Lesson Explainer: Factoring by Grouping Mathematics

In this explainer, we will learn how to factor expressions by grouping.

Let us begin by revisiting the idea of factoring an expression by identifying its highest common factor. Suppose we have 3𝑥+9𝑦6.

Since each term is divisible by 3, we can say that it is a common factor of the expression. Thus, we can factor the expression to get 3(𝑥+3𝑦2).

Since 𝑥+3𝑦2 no longer has any common factors, we can conclude that 3 is the highest common factor, and that this expression is now in its most simplified form. Identifying common factors is a useful skill for factoring, but it may not always be the case that the entire expression has a common factor. Suppose, for instance, we have 2+2𝑥+𝑦+𝑥𝑦.

We can see that this expression has no common factor (other than 1). Suppose, however, we were to group up the terms in the expression as follows: 2+2𝑥+𝑦+𝑥𝑦.FactorofFactorof

We can see that the first two terms share a common factor of 2 and the last two terms share a common factor of 𝑦. If we factored these terms separately, we would get 2(1+𝑥)+𝑦(1+𝑥).

Notice that we now have two terms with a common factor of 1+𝑥. If we factor these terms, we get (1+𝑥)(2+𝑦).

So, even though it was not obvious at first, this expression can be factored into two parentheses. It is important to note, however, that this not the only way to group the terms. Alternatively, by swapping the second and third terms around, we can group them as follows: 2+𝑦+2𝑥+𝑥𝑦.FactorofFactorof

Here, we note that even though the first two terms do no not have a common factor greater than 1, the last two terms have a common factor of 𝑥 and so we have grouped them as shown. Factoring these terms, we get 1(2+𝑦)+𝑥(2+𝑦).

Factoring 2+𝑦 to make 1(2+𝑦) in this way is not strictly necessary, but it allows us to identify the common factor more easily. In particular, we can see that we now have two terms with a common factor of 2+𝑦 (whereas before, 1+𝑥 was the factor). Factoring these terms, we obtain (2+𝑦)(1+𝑥).

This is the same as (1+𝑥)(2+𝑦) since multiplication is commutative. This approach is something we can apply generally as we will now describe.

How To: Factoring by Grouping

Suppose we have an expression with an even number of terms that do not all share a common factor. Then, in certain situations, we can apply the following approach to fully factor the expression.

  • Group the terms into pairs that each share a common factor between them.
  • Factor out the common factor from each pair.
  • Completely factor the expression by identifying a common factor in the factored terms.

Let us consider another example where we can apply this method.

Example 1: Factoring an Expression by Grouping

Factorize fully 𝑧𝑏𝑧𝑥+𝑏𝑥.

Answer

This expression does not have a common factor greater than 1, so we cannot directly factor all of it. However, we can group pairs of terms that share common factors and factor them individually.

Inspecting the first and third terms, we see that they both have a factor of 𝑏. Likewise, the second and fourth terms have a factor of 𝑥. We highlight this below: 𝑧𝑏+𝑏+𝑧𝑥𝑥.FactorofFactorof

Taking 𝑏 and 𝑥 out as factors, we have 𝑏(𝑧+1)𝑥(𝑧+1).

Now, we can see that we have two terms that each have a factor of 𝑧+1. Since 𝑏 is multiplying 𝑧+1 in the first term and 𝑥 is multiplying 𝑧+1 in the second term, we can factor out (𝑏𝑥) to get (𝑧+1)(𝑏𝑥).

As this expression cannot be factored any further, it is in the fully factored form.

An alternate way to get the same result is to group up the first and second terms along with the third and fourth terms as shown: 𝑧𝑏𝑧𝑥+𝑏𝑥.FactorofFactorof

Although the second pair cannot be meaningfully factored, we can factor out 𝑧 from the first two terms to get 𝑧(𝑏𝑥)+1(𝑏𝑥).

Finally, by taking 𝑏𝑥 as a common factor, we obtain the same expression we previously calculated: (𝑏𝑥)(𝑧+1).

Before moving on, let us first address a complication that may occur in the solving of this example. When identifying common factors, we may have found instead 𝑧𝑏+𝑏+𝑧𝑥𝑥.FactorofFactorof

In other words, we might not have immediately realized that 𝑧𝑥𝑥 has a factor of 1. If we had continued with the factoring, we would get 𝑏(𝑧+1)+𝑥(𝑧1).

At this stage, it might seem like it is not possible to continue since the terms do not have the same factor. However, with a little more factoring we can still fix things. In particular, if we make use of the fact that 𝑧1=(𝑧+1), we get 𝑏(𝑧+1)+𝑥(𝑧1)=𝑏(𝑧+1)𝑥(𝑧+1)=(𝑧+1)(𝑏𝑥).

So, we can still get to the same answer.

Our first example included fairly simple factors, but let us recall that it is possible to factor out more than one variable or number from an expression. For instance, we have 9𝑥𝑦𝑧+6𝑥𝑦=3𝑥𝑦(3𝑧+2), where we have factored out the common factor 3𝑥𝑦 from both terms. In the following example, we must keep in mind that the factors we choose may contain more than just one variable.

Example 2: Factoring a Multivariable Expression by Grouping

Factorize fully 9𝑥𝑚4𝑙𝑧+4𝑙𝑥9𝑚𝑧.

Answer

We have been asked to fully factorize the given expression, but, as we can see, there is no common factor among the terms. This shows us that we need to factor by grouping the terms separately.

In order to identify terms that share common factors, in this instance, we can consider the numbers in front of the variables. Specifically, we highlight the ones that match:

Let us consider these pairs of terms one by one. 9𝑥𝑚 and 9𝑚𝑧 have common factors of 9 and 𝑚, so their greatest common factor is 9𝑚. Meanwhile, 4𝑙𝑧 and 4𝑙𝑥 have common factors of 4 and 𝑙, so they have a greatest common factor of 4𝑙. We can rearrange the terms so that the pairs are together by using the commutativity of addition and subtraction. This gives us 9𝑥𝑚9𝑚𝑧+4𝑙𝑥4𝑙𝑧.FactorofFactorof

Then, we can factor the expression as follows: 9𝑚(𝑥𝑧)+4𝑙(𝑥𝑧).

Having factored this expression into two terms, we now see that each term has (𝑥𝑧) as a factor. Hence, we can factor them again, giving us the fully factored form (𝑥𝑧)(9𝑚+4𝑙).

So far, we have considered expressions where no variable has been raised to a power. Sometimes, we may also be able to factor quadratic or cubic expressions by grouping terms that we can factor. For instance, suppose we have 𝑥+3𝑥+2𝑥+6.

Like previous examples, this expression has no common factor. Ordinarily, since this is a cubic expression, it would be difficult for us to factor it. However, there is something about the coefficients of this expression that suggests it can be factored by grouping terms. In particular, we can see that the ratio between the coefficients of the first two terms and the last two terms is the same:

It may not be immediately clear why this is significant, so let us first continue by applying the approach from before and grouping terms together that can be factored. If we do this, we find that 𝑥+3𝑥+2𝑥+6.FactorofFactorof

Thus, factoring out 𝑥 from the first two terms and 2 from the second two terms, we get 𝑥(𝑥+3)+2(𝑥+3).

Now, we see that we have two terms that both have a factor of 𝑥+3. Note that 𝑥+3 has the same ratio between the coefficients of the terms that we noticed earlier: 13. This is because the ratio between the pairs of terms was preserved when we factored them. Continuing onward, we can factor 𝑥+3 out to get (𝑥+3)𝑥+2.

The term 𝑥+2 cannot be factored further; therefore, we have managed to fully factor this expression.

It is important to realise that we cannot always use this approach to factor cubic expressions. For instance, suppose that our expression was 𝑥+4𝑥2𝑥3.

However, let us consider the ratios between the coefficients of the terms in this expression:

Since the ratios are not the same, this implies we cannot use factoring by grouping to factor the expression. Indeed, if we applied the above approach and factored the groups of terms, we would get 𝑥(𝑥+4)(2𝑥+3).

However, since 2𝑥+3 does not share a factor with 𝑥(𝑥+4), we cannot proceed further.

Despite the fact that factoring by grouping is not always an option for cubic expressions, it is very beneficial to use in specific situations. Let us see another example where we can use this technique.

Example 3: Factoring a Cubic Expression by Grouping

Factorize fully 𝑥2𝑥+5𝑥10.

Answer

The first thing we notice is that the expression is cubic since the largest exponent of a variable in any nonzero term is 3. Ordinarily, there would be no easy way to factor this expression fully, but we notice that there is a common ratio between the coefficients of the first two terms and the last two terms:

This suggests that we can factor the expression by grouping the first two and the last two terms together. If we group them in this way, we have 𝑥2𝑥+5𝑥10.FactorofFactorof

Since we have a factor of 𝑥 in the first two terms and 5 in the last two terms, we can factor them to get 𝑥(𝑥2)+5(𝑥2).

Notice that we now have two terms with a common factor of 𝑥2. Thus, we factor this out to get the fully factored form: 𝑥+5(𝑥2).

In the previous example, we showed one way to factor a cubic expression by grouping pairs of terms and factoring them. One other way of factoring a cubic expression is identifying when it is a difference or sum of two cubes. Recall that the difference of two cubes can be written as 𝑎𝑏=(𝑎𝑏)𝑎+𝑎𝑏+𝑏.

Similarly, the sum of two cubes can be written as 𝑎+𝑏=(𝑎+𝑏)𝑎𝑎𝑏+𝑏.

In some cases, we can group terms so that parts of an expression can be factored using the above formulas. Let us see an example of this.

Example 4: Factoring an Expression by Grouping Terms and Using the Sum of Cubes Formula

Factorize fully 𝑎+𝑏+𝑎+𝑏.

Answer

Let us consider how we can go about factoring this expression. The first thing we might notice is that the first two terms are the sum of two cubes. Recall that we have a formula for factoring the sum of two cubes: 𝑎+𝑏=(𝑎+𝑏)𝑎𝑎𝑏+𝑏.

The given expression has more terms than just the first two, but if we group the terms, then we can factor them separately. This is to get the following: 𝑎+𝑏+𝑎+𝑏.SumofcubesFactorof

Using the above formula, this then becomes (𝑎+𝑏)𝑎𝑎𝑏+𝑏+1(𝑎+𝑏).

Notice that we now have two terms with a common factor of 𝑎+𝑏, which means that we can factor 𝑎+𝑏 from both terms. Namely, this means we are combining the underlined terms below into one parenthesis: (𝑎+𝑏)𝑎𝑎𝑏+𝑏+1(𝑎+𝑏).

Doing this, we get (𝑎+𝑏)𝑎𝑎𝑏+𝑏+1.

As the expression in the second parenthesis cannot be factored any further, we conclude that the cubic expression has been fully factored.

For our final example, we will further investigate how we can combine factoring by grouping with other methods of factoring.

Example 5: Factoring an Expression by Recognizing a Perfect Square Trinomial

Factorize fully 1𝑥+14𝑥𝑦49𝑦.

Answer

Here, we have been given a polynomial in 𝑥 and 𝑦 with four terms. Although it does not immediately appear to be in a form that we can factor, we can potentially group it into parts and factor each part separately. In fact, we can see that, if we do not include the 1, the rest of the expression takes on a familiar form: 1+𝑥+14𝑥𝑦49𝑦.Perfectsquaretrinomial

We can see that this trinomial is the result of a perfect square because of the coefficients: namely, 49 is 7 and 14 is 2×7.

To make this more obvious, we can begin by factoring the negative sign out to get 1𝑥14𝑥𝑦+49𝑦.

Recall that perfect square trinomials of the above form can be factored as follows: 𝑎2𝑎𝑏+𝑏=(𝑎𝑏).

We can use the above expression to directly factor the expression in parentheses. Substituting 𝑎 for 𝑥, and 𝑏 for 7𝑦, we can see that this becomes 1(𝑥7𝑦).

To fully factor the expression, we need one more step. Since (𝑥7𝑦) is the square of 𝑥7𝑦, we note that the above expression is a difference of two squares. In particular, recall that 1𝑎 can be factored using 1𝑎=(1+𝑎)(1𝑎).

If we let 𝑎=𝑥7𝑦, then we can use this formula to obtain the fully factored form of the expression: 1(𝑥7𝑦)=(1+(𝑥7𝑦))(1(𝑥7𝑦))=(1+𝑥7𝑦)(1𝑥+7𝑦).

Let us finish by considering the main things we have learned in this explainer.

Key Points

  • We can factor expressions that lack a common factor by grouping them into pairs that share common factors.
  • We can also factor cubic expressions that have a shared ratio between the terms using factoring by grouping.
  • If we see that we can apply a factoring formula to a part of an expression, such as the difference or the sum of cubes, the difference of squares, or perfect square trinomials, then we can group that expression and factor it separately.

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