Factoring
Factoring is just another way of looking at the distributive property. In multiplication, we can
“distribute through” a number or variable to all of the inside numbers/variables like so:
3x(2x2 + 4x + 5) = 6x3 + 12x2 + 15x.
But, one thing to keep in mind is that if we can forward, then we can also go backwards! In
other words, we can go from 6x3 + 12x2 + 15x to 3x(2x2 + 4x + 5). That’s called “factoring”.
In factoring out a monomial (or greatest common factor), I usually will break my problem
down into parts. Let’s check out this problem.
Example 1: Factor 4x3 + 12x2 – 8x5:
The first part we look at will be the coefficients. (Those are the numbers in front of the variable
and/or the constant.) We factor out the largest number that will “go into” 4, 12, and 8. In this case,
that number is 4.
The second thing we look at is the variable. If there’s more than one variable, then we
look at them individually. We want to factor out the most x’s that we can without going over on any
of them, and we have to take the same number of x’s out of all three terms. So, the most we can
factor out of all three terms is two x’s: x2.
After we figure out what’s going to be taken out, we have to see what’s going to be left inside the
parentheses. Actually, we’re dividing when we factor. Remember, when we multiply back through,
we’d better get what we started with!
Okay, let’s factor: 4x3 + 12x2 – 8x5 = _____(________________)
Practice – Factor out a monomial.
1) 12x4 – 18x8 + 36x5
2) 2xy – 6x2y3 + 10x3y
*Note: when you factor out
an entire monomial, we must
leave a place holder because
we are actually dividing. That
place holder is “1”.
Factoring by Grouping: Factoring by grouping gets its name because we’re going to group pairs
of terms together to factor them. Notice that we need an even number of terms to pair terms up.
Example 2: Factor 2x3 – 4x2 + 3x – 6.
We’re going to group the first two terms together and then group the last two terms together. The
idea is to factor out everything we can just in the first grouping, and then factor out everything we
can in the second grouping.
2x3 – 4x2 + 3x – 6
_____(_________) + _____(_________).
It’s not recommended that we put
parentheses around our groupings
because that can potentially mess up our
equation if there’s a negative involved,
so I’m just going to underline them.
*Note – in the second grouping, always take the sign of the third term in the factoring. In this case,
we’d factor out a positive because the 3x is positive.
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See how the parentheses have the same expression in them after factoring? So, what do we do
with it now? Okay, let’s put that on a back burner for a minute.
How would we factor the following?
2x2y + 3y
2x2
2x2
+3
+3
Notice how it doesn’t matter at all what it is that we are factoring out as long as it’s exactly the same
thing and it’s something that’s being multiplied.
The factoring by grouping problem is very much the same. We have
2x2 • (something) + 3 • (something)
No, matter what that “something” is, we can factor it out, as long as it’s the same. So, in both of our
parentheses (bubbles), we have “x – 2”. So, we can factor out our entire parentheses (factor out
the bubble).
Then we have to see what’s left over after we factor that out.
_____(__________) + _____(__________) =
(x – 2) (_____________)
So, how can we tell if factoring by grouping will work without having to do all of the work?
Let’s look at our example problem:
2x3 – 4x2 + 3x – 6
Use the example problem to fill in the box
on the right.
The product of the 1st and 4th terms
= ________________
The product of the 2nd and 3rd terms
= ________________
If the product of the first and last (outer) terms = the product of the middle two (inner) terms,
then factoring by grouping will work!
Not only is this a time-saving idea when it comes to factoring by grouping, but it’s also the idea
behind why something called the “a•c Method” of factoring trinomials works. That’s why we’re
interested in pursuing this line of thinking. 
So, what happens when the product of the outer terms is not equal to the product of the inner
terms? Sometimes the terms can be rearranged to work right, but many times they can’t.
Let’s take our original problem and make one little adjustment: change one of the signs.
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Example 3: Factor 2x3 – 4x2 + 3x + 6
The product of the 1st and 4th terms
Well, according to the box over to the right, our
products don’t equate, and it doesn’t look like we
can rearrange things to make them equate.
= ________________
The product of the 2nd and 3rd terms
Let’s try to factor it anyways, just to see what happens:
= ________________
2x3 – 4x2 + 3x + 6
2x2(x – 2) + 3(x + 2)
The numbers in parentheses aren’t the same, so we can’t factor
anything out! This is prime. Anytime something can’t be factored,
we say that it’s prime.
Practice - Checking for Factoring by Grouping
Check the following polynomials to see if factoring by grouping can be used. Then circle “Yes” or
“No” if factoring by grouping can or cannot be used, respectively.
1) 5x3 – 10x2 + 7x – 14
2) 3x3 + x2 + 6x + 2
3) 4x3 – 6x2 + 5x – 3
Outer terms = _________
Outer terms = _________
Outer terms = _________
Inner terms = _________
Inner terms = _________
Inner terms = _________
Yes
No
Yes
No
Yes
No
4) 8x3 + 6x2 + 12x – 9
5) 12x3 + 9x2 + 16x + 12
6) x3 – 7x2 – 4x + 28
Outer terms = _________
Outer terms = _________
Outer terms = _________
Inner terms = _________
Inner terms = _________
Inner terms = _________
Yes
No
Yes
No
Answers: 1) –70x3, –70x3, Yes
2) 6x3, 6x3, Yes
5) 144x3, 144x3, Yes
6) 28x3, 28x3, Yes
Yes
3) –12x3, –30x3, No
Practice – Factor by Grouping
1) 4x3 – 7x2 + 8x – 14
2) 8x3 – 3x2 – 16x + 6
3) 3a – 6b + 7a2 – 14ab
4) 6x2 + 3x – 4x – 2
No
4) –72x3, 72x3, No
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Factoring Trinomials:
I know this is everyone’s favorite part! 
Okay, remember when you were taught how to factor something like this: x2 + 4x + 3 ?
Most times, a teacher will say
“find the factors of the last term that add to give us the middle coefficient”,
or something similar to that effect. If we do that, here’s what we get:
x2 + 4x + 3
Factors of
3
1
3
–1
–3
 +1 + 3 = +4
 –1 + –3 = –4
Then we’re told that we’re starting off with (x ______)(x ______) and we stick the +1 and +3 in
the spaces because +1 + 3 = +4, which is our middle coefficient.
So, x2 + 4x + 3 = (_________)(_________)
This looks a lot like magic instead of mathematics. Instead of the “magic” standpoint, let’s examine
the same problem from a “factoring by grouping” standpoint.
For factoring by grouping, we need to have four terms, not three. That means we need to break up
one of the terms into two pieces, and it’s going to be the middle term. Why the middle term? Well,
multiply out the following:
(x + 1)(x + 3)
= ____________________ =
x2 + 4x + 3
Which term is it that starts out being in two pieces before we combine them? It’s the middle term.
That’s why the middle term is the one we’re breaking up. Remember, factoring is just working
multiplication in reverse.
So, the two underlined numbers below must add to give us “4x”. Also, they must be “like terms” with
the “4x”, so they must both have a singleton “x” in them.
x2 + 4x + 3
x2 + ______x + ______x + 3
x2 +
_ _+ ______ + 3
x(_______) + 3(_______)
 But, how do we break the “4x” up?
Factors of
3
1
3
-1
-3
(_________)(__________)
See, it’s the exact same thing as before! Someone
down the line noticed a pattern and created a
shortcut, that’s all! The reason that we need to go
through that is because it’s the basis of why the
“a•c Method” works on things where the coefficient
of “x2” isn’t 1.
Product of outer terms =___________
So, the product of the inner terms
must give us the same thing!
That’s why we’re looking for factors of
“3”. The x2 is taken care of by the
middle terms having a single “x” each.
The two middle terms must add to
give us “4x”, so that’s why we look for
“factors of 3 that add to give us 4”.
Let’s work several like the above problem just to get used to the process.
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Practice – Factor the following trinomials.
1) x2 + 7x – 30
Outer terms:_______
2)
So, Inner terms: ______
x2 – 4x – 12
Outer terms:______
So, Inner terms: _____
The A•C Method, or Factoring Trinomials using Grouping
Previously, we learned how to factor polynomials using grouping and then how to factor quadratic
trinomials with a leading coefficient of “1” by using factoring by grouping. One important thing to
remember is that when using factoring by grouping, the product of the outer two terms equals the
product of the inner two terms. This is the crux of how the “a•c method” works.
The “a•c method” works exactly the same way for all quadratic trinomials (trinomials of the form
ax2 + bx + c, which is the reason it’s called the “a•c method”. The first term times the last term –
ignoring the variables – is a•c.)
The shortcut that we can use for the problems when a = 1 does not work if the “a” value in the
quadratic isn’t a “1”, but the “a•c method” does which is what makes the “a•c method” nice: it’s
consistent. It always works. Always! Yay!
Example 1: Factor 6x2 – 7x – 3.
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6x – 7x – 3
6x2 ______x
_________
Factors
of _____
Product of the Outer Terms
is ___________,
_______x – 3
so the product of the Inner
Terms must also be
_________
_____________.
_____(_________)
______(_________)
Which of these pairs adds to be –7?
(_________)(__________)
This method also works with the difference of squares, “backwards” trinomials, and multi-variable
set-ups. It also prevents us from having to deal with the “trial and error” method that sometimes
takes up a lot of time and can seem random at best.
Next, we look at some examples of other ways the a•c-method can be used. All are quadratic
variations.
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Difference of Squares Example: Factor 9x2 – 4.
Factors
of _____
9x2 – 4
9x2 + ______ – 4
9x ________ ________ – 4
______(__________)
is ___________,
so the product of the Inner
Terms must also be
2
___________
Product of the Outer Terms
_____________.
___________
______(__________)
(_________)(__________)
“Backwards” Trinomial Example: Factor 12 + 5x – 2x2
12 + 5x – 2x2
2
12 ________ ________ – 2x
___________
______(__________)
Factors
of _____
Product of the Outer Terms
is ___________,
so the product of the Inner
Terms must also be
___________
______(__________)
_____________.
(_________)(__________)
Multi-Variable Trinomial Example: Factor 15x2 + 11xy + 2y2.
15x2 + 11xy + 2y2
15x2 _____xy
___________
______(__________)
____ xy + 2y2
___________
______(__________)
Factors
of _____
Product of the Outer Terms
is __________,
So the product of the inner
two terms must also be
__________.
(_________)(__________)
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Practice – Factor the following:
1) 6x2 – 11x – 10
2) 15x2 – 19xy + 6y2
3) 10 + 14x – 12x2
4) 25x2 – 36
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