Click on the following links to move to different types of factoring on this page.
Factor out the GCF
Factoring 4-term Polynomials by Grouping
Factoring Trinomials with Leading Coefficient of 1
Factoring trinomials with Leading Coefficient not 1
Factoring Special Cases

Factoring Basics

Factors are the building blocks of multiplication. They are the integers that you can multiply together to produce another integer: [latex]2[/latex] and [latex]10[/latex] are factors of [latex]20[/latex], as are [latex]4, 5, 1, 20[/latex]. To factor an integer is to rewrite it as a product. [latex]20=4\cdot{5}[/latex] or [latex]20=1\cdot{20}[/latex]. In algebra, we use the word factor as both a noun – something being multiplied – and as a verb – the action of rewriting a sum or difference as a product. Factoring is very helpful in simplifying expressions and solving equations involving polynomials.

The greatest common factor (GCF) of two integers is the largest common factor to each integer. For instance, [latex]4[/latex] is the GCF of [latex]16[/latex] and [latex]20[/latex] because it is the largest integer that divides evenly into both [latex]16[/latex] and [latex]20[/latex]. The GCF of polynomials works the same way: [latex]4x[/latex] is the GCF of [latex]16x[/latex] and [latex]20{x}^{2}[/latex] because it is the largest polynomial that divides evenly into both [latex]16x[/latex] and [latex]20{x}^{2}[/latex].

When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables.

In the example above, the monomials have the factors [latex]5[/latex], $$b$$, and $$b$$ in common, which means their greatest common factor is [latex]5\cdot{b}\cdot{b}[/latex], or simply [latex]5b^{2}[/latex].

The video that follows gives an example of finding the greatest common factor of two monomials with only one variable.

Sometimes you may encounter a polynomial with more than one variable, so it is important to check whether both variables are part of the GCF. In the next example, we find the GCF of two terms which both contain two variables.

The video that follows shows another example of finding the greatest common factor of two monomials with more than one variable.

You might be picking up on a useful shortcut for the variables. Notice that (assuming the variable is in common) we always select the smaller power of the variable.

Factoring out the Greatest Common Factor (GCF)

Now that you have practiced identifying the GCF of terms with one and two variables, we can apply this idea to factoring the GCF out of a polynomial. Notice that the instructions are now “Factor” instead of “Find the greatest common factor.”

To factor a polynomial, first identify the greatest common factor of the terms. You can then use the distributive property to rewrite the polynomial in factored form.

The factored form of the polynomial [latex]25b^{3}+10b^{2}[/latex] is [latex]5b^{2}\left(5b+2\right)[/latex]. You can check this by doing the multiplication. [latex]5b^{2}\left(5b+2\right)=25b^{3}+10b^{2}[/latex].

Note that if you do not factor the greatest common factor at first, you can continue factoring, rather than start all over.

For example:

[latex]\begin{array}{l}25b^{3}+10b^{2}=5\left(5b^{3}+2b^{2}\right)\,\,\,\,\,\,\,\,\,\,\,\text{Factor out }5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=5b^{2}\left(5b+2\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Factor out }b^{2}\end{array}[/latex]

Notice that you arrive at the same simplified form whether you factor out the GCF immediately or if you pull out factors individually.

In the following video, we show two more examples of how to find and factor the GCF from binomials.

We will show a few examples of finding the GCF of a polynomial with several terms and sometimes with two variables. No matter how many terms the polynomial has, you can use the same technique described above to factor out its GCF.

In the following video, you will see two more example of how to find and factor out the greatest common factor of a polynomial.

In the next example, the leading coefficient is negative. In this case, it is convention to include the negative with the GCF. In addition, we will see that this can assist us in other ways as well.

Another interesting situation we may encounter is where the GCF could even include an entire expression containing multiple terms, as shown in the next example.

Factor by Grouping

Factoring by grouping is a technique that allows us to factor a polynomial whose terms don’t all share a GCF. In the following example, we will introduce you to the technique. Remember, some of the main reasons we want to be able to factor well  is because it will help solve polynomial equations and graph quadratic functions more easily.

Notice that with this process, we have been able to factor a four-term polynomial into the product of two binomials.

This process is called the grouping technique . Broken down into individual steps, here’s how we apply this technique to a four-term polynomial (you can also follow this process in the example below).

It is worth noting that grouping can be used for polynomials with more than four terms, but the steps above and our examples will focus on four-term polynomials.  Let’s try factoring a few more. Note how there is a now a coefficient in front of the $$x^2$$ term. We will just consider this another factor when we are finding the GCF.

The following video provides another example of factor by grouping.

In the following video we present another example of factor by grouping.

Factoring Trinomials with Leading Coefficient of 1

Trinomials are polynomials with three terms. We are going to show you a method for factoring a trinomial whose leading coefficient is [latex]1[/latex].  Although we should always begin by looking for a GCF, factoring out the GCF is not the only way that trinomials can be factored. The trinomial [latex]{x}^{2}+5x+6[/latex] has a GCF of [latex]1[/latex], but it can be written as the product of the binomial factors.

The last example in the previous section asked us to graph $$g(x)=x^2+5x+6$$. It appeared that it would be easier to find the $$x$$-intercept(s), vertex, and axis of symmetry if we could change the function to intercept form. How do we change this function from general form $$f(x)=ax^2+bx+c$$ to intercept (factored) form $$f(x)=(x-p)(x-q)$$? We need to learn how to factor trinomials.

We will start by factoring the polynomial $$1x^2+5x+6$$.

Let’s identify the values for $$a$$, $$b$$, and $$c$$. Looking at the given polynomial, $$a=1$$, $$b=5$$, and $$c=6$$.

We need to find two integers $$m$$ and $$n$$ whose product is $$c$$ and whose sum is $$b$$. In other words, let’s think of two integers that multiply to be $$6$$ and add to be $$5$$. The factors of 6 are

Which pair of factors add to be $$5$$? The only way to get a sum of 5 is by adding the $$2$$ and $$3$$ ($$2+3=5$$). We have identified $$m$$ as $$2$$ and $$n$$ as $$3$$.

$$\begin{align} ax^2+bx+c &= (x+m)(x+n)\\ &= (x+2)(x+3) \end{align}$$

NOTE: The answer could also be written as $$(x+3)(x+2)$$. We can change the order of the factors since multiplication is commutative.

To check if it is factored correctly, multiply the two binomials:

[latex]\begin{align}\left(x+2\right)\left(x+3\right) &= x^2+3x+2x+6\\ &= x^2+5x+6 \end{align}[/latex]

Let us put this idea to practice with the following example.

In the following video, we present two more examples of factoring a trinomial with a leading coefficient of $$1$$.

To summarize our process, consider the following steps:

In our next example, we show that when $$c$$ is negative, either $$m$$ or $$n$$ will be negative.

In the last example, the $$b$$ term was negative and the $$c$$ term was positive. This will always mean that if it can be factored, $$m$$ and $$n$$ will both be negative.

Factoring trinomials with Leading Coefficient not 1

Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by separating the $$x$$ term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[/latex] can be factored as [latex]\left(2x+3\right)\left(x+1\right)[/latex] using this process. We begin by rewriting the original expression as a 4-term polynomial [latex]2{x}^{2}+2x+3x+3[/latex]. The process to do this is explained in the next example. We will  use factor by grouping to complete the factorization of the 4-term polynomial resulting in [latex]2x\left(x+1\right)+3\left(x+1\right)[/latex]. We then pull out the GCF of [latex]\left(x+1\right)[/latex] to find the factored expression.

The following video has another example about factoring trinomials when the leading coefficient is not 1.

In some situations, $$a$$ is negative, as in $$-4h^2+11h+3$$. It often makes sense to factor out $$-1$$ as the first step, making the polynomial easier to factor.

Factoring Special Cases

Factoring a Perfect Square Trinomial

What is a perfect square? A perfect square is the product of an integer multiplied by itself. For example, $$4 \cdot 4 = 16$$ So, $$16$$ is a perfect square. Other examples of perfect squares are, $$1, 4, 9, 16, 25, 36, 49, 64, 81$$, and $$100$$. A perfect square trinomial is a trinomial that can be written in factored form as a binomial squared. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.

In the following video, we provide another short description of perfect square trinomials and shows how to factor them using a formula.

Factoring a Difference of Squares

A difference of squares can be factored into two binomials but has only two terms.  Let’s start from the product of two special binomials to see the pattern.

Consider the product of the following two binomials, which are identical except for the operation: $$(x-2)(x+2)$$. If we multiply them together, the middle term zeros out.

Multiply. $$(x-2)(x+2)$$

$$\begin{align} &= x^2-2x+2x-2^2\\ &= x^2-2^2\\ &= x^2-4 \end{align}$$

The polynomial $$x^2-4$$ is called a difference of squares because each term is a perfect square and the operation connecting them is subtraction. A difference of squares will always factor in the following way:

[latex]\\[/latex]

We can use this equation to factor any differences of squares.

 Summary

In all factoring problems, we first look for a common factor that can be factored out of all terms. After achieving this (if possible), we have now learned special factoring techniques in this section.

If a trinomial is of the form of a perfect square trinomial, it factors as