but 8 WILL NOT
DIVIDE 1 without remainder! Therefore there is no numerical GCF.
KEY 2) x^{13 } is the lowest variable exponential but x is NOT IN COMMON to
all terms. Therefore there is NO variable GCF!
KEY 3) The is NO GCF!
answer:
SECTION 18.2: FACTORING
TRINOMIALS, IN THE FORM OF
ax^{2 }+
bx + c, where a, b, and c are
integers and a = +1, INTO TWO BINOMIALS
You learned how to multiply binomials like (x + 3)(x +2). The result of this
multiplication is
x^{2 }+ 5x + 6.
Now what if I asked you "what 2 binomials would you have to multiply together
to get the trinomial
x^{2}+ 5x + 6?" You are being asked to think in "reverse."
FACTORING A TRINOMIAL INTO TWO BINOMIALS
MEANS 
REWRITE THE TRINOMIAL AS TWO
BINOMIALS MULTIPLIED TOGETHER.

Let's go through an example of
factoring a trinomial, in the form of where a = +1 and b and c are integers, into two binomials.
I will show you a "recipe" for factoring!
Example 1) Factor
Comment: notice that the coefficient of
x is 1. For my trick to work
x must have 1 as a coefficient.
Step 1) Place a rectangle around the 9 and the operation in front of 9.
Think of the operation in front of the 9 as the "sign" of 9.
In this case +9.
Step 2) Place a rectangle around the 18 and the operation in front of 18.
Think of the operation in front of the 18 as the "sign" of 18. In this case
+18.
Step 3) Find a pair of factors of +18 that when multiplied will produce +18
but when added will produce +9.
+ 6 and + 3. Why? (6)(3) = 18 and 6 + 3 = 9
Step 4) Your answer will be (x + 6)(x + 3)
comment: the order of the +6 and + 3 is irrelevant.
CHECK: (x + 6)(x + 3) = x(x) + x(3) + 6(x) + 6(3)
= x + 3x + 6x + 18 =
x + 9x + 18
Example 2) Factor
x + 13x + 30
Step 1) Place a rectangle around the 13 and the operation in front of 13.
Think of the operation in front of the 13 as the "sign" of 13.
In this case 13 is +13.
Step 2) Place a rectangle around the 30 and the operation in front of 30.
Think of the operation in front of the 30 as the "sign" of 30. In this case 30
is +30.
Step 3) Find a pair of factors of +30 that when multiplied will produce +30
but when added will produce +13.
+ 10 and + 3. Why? (10)(3) = 30 and 10 + 3 = 13
Step 4) Your answer will be (x + 10)(x + 3)
CHECK: (x + 10)(x + 3) = x(x) + x(3) + 10(x) + 10(3)
= x + 3x + 10x + 30 =
x + 13x + 30
Example 3) Factor
x  3x  28
Step 1) Place a rectangle around the 3 and the operation in front of 3.
Think of the operation in front of the 3 as the "sign" of 3.
In this case 3 is 3.
Step 2) Place a rectangle around the 28 and the operation in front of 28.
Think of the operation in front of the 28 as the "sign" of 28. In this case 28
is 28.
Step 3) Find a pair of factors of 28 that when multiplied will produce 28
but when added will produce 3.
 7 and + 4. Why? (7)(4) = 28 and 7 + 4 = 3
Step 4) Your answer will be (x  7)(x + 4)
CHECK:
(x  7)(x + 4) = (x + 7)(x + 4) = x(x) + x(4) + 7(x)
+ 7(4) =
x + 4x + 7x + 28 =
x + 3x + 28 =
x  3x  28
SOME COMMENTS ON MULTIPLYING BINOMIALS 
WHEN MULTIPLYING BINOMIALS BY BINOMIALS
I CHANGED ALL SUBTRACTION TO ADDITION.
YOU MUST START LEARNING TO DO THIS IN YOUR HEAD! 
Example 4) Factor
I plan to skip some of the steps given the fact that I have done so many
examples.
 6 and  4 will be the proper numbers. and
Your answer will be .
CHECK:
=
=
=
SECTION 18.3: FACTORING
TRINOMIALS IN THE FORM OF,
ax^{2} +
bx + c, WHERE a, b, and c ARE INTEGERS AND a IS NOT EQUAL TO 1
Consider the following trinomial .
Notice the lead coefficient is NOT equal to 1. We MUST use a different
technique to factor this trinomial. I believe the technique that I am
introducing to you is the easiest way to factor.
Example) Factor
Step 1) Factor the coefficient of
x,
in this case 10, into all the factor pairs that will result in a product of
this coefficient: order is irrelevant. Do the same for the "numerical
part, i.e. the number that does not have a variable factor", in this
case 3, but also swap the order of the factors. DO NOT WORRY ABOUT THE SIGN OF 3.
Step 2) Pair up the group 1 factor pairs with the group 2 factor pairs.
(5)(2)


(3)(1) 
(5)(2) 

(1)(3) 
(10)(1) 

(3)(1) 
(10)(1) 

(1)(3) 
HINT: IF YOU WANT TO KNOW HOW MANY PAIRINGS YOU WILL HAVE MULTIPLY
THE NUMBER OF PAIRS IN THE FIRST GROUP, IN THIS CASE 2, BY THE
NUMBER OF PAIRS IN THE SECOND GROUP, ALSO TWO.
SO THE TOTAL NUMBER OF FACTOR PAIRS WILL BE 2 x 2 = 4.
Step 3) You will multiply the outer and inner factor pairs producing two
numbers called the "result pair." See the image below:
Step 4) Look at the OPERATION of the third term of your
trinomial you are factoring. If the third term is subtraction you must
find a result pair that subtracts  subtract smaller number
from the larger  to the coefficient of the second term. If the third term is addition you must find a result pair that adds to the
coefficient of the second term.
Since 10x^{2}  13 x  3 has a third term, the numerical part, with an operation of subtraction, THE TRICK
is that we must look for the result pair that will subtract to 13, the coefficient of of the second term.
YOU HAVE FOUND THE CORRECT FACTOR PAIR: (5)(2) and (1)(3).
Notice how I DO NOT change the order of the factor pairs as
they were written above  THIS IS IMPORTANT!
Step 5) The first factor pair, (5)(2), goes into the first position of each
binomial in the original order the factor pairs were written; the second
factor pair,(1)(3), goes into the second position of each binomial, in the
original order the factor pairs were written.
The result being: (5 1)(2 3)
Step 6) Place the x's in the first position of each binomial. Then determine
your operations in the binomials that will produce when the binomials are multiplied.
(5x 1)(2x 3)
and
(5x + 1)(2x  3) will produce 10x^{2}  13x  3
Therefore, factors into the two binomials
Check:
COMMENT: STUDENTS ALWAYS ASK ME HOW TO DETERMINE THE OPERATIONS THAT GO INTO THE
BINOMIALS. FOR THE TIME BEING I WILL MAKE YOU FIGURE OUT THE OPERATIONS ON
YOUR OWN. BUT THERE IS A "TRICK" TO DETERMINE THE OPERATIONS
AS WELL.
example 1) Factor
(6x 5)(2x 3 )
with operations
(6x  5)(2x  3 )
S.A.T. HINT: In general the factor pairs that have "large
distances" between them, e.g. (12)(1) and (15)(1), will probably not be one of
your answers. So in the above example you can initially "forget" about (12)(1) and (15)(1). BUT, I cannot guarantee
that these factor pairs are not going to be in your final answer.
example 2) Factor
group 1 factor pairs 

group 2 factor pairs 
(6)(3) 

(7)(5) (5)(7) 
(9)(2) 

(35)(1) (1)(35) 
(18)(1) 


I'll eliminate (18)(1) and (35)(1) and (1)(35) from my options. REMEMBER MY HINT: Big Gaps Between Factors Can Usually Be Weeded Out.
Since the operation of the third term is positive my inner and outer products
must add to 57.
I believe (6)(3) and (5)(7) are the correct factor pairs.
(6 5)(3 7)
answer: (6x  5)(3x  7)
Again, it is up to you to determine the correct operations in the binomials.
SECTION 18.4: WHAT IS A
PERFECT SQUARE?
A "perfect square" is a "positive integer" or a "math expression" that is the result of a positive
integer squared or a math expression squared.
SECTION 18.5: WHAT IS THE
DIFFERENCE OF TWO SQUARES?
What is similar about ex. 1 through ex. 3 below?
ex. 1)
x^{2} 16
ex. 2)
4d^{4} 16
ex. 3)
9k^{2}
100m^{4}
Explanation:
 Each of the above are made up of two
terms, i.e. binomials, that are subtracted.
 The terms in each binomial are "perfect squares".
ex. 1)
x^{2} 16 =
(x)^{2} 
(4)
ex. 2)
4d  16 =
(2d) 
(4)
ex. 3)
9k 
100m =
(3k) 
(10m)
SECTION 18.6: FACTORING THE
DIFFERENCE OF TWO SQUARES
ex 1)
ex 2)
ex. 3)
SECTION 18.7: WHAT IS A
PERFECT CUBE?
A "perfect cube" is a "positive integer" or
a "math expression" that is the result of a positive integer
cubed or a math expression cubed.
SECTION 18.8: FACTORING THE
SUM OR DIFFERENCE OF TWO PERFECT CUBES
Here are some examples of a "DIFFERENCE OF TWO CUBES."
ex. 1)
ex. 2)
ex.
3)
Notice that each term in the
above examples is a perfect cube. Now I will write the second term in the above examples as a number cubed.
ex. 1)
ex. 2)
ex.
3)
Here are some examples of a "SUM OF TWO CUBES." Now I will write the second term in the above examples as a number cubed.
ex. 1)
ex. 2)
ex.
3)
Notice that each term in the
above examples is a perfect cube.
Now I will write the second term in the above examples as a number cubed.
ex. 1)
ex. 2)
ex.
3)
There is a very simple pattern recognition rule for factoring
a sum or a difference of two cubes.
RULES FOR FACTORING A SUM OR DIFFERENCE OF TWO CUBES 
AND

Examples) Factor the following using the rule :
ex. 1)
step 1)
a = x and b = 2
step 2)
ex. 2)
step 1)
a = x and b = 3
step 2)
ex. 3)
step 1)
a = 2x and b = 1
step 2)
Examples: Factor the following using the rule :
ex. 1)
step 1)
a = x and b = 2
step 2)
ex. 2)
step 1)
a = x and b = 3
step 2)
ex. 3)
step 1)
a = 2x and b = 1
step 2)
MORE PROBLEMS OVER
CHAPTER 18
Factor the following problems:
ex. 1) 

ans: 
ex. 2) 

ans: 
ex. 3) 

ans: 
ex. 4) 

ans: 
ex. 5) 

ans: 
ex. 6) 

ans: 
ex. 7) 

ans: 
ex. 8) 

ans: 
ex. 9) 

ans: 
ex. 10) 

ans: 
