Four terms with nothing in common — pair them up and a shared bracket appears.
Factor
x³ + 3x² + 2x + 6
Group in pairs: x²(x + 3) + 2(x + 3). Both halves now carry the same bracket, so it becomes (x + 3)(x² + 2). C swaps the roles — the shared bracket is the one you pulled out, and the leftovers form the other factor.
Nothing divides all four, so there is no common factor to pull out.
two pairs, taken in order
Two terms suggests a difference of squares or cubes. Three suggests a trinomial. Four almost always means grouping.
x³ + 3x² = x²(x + 3)
first pair
2x + 6 = 2(x + 3)
second pair
x²(x + 3) + 2(x + 3)
the brackets match
Matching brackets is the sign it is working. If they differ, something needs fixing.
x²(x + 3) + 2(x + 3)
treat (x + 3) as one object
(x + 3)(x² + 2)
what is left behind
The second bracket collects the outside pieces: the x² and the 2. It is the same move as factoring x²a + 2a into a(x² + 2).
Two fixes: flip a sign, or reorder the terms.
ax − ay + by − bx
the expression
a(x − y) + b(y − x)
brackets are reversed
a(x − y) − b(x − y)
factor out −b instead
(x − y)(a − b)
now they match
Reversed brackets are not a dead end. Pulling out a negative flips the order, and the two brackets line up.
For 2x² + 3y + 6x + xy, no adjacent pair helps — reorder to 2x² + 6x + xy + 3y first.
Splitting a trinomial's middle term turns three terms into four.
2x² − 5x − 3
ac = −6, pair +1 and −6
2x² + x − 6x − 3
now four terms
x(2x + 1) − 3(2x + 1)
group
(2x + 1)(x − 3)
same finish
Which is why grouping is worth being fluent in — it is the engine inside the factoring method you already use on quadratics.
Factor completely: x³ + 2x² − 25x − 50. Grouping often leaves a bracket that still factors.
x²(x + 2) − 25(x + 2)
take −25 from the second pair
(x + 2)(x² − 25)
not finished yet
(x + 2)(x − 5)(x + 5)
x² − 25 is a difference of squares
Three linear factors also give three roots: x³ + 2x² − 25x − 50 = 0 at x = −2, 5 and −5. Questions that ask “which is a factor” or “sum of the roots” are this same factoring with one more line.
Factor completely: 2x³ + x² − 2x − 1.
x²(2x + 1) − 1(2x + 1)
the second pair shares only −1
(2x + 1)(x² − 1)
the −1 stays behind
(2x + 1)(x − 1)(x + 1)
x = 2: 16 + 4 − 4 − 1 = 15 = 5 · 1 · 3 ✓
Different: when a pair looks like it has no common factor, it is 1 or −1 — write it in so it is not lost.
Writing x²(2x + 1) − (2x + 1) = (2x + 1)x² drops the −1 that was holding a place. Multiply back: x²(2x + 1) alone has no −2x or −1.
Factor 4x² + 3y + 12x + xy.
4x² + 3y: no common factor
pairs in the given order fail
(4x² + 12x) + (xy + 3y)
reorder: x² with x, y with y
4x(x + 3) + y(x + 3)
the brackets match
(x + 3)(4x + y)
x = y = 1: 20 = 4 · 5 ✓
Different: the terms arrive in an order that hides the grouping; pair terms that share a letter.
Pairs in order do not always work. Group the squares together: factor a² − b² + 5a + 5b.
(a² − b²) + (5a + 5b)
squares with squares
(a − b)(a + b) + 5(a + b)
(a + b) appears twice
(a + b)(a − b + 5)
take it out
Three plus one: when three terms form a perfect square trinomial, the fourth is usually a square being subtracted. Factor x² − 10x + 25 − 4y².
(x² − 10x + 25) − 4y²
group the first three
(x − 5)² − (2y)²
a difference of two squares
(x − 5 − 2y)(x − 5 + 2y)
A² − B² = (A − B)(A + B)
If the three square terms come last, as in 4y² − x² + 10x − 25, take out a minus: 4y² − (x² − 10x + 25) = (2y)² − (x − 5)². Every sign inside the bracket flips — miss one and the trinomial is no longer a perfect square.
Four-term items are usually “factor completely”, “which is NOT a factor”, or a cubic equation to solve. Group, then always check the leftover bracket for a difference of squares. To test a choice like (x − 5) quickly, put x = 5 into the original: a factor makes it zero.
Factor
2x³ − 5x² + 6x − 15
x²(2x − 5) + 3(2x − 5) exposes the shared bracket, giving (2x − 5)(x² + 3). The second pair must come out as +3(2x − 5) — pulling out −3 would leave (5 − 2x) and the brackets would no longer match.
