A two-term polynomial is a narrow case. Once you have pulled out any common factor, only three identities can split a binomial into a product, and you can scan for them in seconds. If none fit, the binomial almost certainly does not factor over the reals — the right move is to stop, not invent something.
Scanning order matters. Squares come first because they are the cleanest and most common pattern. Cubes come second. Anything else is a dead end.
The three binomial identities
These are the only three patterns that turn a two-term expression into a product over the reals:
Memorise all three as a unit. The cubes are not symmetric — the linear factor takes the same sign as the original expression, and the quadratic factor has the opposite middle sign.
NOT factorable over the reals
There is a fourth pattern that looks like it should fit but does not:
A sum of squares cannot be written as a product of two real polynomials of lower degree. If you try, you end up with complex numbers (a + bi)(a - bi) — not allowed in real factoring. See the sibling on the x^2 + y^2 misconception for the longer story.
Recognition algorithm (after GCF)
Once any common factor is pulled out front, walk down this checklist:
- Both terms perfect squares and sign between them minus? Difference of squares.
- Both terms perfect cubes and sign minus? Difference of cubes.
- Both terms perfect cubes and sign plus? Sum of cubes.
- None of the above? Probably does not factor over the reals. Stop.
Walk the list once, top to bottom. No backtracking, no guessing.
Recognising squares
Perfect squares of small integers — these should be instant:
For variables, a square is any even exponent: x^2, x^4, x^6. The square root halves the exponent: \sqrt{x^6} = x^3. Mixed terms work the same way: 25x^2 = (5x)^2, 49x^4 = (7x^2)^2.
Recognising cubes
Perfect cubes of small integers — keep this list in your head too:
For variables, a cube is any exponent that is a multiple of 3: x^3, x^6, x^9. The cube root divides the exponent by 3: \sqrt[3]{x^9} = x^3.
Notice that 64 appears in both lists (8^2 and 4^3), and x^6 appears in both ((x^3)^2 and (x^2)^3). This double-citizenship causes the only real ambiguity in two-term factoring.
Overlap: when x^6 is both a square and a cube
Consider x^6 - 1. You can read it two ways:
- As a difference of squares: (x^3)^2 - 1^2 = (x^3 - 1)(x^3 + 1).
- As a difference of cubes: (x^2)^3 - 1^3 = (x^2 - 1)(x^4 + x^2 + 1).
Both are valid first moves, but the squares-first path produces cleaner factors that break down further:
The cubes-first path leaves (x^2 - 1)(x^4 + x^2 + 1), where the second factor is harder to crack. That is why the algorithm puts squares before cubes.
Worked example 1 — difference of squares
Both terms are perfect squares (x^2 and 4^2), the sign is minus. Difference of squares applies with a = x, b = 4:
Worked example 2 — difference of squares with a coefficient
25x^2 = (5x)^2 and 9 = 3^2. Both squares, sign is minus, so a = 5x and b = 3:
The coefficient does not change the pattern — it just changes what a is.
Worked example 3 — difference of cubes
Both perfect cubes (x^3 and 2^3), sign is minus. Difference of cubes with a = x, b = 2:
The linear factor takes the minus, and the quadratic factor has +ab as its middle term.
Worked example 4 — sum of cubes
Both cubes (x^3 and 3^3), sign is plus. Sum of cubes with a = x, b = 3:
Now the linear factor takes the plus, and the quadratic factor has -ab in the middle. Notice the sign flip between the two factors — this is the easiest place to slip up.
Worked example 5 — neither pattern
Both perfect squares, sign is plus — sum of squares. Does not factor over the reals. Stop. Do not write fake products like (x + 2)(x + 2) — that expands to x^2 + 4x + 4, not x^2 + 4. The honest answer is "irreducible over \mathbb{R}" and you move on.
Worked example 6 — overlap, fully factored
x^6 = (x^3)^2 and 64 = 8^2, so this is a difference of squares first:
Now each factor is itself a sum or difference of cubes:
Combining everything:
Four factors in total. Squares first delivered the cleaner decomposition.
Common confusions
- "Two terms means difference of squares." Only when both terms are perfect squares and the sign is minus. x^2 + 5 has two terms but is not a difference of squares — 5 is not a perfect square and the sign is plus.
- "A sum of squares factors with a sign flip." No. a^2 + b^2 does not equal (a + b)(a - b); that would be a^2 - b^2. The sum of squares stays put over the reals.
- "a^2 - b^2 and a - b are the same pattern." They are not. a - b is a linear binomial with no further factoring. The squares identity needs the squared terms — without them, there is nothing to split.
Recognition drill
For each binomial, name the pattern (or "none") and write the factored form:
- x^2 - 25 → difference of squares → (x - 5)(x + 5).
- x^2 + 25 → sum of squares → not factorable over the reals.
- x^3 - 125 → difference of cubes → (x - 5)(x^2 + 5x + 25).
- x^3 + 64 → sum of cubes → (x + 4)(x^2 - 4x + 16).
- 4x^2 - 9 → difference of squares → (2x - 3)(2x + 3).
- x^4 - 1 → difference of squares → (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1).
- x^6 + 8 → sum of cubes (treat x^6 as (x^2)^3 and 8 as 2^3) → (x^2 + 2)(x^4 - 2x^2 + 4).
Run through all seven without looking back. When the routing feels mechanical, you have internalised the pattern.
GCF first, always
Before any of this, pull out the common factor. The identities are tuned for clean binomials.
2x^2 - 18 on its own is not a difference of squares — 2x^2 is not a perfect square. Pull out the 2 and the inside becomes the textbook pattern. Same with cubes: 5x^3 - 40 = 5(x^3 - 8) = 5(x - 2)(x^2 + 2x + 4).
Closing
Two terms equals a three-identity scan. Squares first, then cubes, then cubes with the sign flipped. Pull the GCF out before you start. If none fit — sum of squares, or terms that are not perfect powers — do not force it. Write "irreducible over \mathbb{R}" and move on. Forcing a factoring that does not exist is the most common way to lose marks on a binomial question.