Factoring by Grouping: Complete Guide with Examples
A teaching-style walkthrough of the method, with five practice problems and answers at the end.
What Problem Is This Method Solving?
Most factoring techniques are built for a specific shape: three terms for a trinomial, two terms for a difference of squares. Grouping exists for the shape those methods can’t touch — four terms, with no single factor common to all of them.
The insight behind it is that a four-term polynomial can often be split into two pairs, where each pair shares its own separate factor, and after pulling those out, both pairs happen to leave behind the exact same expression. That shared leftover becomes one factor of the final answer.
The Method, One Step at a Time
1. Pair up the terms. Group the first two terms together and the last two together.
2. Factor each pair separately. Pull the greatest common factor out of each pair on its own.
3. Compare what’s left. If both pairs left behind an identical expression in parentheses, that’s your signal the grouping worked.
4. Factor out the shared piece. That common expression becomes one factor; the two things you originally pulled out combine to form the other.
Worked Example — Positive Terms Throughout
Pair it: (x³ + 4x²) + (3x + 12). First pair’s GCF is x², leaving x²(x + 4). Second pair’s GCF is 3, leaving 3(x + 4). Both left behind (x + 4) — factor it out: (x + 4)(x² + 3).
Worked Example — A Coefficient to Track
Pair it: (2x³ + 6x²) + (5x + 15). First pair’s GCF is 2x², leaving 2x²(x + 3). Second pair’s GCF is 5, leaving 5(x + 3). Shared binomial: (x + 3). Result: (x + 3)(2x² + 5).
Notice the two GCFs here, 2x² and 5, don’t need to match each other at all — only the binomial left behind after dividing them out has to match.
Worked Example — Subtraction Throughout
Pair it: (x³ − 2x²) + (−9x + 18). First pair’s GCF is x², leaving x²(x − 2). Second pair’s GCF is −9 (not 9 — the leading sign of the pair decides this), leaving −9(x − 2). Shared binomial: (x − 2). Result: (x − 2)(x² − 9).
Since (x² − 9) is a difference of squares, this doesn’t have to stop here — it factors further into (x − 3)(x + 3), giving a fully factored form of (x − 2)(x − 3)(x + 3).
Practice Problems
Try these on paper first, then check the answers underneath.
- x³ + 5x² + 2x + 10
- 3x³ − 6x² + 4x − 8
- x³ + x² − 4x − 4
- 4x³ + 12x² − 3x − 9
- x³ − 3x² − x + 3
Answers: (1) (x + 5)(x² + 2) — (2) (3x² + 4)(x − 2) — (3) (x + 1)(x² − 4), which factors further to (x + 1)(x − 2)(x + 2) — (4) (4x² − 3)(x + 3) — (5) (x − 3)(x² − 1), which factors further to (x − 3)(x − 1)(x + 1).
