Math101 · Independent Ontario learning library
Math101learn.math101.caFactoring by Grouping
Factoring by grouping rewrites a polynomial as groups with a shared binomial or other common factor. For four terms, pair terms, factor the GCF from each pair, then factor the repeated bracket.
Grouping factors polynomials that do not have a single GCF and reveals the structure behind many cubic and trinomial factorizations.
Intuition and core definition
Factoring by grouping rewrites a polynomial as groups with a shared binomial or other common factor. For four terms, pair terms, factor the GCF from each pair, then factor the repeated bracket. The method is valid because it applies the distributive property twice in reverse.
Notation, language, and conditions
$ax+ay+bx+by=a(x+y)+b(x+y)=(a+b)(x+y)$. A negative GCF may be chosen from one group so the bracket expressions match exactly. Grouping also supports splitting the middle term of a trinomial after finding numbers with a required product and sum.
Why this idea matters
Factoring by grouping exposes a shared binomial by first creating common factors within carefully chosen pairs of terms.
A dependable method
- Remove a GCF common to all terms.
- Arrange and group terms so each group has a useful GCF.
- Factor the GCF from each group, taking a negative factor if needed.
- Confirm the remaining group factors are identical.
- Factor that shared expression and expand to verify.
Worked example
Representations and interpretation
A two-row area box places terms so row or column GCFs label sides. The shared binomial is visible as a repeated dimension, and the remaining GCFs form the other factor.
Reasoning about variations
The first grouping may fail even when another ordering works. For $ax+ay+bx+by$, adjacent grouping works; for a rearranged expression, reorder terms before grouping, since addition is commutative.
Common mistakes
How to check your work
- Expand the final product term by term.
- Compare leading term, constant term, and total term count.
- Verify the alleged common bracket matches exactly in both groups.
Practice
- Factor $x^3+3x^2+2x+6$.
- Factor $5a^2-10a+3a-6$.
- What should you try if two grouped brackets are opposites?
Answers and brief solutions
Show answers
- $(x+3)(x^2+2)$ $x^2(x+3)+2(x+3)$.
- $(a-2)(5a+3)$ $5a(a-2)+3(a-2)$.
- Factor a negative from one group This can make the bracket factors identical.
Synthesis and transfer
A four-term area expression can be rearranged into two strips with one common width; factoring each strip makes that repeated dimension visible as the final binomial factor.
For $6x^2+9x+4x+6$, group the first two and last two terms to obtain $3x(2x+3)+2(2x+3)$. The shared binomial then gives $(3x+2)(2x+3)$. A different initial grouping may fail to expose a match, so rearranging terms is allowed because addition is commutative. The method succeeds only when the residual factors are identical up to a sign; factoring out a negative can repair an apparent mismatch. Expanding the two final binomials reconstructs all four original terms and confirms that neither a sign nor a term was lost during grouping.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Factor $x^3+3x^2+2x+6$.
- $x^2(x+3)+2(x+3)$.
End of lesson
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