When you see a polynomial with exactly 4 terms, treat that as a signal. The number 4 is not random. It is the natural output of multiplying two binomials — (a + b)(c + d) expands into 4 terms — so when you face 4 terms going the other way, you are almost certainly trying to undo that expansion. The tool for that is grouping.

Grouping is your default first move when you count 4 terms. It works the vast majority of the time. The rare cases where it fails are usually disguised difference-of-squares patterns or expressions that simply do not factor over rationals. Either way, you start with grouping. Here is the procedure, the standard cases, and the escape hatches.

The grouping procedure

The whole method is four small steps:

  1. Arrange the 4 terms in a useful order — sometimes the given order already works; sometimes you need to swap terms around so that pairs share a common factor.
  2. Split into two pairs — usually the first two terms form one pair, the last two form the other.
  3. Factor the common term out of each pair — pull the GCF (greatest common factor) from each pair separately.
  4. If the remaining brackets match, factor them out — the matching bracket is your common binomial factor; the leading factors form the other binomial.

That is it. You will use this pattern hundreds of times.

Worked example — standard case

Factor ax + ay + bx + by.

Pair them as written:

(ax + ay) + (bx + by)

Factor each pair. From the first pair, a is common. From the second, b is common:

a(x + y) + b(x + y)

Both brackets are (x + y). That is the signal that grouping has worked. Pull (x + y) out:

(a + b)(x + y)

Done. Two binomials, just like the expansion you reversed.

Worked example 2 — classic

Factor x³ + x² + 2x + 2.

Pair the first two and last two:

(x^3 + x^2) + (2x + 2)

From the first pair, x² is common. From the second, 2 is common:

x^2(x + 1) + 2(x + 1)

Both brackets read (x + 1). Pull it out:

(x^2 + 2)(x + 1)

This is the textbook pattern — and it is exactly what most exam questions look like.

When the obvious pairing doesn't work — reorder

Sometimes the order you are given is unfriendly. Take ax + by + bx + ay.

If you pair as written: (ax + by) + (bx + ay). From the first pair you cannot pull anything useful — a divides ax but not by. Dead end.

Now reorder. Group the a terms together and the b terms together:

(ax + ay) + (bx + by)
a(x + y) + b(x + y)
(a + b)(x + y)

The expression was always factorable. The pairing just had to be smarter. If the first pairing fails, reorder before giving up.

Worked example 3 — requires reordering

Factor 2x³ + 5x - 6x² - 15.

The given order mixes powers and constants. Pair commensurate terms — group the cubic with the quadratic, and the linear with the constant:

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

From the first pair, 2x² is common. From the second, 5 is common:

2x^2(x - 3) + 5(x - 3)

Brackets match. Pull out (x - 3):

(2x^2 + 5)(x - 3)

Notice the rearrangement was not arbitrary — you grouped terms that shared a polynomial factor.

When grouping FAILS completely

Sometimes nothing works. Try x⁴ + x³ + x + 2.

Pair as written: (x⁴ + x³) + (x + 2).

x^3(x + 1) + 1(x + 2)

Brackets are (x + 1) and (x + 2). They do not match.

Try reordering. Pair (x⁴ + x) + (x³ + 2):

x(x^3 + 1) + (x^3 + 2)

Different cubic brackets. No match.

Try (x⁴ + 2) + (x³ + x):

(x^4 + 2) + x(x^2 + 1)

The first piece does not factor nicely. No grouping path works. This polynomial likely does not factor over rationals, or needs a more advanced method (like the rational roots theorem). Grouping was the right thing to try first; you just confirmed it does not apply.

Alternative — the "3 terms + 1" pattern

Some 4-term expressions are not 2+2 groupings. They are 3+1, where three terms form a perfect square and the fourth is a square on its own.

Take x² + 2xy + y² - 9. The first three terms are (x + y)². The fourth is 9 = 3². So:

(x + y)^2 - 9 = (x + y)^2 - 3^2

That is a difference of squares:

(x + y - 3)(x + y + 3)

Whenever you see 4 terms and three of them look like a perfect-square trinomial, this is your pattern. Group as 3+1, not 2+2.

Worked example 4 — the disguised difference of squares

Factor x² - y² + 6x + 9.

This looks like 4 random terms. But rearrange:

(x^2 + 6x + 9) - y^2

The first three are (x + 3)². So:

(x + 3)^2 - y^2

Difference of squares:

(x + 3 - y)(x + 3 + y)

The lesson — when 2+2 grouping fails, scan for a perfect-square trinomial hiding inside.

Why 4 terms especially prefers grouping

There is a reason 4 terms and grouping go together. Most 4-term polynomials in your textbook came from expanding a product of two binomials:

(ax + b)(cx + d) = acx^2 + adx + bcx + bd

That is 4 terms before you collect like terms. Grouping is literally the reverse operation — you are undoing the FOIL expansion. So 4 terms is the natural fingerprint of "two binomials multiplied", and grouping is how you recover those binomials.

This is also why grouping fails sometimes — not every 4-term polynomial came from a clean binomial product. But most do.

Exception — 4 terms that are already GCF-pullable

Always check the GCF before you start grouping. Take 2x + 4y + 6z + 8w.

Every term is divisible by 2:

2(x + 2y + 3z + 4w)

The inside still has 4 terms, but no common binomial factor — so no further grouping. Sometimes pulling the GCF is the entire factorization.

The rule — pull the GCF first. If what remains still has 4 terms, then try grouping.

Recognition drill

Quick reps. For each, identify the pairing and factor:

6x + 4y + 9x² + 6xy — reorder to put quadratic and mixed-quadratic together:

(9x^2 + 6xy) + (6x + 4y) = 3x(3x + 2y) + 2(3x + 2y) = (3x + 2)(3x + 2y)

a³ - a² + a - 1 — pair as given:

(a^3 - a^2) + (a - 1) = a^2(a - 1) + 1(a - 1) = (a^2 + 1)(a - 1)

x² + 3x - 2x - 6 — pair as given, watch the signs:

(x^2 + 3x) + (-2x - 6) = x(x + 3) - 2(x + 3) = (x - 2)(x + 3)

pq + p + q + 1 — pair as given:

(pq + p) + (q + 1) = p(q + 1) + 1(q + 1) = (p + 1)(q + 1)

Each took under 30 seconds because the pattern is the same every time.

Common confusions

"Grouping requires a specific order." No — you choose the pairing. If (first two) + (last two) fails, try (first and third) + (second and fourth). The expression does not care; only the pairing matters.

"If the first pairing fails, the polynomial cannot be factored." Wrong. Try other pairings. Try the 3+1 perfect-square pattern. Only after exhausting those should you suspect the polynomial does not factor over rationals.

"Grouping always works on 4 terms." Also wrong, but rarely. About 90% of exam-style 4-term expressions yield to grouping. The remaining 10% are usually disguised difference-of-squares (the 3+1 pattern) or genuinely irreducible.

The bottom line

4 terms → grouping first. The procedure is mechanical — pair, factor each pair, check that the brackets match, pull the bracket out. If the obvious pairing fails, reorder. If grouping still fails, scan for a perfect-square trinomial hiding inside (the 3+1 pattern). About 90% of the time, one of these two paths works. The other 10% of the time, the polynomial probably does not factor over the rationals, and you reach for heavier tools — but grouping was still the right thing to try first.

When in doubt, count the terms. If you see 4, your hand should already be moving toward a pair of brackets.