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Factoring Four-Term Polynomials by Grouping

Welcome back. In the last lesson, you factored sums and differences of cubes by recognizing a two-term special-product pattern and by extracting a GCF first when needed. Factoring by grouping is more flexible: instead of spotting one identity immediately, you reorganize a four-term polynomial so that the distributive property reveals a shared binomial.

By the end of this lesson, you should be able to factor a four-term polynomial by grouping, handle negative signs correctly, decide when rearranging terms is useful, and check whether your final answer is fully factored.

The idea behind grouping

Factoring by grouping is just the distributive property used in reverse. The underlying identity is

a c + a d + b c + b d = (a + b) (c + d) .

A four-term polynomial may not have one factor shared by all four terms. But after splitting it into two pairs, each pair can have its own GCF. If those two GCF factorizations contain the same binomial, that binomial can then be factored out.

In its most useful form, the pattern looks like this:

A (B) + C (B) = (A + C) (B) .

The goal of grouping is therefore not merely “make two groups.” The goal is:

Factor each group so that the same binomial appears in both.

Before grouping, maintain the factoring habit established earlier:

Check for a GCF shared by every term.
If there is one, factor it out first.
Then group the remaining four terms.

This brief reading formalizes the method and the identity that justifies it.

4.2: Factoring by Grouping - Mathematics LibreTexts

Read “4.2: Factoring by Grouping” from Mathematics LibreTexts for a concise definition of the method and its algebraic basis.

In the section “Definitions and Theorems,” read the definition and grouping identity. Focus on why the repeated binomial factor is the decisive feature, rather than treating the procedure as an arbitrary arrangement of parentheses.

The standard procedure

For a polynomial already arranged in descending powers, start by pairing adjacent terms:

(first two terms) + (last two terms) .

Then follow this routine:

Group the polynomial into two pairs.
Factor the GCF from each pair.
Compare the binomials left inside the parentheses.
If they match exactly, factor out that common binomial.
Inspect the remaining factor for any further GCF or special pattern.
Verify by expanding, particularly if signs were involved.

Consider

6 x^{3} + 18 x^{2} + 5 x + 15.

No nontrivial factor is shared by all four terms. Group adjacent terms:

(6 x^{3} + 18 x^{2}) + (5 x + 15) .

Now factor the GCF from each group:

6 x^{2} (x + 3) + 5 (x + 3) .

Both terms contain $x + 3$ . Treat it as one common factor:

6 x^{2} (x + 3) + 5 (x + 3) = (x + 3) (6 x^{2} + 5) .

6 x^{3} + 18 x^{2} + 5 x + 15 = (x + 3) (6 x^{2} + 5) .

A quick expansion checks the work:

(x + 3) (6 x^{2} + 5) = 6 x^{3} + 18 x^{2} + 5 x + 15.

Notice the logic: the first group produced $6 x^{2} (x + 3)$ , and the second produced $5 (x + 3)$ . Once the common factor $x + 3$ was exposed, the final step was no different from factoring

6 x^{2} B + 5 B .

The following video gives two compact examples, including one where a negative GCF is necessary.

Factor By Grouping Polynomials - 4 Terms, Trinomials - 3 Terms, Algebra 2

Watch “Factor By Grouping Polynomials - 4 Terms, Trinomials - 3 Terms, Algebra 2” from The Organic Chemistry Tutor. It shows the repeated-binomial goal clearly and reinforces the role of each group’s GCF.

Watch the first example, following how the two GCF factorizations both produce x+4. Then watch the cubic example, paying close attention to why a negative factor is extracted from the second group. The coefficient-ratio observation is a quick diagnostic for these arranged examples, but the matching binomial is the actual test.

Negative signs: preserve the polynomial before factoring it

Most grouping errors are sign errors. The essential rule is:

If factoring out a negative GCF is what makes the two binomials match, factor out the negative GCF.

For example, factor

2 x^{3} + 6 x^{2} - 5 x - 15.

Include the sign of the third term inside the second group:

(2 x^{3} + 6 x^{2}) + (- 5 x - 15) .

Now factor each pair:

2 x^{2} (x + 3) - 5 (x + 3) .

The shared binomial is $x + 3$ , so

2 x^{2} (x + 3) - 5 (x + 3) = (x + 3) (2 x^{2} - 5) .

Therefore,

2 x^{3} + 6 x^{2} - 5 x - 15 = (x + 3) (2 x^{2} - 5) .

Why not factor $5$ from the second group? Doing so gives

5 (- x - 3),

which does not match $x + 3$ . It is algebraically valid, but it hides the common factor instead of revealing it.

There is a related parentheses issue. Writing

(2 x^{3} + 6 x^{2}) - (5 x + 15)

is valid, because distributing the minus sign gives $- 5 x - 15$ . But writing

(2 x^{3} + 6 x^{2}) - (5 x - 15)

would change the original polynomial: it would produce $- 5 x + 15$ . Grouping must preserve every term and every sign exactly.

Rearranging terms can make the structure visible

Polynomials can be written in any order because addition is commutative. But grouping is easiest when the terms are arranged strategically, usually in descending order of degree.

This diagram rearranges $3x^2+6+2x+x^3$ into descending order, groups adjacent pairs, factors a GCF from each pair, and then factors the shared binomial $x+3$.

The diagram’s polynomial begins as

3 x^{2} + 6 + 2 x + x^{3} .

The first two terms as written do not suggest a useful GCF. Sort it first:

x^{3} + 3 x^{2} + 2 x + 6.

Then group:

(x^{3} + 3 x^{2}) + (2 x + 6) .

Factor each group:

x^{2} (x + 3) + 2 (x + 3) .

Now factor the common binomial:

x^{2} (x + 3) + 2 (x + 3) = (x^{2} + 2) (x + 3) .

Thus,

3 x^{2} + 6 + 2 x + x^{3} = (x^{2} + 2) (x + 3) .

Sorting was not a cosmetic step. It placed terms with compatible factors beside one another.

Grouping also works with more than one variable. For instance,

15 x y - 10 x + 6 y - 4

can be grouped as

(15 x y - 10 x) + (6 y - 4) .

Factor the GCF from each group:

5 x (3 y - 2) + 2 (3 y - 2) .

The shared binomial is $3 y - 2$ , so

15 x y - 10 x + 6 y - 4 = (3 y - 2) (5 x + 2) .

Variables are handled exactly like numerical factors: they are included in a group’s GCF only when they divide every term in that group.

When grouping stops, and when it does not work

A successful grouping factorization is not always the final answer. After factoring, inspect every factor using the tools from earlier lessons.

For example,

x^{3} + 3 x^{2} - 4 x - 12

groups as

(x^{3} + 3 x^{2}) + (- 4 x - 12) .

Factoring each group gives

x^{2} (x + 3) - 4 (x + 3) .

So the first grouping result is

(x + 3) (x^{2} - 4) .

But $x^{2} - 4$ is a difference of squares:

x^{2} - 4 = (x - 2) (x + 2) .

Therefore the complete factorization is

x^{3} + 3 x^{2} - 4 x - 12 = (x + 3) (x - 2) (x + 2) .

On the other hand, not every four-term polynomial factors by grouping. Consider

2 x^{2} + 3 x + 4 x + 12.

Grouping adjacent terms gives

(2 x^{2} + 3 x) + (4 x + 12) .

Factoring the groups produces

x (2 x + 3) + 4 (x + 3) .

The binomials $2 x + 3$ and $x + 3$ do not match, so the method has not succeeded. Rearranging may occasionally provide a better pairing, but one failed attempt does not license changing terms or signs just to force a match.

Use this diagnostic distinction:

Result after factoring the two groups	Meaning
The same binomial appears in both groups	Factor out that binomial.
The binomials differ only by an overall negative sign	Factor a negative GCF from one group so they match.
The binomials are genuinely different	Try a sensible rearrangement if available; otherwise grouping does not apply.
A product has been obtained	Check each factor for GCFs and known special patterns.

Key takeaways

Factoring by grouping is reverse distribution applied twice:

A (B) + C (B) = (A + C) (B) .

For four-term polynomials:

factor out any global GCF first;
group terms in compatible pairs, often after sorting by degree;
factor the GCF from each pair;
look for an identical binomial factor;
factor out that common binomial;
use a negative GCF when necessary to preserve matching signs;
continue factoring if the resulting factors contain a GCF, difference of squares, or another known pattern;
verify by expansion when signs or rearrangement make the algebra less transparent.

Next, you will factor expressions with quadratic form by making a substitution. That method recognizes a polynomial behaving like a simpler quadratic even when its exponents are larger.

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Factoring Sums and Differences of Cubes

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Factoring Quadratics by Substitution

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