Split a four-term polynomial into two pairs, factor each, and pull out what they share.
A student at Jericho High School was asked to factor x3+3x2+2x+6 on a test. She tried every shortcut she knew — difference of squares, perfect square trinomial, simple GCF — and none of them worked. She left it blank. But there was a method she had not learned yet: factoring by grouping. It turns out this polynomial factors cleanly into (x2+2)(x+3), and the method to get there is completely systematic.
Factoring by grouping works on polynomials with four terms. The idea is to split the four terms into two pairs, factor the GCF out of each pair separately, and then check whether what remains inside the parentheses is identical for both pairs. If it is, you factor that shared binomial out as a single unit.
Here is the general structure. Given a four-term polynomial, you group it as:
(ax3+bx2)+(cx+d)
Then you pull the GCF from each group:
x2(ax+b)+c(x+),
That general form gets messy fast, so it is easier to just walk through examples. The key question you are always asking yourself is: after I factor each pair, do the two parenthetical expressions match?
Factor x³ + 3x² + 2x + 6
(x3+3x2)+(2x+6)
Split into two pairs. The first two terms go together, the last two go together.
x2(x+3)+2(x+3)
Factor the GCF from each pair. x² comes out of the first, 2 comes out of the second. The thing left in each parenthesis is (x + 3) — they match.
(x2+2)(x+3)
Since both pairs share (x + 3), factor it out. The x² and the 2 become one new factor.
(x2+2)(x+3)✓
Done. You can verify by expanding: x³ + 3x² + 2x + 6.
That matching step is what makes or breaks the method. If the two binomials do not match after you factor each group, try rearranging the original four terms. Sometimes the order the problem gives you is not the order that leads to a match.
Factor 2x³ - x² + 6x - 3
(2x3−x2)+(6x−3)
Group the first pair and the second pair.
x2(2x−1)+3(2x−1)
GCF of the first group is x², GCF of the second is 3. Both groups now show (2x - 1) — they match.
(x2+3)(2x−1)
Factor out the shared binomial (2x - 1). The leftover pieces, x² and 3, form the other factor.
(x2+3)(2x−1)✓
Done. Multiply out to check: 2x³ - x² + 6x - 3.
One thing to watch for: when the second pair has a negative leading term, you need to factor out a negative GCF. This keeps the binomials identical. If you factor out a positive from the second pair and the signs inside do not match the first pair, try factoring out a negative instead.
Factor x³ - 5x² - 3x + 15
(x3−5x2)+(−3x+15)
Split into pairs. Keep the negative with the third term.
x2(x−5)+(−3)(x−5)
GCF of the first group is x². GCF of the second group is -3, not 3 — pulling out -3 turns -3x + 15 into -3(x - 5), which matches the first pair.
(x2−3)(x−5)
Factor out the shared binomial (x - 5). The remaining pieces are x² and -3.
(x2−3)(x−5)✓
Check by expanding: x³ - 5x² - 3x + 15.
Sometimes the polynomial looks like a trinomial but has a hidden grouping. If a problem gives you something like x3+x2+x+1, the grouping still works — do not let the uniform coefficients throw you off.
Interactive graph — scroll to zoom, drag to pan
The two expressions in the graph are identical — the red and blue curves lie exactly on top of each other. That overlap confirms that factoring by grouping does not change the polynomial, only the way it is written.
Factor each polynomial completely using grouping.
Practice Questions
x3+2x2+3x+6
3x3−6x2+x−2
x3+4x2−2x−8
Regents Corner
On the Algebra II Regents, factoring by grouping appears on Part II and Part III, where partial credit is available. The grader needs to see your grouping step and your GCF step — skipping straight to the answer without showing the intermediate factored pairs will cost you a point even if your final answer is correct.
When the second pair starts with a negative term, students factor out a positive GCF and write something like x²(x - 5) + 3(-x + 5). The binomials (x - 5) and (-x + 5) look similar but are not the same — they are opposites, and the method fails. Factor out -3 instead to get -3(x - 5), which matches. Any time your binomials look like negatives of each other, that is the fix.