Factoring Completely Lessons
Introduction
Previous factoring lessons each focused on factoring a polynomial using a single pattern such as
Greatest Common Factor
Example: 3x^{2} + 9x^{3} + 12x^{4} factored into 3x^{2}(1 + 3x + 4x^{2})
Difference Between Two Squares
Example: y^{2} – 9 factored into (y + 3)(y – 3)
Trinomial
Example: x^{2} – 2x – 3 factored into (x + 1)(x – 3)
The lessons linked above give systematic techniques to factor certain types of polynomials. In practice, solving equations using factoring often requires the use of a more complex process called “Factoring Completely”. This lesson explains how to factor
completely by combining the three basic techniques listed above.
First, lets take a closer look at why we need the Factoring Completely process. Examine the expression below:
If we simplify this expression, we get:
(x
^{2} + 1)(x^{2} – 1)
(x
^{4} + x^{2} – x^{2} – 1)
x
^{4} – 1
Now you should recognize this expression as a difference between two squares. Using the technique presented in the
Difference Between Two Squares lesson, we can factor this into
But notice that this expression is not the same as the factored expression that we started with. We need another step to factor this into the expression that we started with. This is where Factoring Completely comes in.
The Process
Factoring completely is a three step process:
 Factor a GCF from the expression, if possible.
 Factor a Trinomial, if possible.
 Factor a Difference Between Two Squares as many times as possible.
First Example
Let’s see how this applies to our initial example:
Step 1
Step one is to factor a GCF. Since the GCF of x^{4} and 1 is 1, we skip this step.
Step 2
Since the expression only has two terms, we cannot factor a trinomial.
Step 3
Factoring (x^{4} – 1) as a difference between two squares results in
Now be sure to remember the key phrase “as many times as possible.” We must now look to see if there is anywhere else we can factor another Difference Between Two Squares. In (x^{2} + 1), both terms are positive, so this cannot be factored. However,
in (x^{2} – 1), the second term is negative, and both terms are perfect squares otherwise. So (x^{2} – 1) factors into (x + 1)(x – 1). As a result, our example expression is finally factored into
which is factored completely.
How did this differ from our first (and failed) attempt to factor the example? When Factoring Completely, we used the Difference Between Two Squares method more than once.
Second Example
Let’s try another example which requires factoring in steps 1 and 2:
Again, the three steps in Factoring Completely are:
 Factor a GCF from the expression, if possible.
 Factor a Trinomial, if possible.
 Factor a Difference Between Two Squares as many times as possible.
Step 1
We see that the terms in our example have a greatest common factor of 5x. As instructed, we will factor out this GCF:
Step 2
We see that (x^{2} – 2x – 3) is a factorable trinomial, so we factor it:
Proceeding to Step 3, we can look over our expression and see that neither 5x, nor (x + 1), nor (x – 3) can be factored as a difference between two squares. We have factored 5x
^{3} – 10x^{2} – 15x completely.
Final Example
For our final example, we will make use of all three Factoring Completely steps.
Step 1
We factor out a Greatest Common Factor of 3.
Step 2
Step 3
Finally, we identify (4x^{2} – 9) as a binomial that can be factored into (2x + 3)(2x – 3). So the completely factored result is
Factoring Completely Resources
Expression Factoring Calculator Practice Problems / Worksheet

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