In short
Three rules to decide:
- Matches a known identity? ((a\pm b)^2, (a+b)(a-b), (a\pm b)^3) — use the identity. Faster, fewer slip errors.
- Generic factors that don't fit any identity? ((2x+3)(x+5), etc.) — just FOIL or use the box method. No shortcut available.
- High power on a binomial? ((x-3)^4, (2y+1)^5) — Pascal's triangle plus alternating signs crushes brute expansion.
You sit down with a problem: expand (2x + 5)^2. You have two paths. Path one — write it as (2x+5)(2x+5) and FOIL: First, Outer, Inner, Last. Path two — recognise the pattern (a+b)^2 and write a^2 + 2ab + b^2 directly. Both give the same answer. One takes 5 seconds; the other takes 20. In an exam where every minute matters, knowing when each tool wins is half the battle.
This article gives you a decision tree. Look at the expression. Does it match an identity you know? Use it. Does it not? FOIL. Is the power 3 or higher? Pascal's triangle. That's it.
Why the choice matters
Time and accuracy. A boards or JEE aspirant solves around 30 algebraic manipulations per paper. If each one takes 30 seconds longer than necessary, that's 15 minutes lost — enough to attempt three more questions. Why: identities collapse a multi-step distribution into a one-line pattern match, so your hand never has to track four separate products.
Slip errors are the second cost. Every time you write a - sign, every time you carry an exponent, every time you combine like terms, there's a small chance of a mistake. FOIL has four products to track; the identity (a-b)^2 = a^2 - 2ab + b^2 has three terms with fixed signs. Why: fewer intermediate steps means fewer places to drop a sign or miscopy a coefficient.
The decision tree
The three identity patterns to recognise instantly
Train your eye to spot these shapes the way you spot a familiar face in a crowd:
- Square of a binomial: (a \pm b)^2 — anything raised to the power 2 with two terms inside the bracket.
- Difference of squares disguise: (a + b)(a - b) — two binomials with the same terms but opposite middle signs.
- Cube of a binomial: (a \pm b)^3 — same structure as the square but power 3.
If your expression matches any of these, the identity wins every time. If it doesn't, move on.
Three worked comparisons
Example 1 — Identity wins clearly
Expand (2x + 5)^2.
The FOIL route. Write it as (2x+5)(2x+5). Then:
- First: 2x \cdot 2x = 4x^2
- Outer: 2x \cdot 5 = 10x
- Inner: 5 \cdot 2x = 10x
- Last: 5 \cdot 5 = 25
Sum: 4x^2 + 10x + 10x + 25 = 4x^2 + 20x + 25. Took about 20 seconds, four products to track, two like terms to combine.
The identity route. Pattern-match: (a+b)^2 with a = 2x, b = 5. Apply a^2 + 2ab + b^2:
About 5 seconds. Why: you wrote three terms directly from the formula instead of generating four products and merging two of them.
Same answer, four times faster.
Example 2 — No identity fits, just FOIL
Expand (2x + 3)(x + 5).
Look at it. The two binomials have different leading coefficients (2x vs x) and different constants (3 vs 5). It's not (a+b)^2 — the two factors aren't even the same. It's not (a+b)(a-b) — both signs are positive. No identity fits.
So FOIL, with no apology:
- First: 2x \cdot x = 2x^2
- Outer: 2x \cdot 5 = 10x
- Inner: 3 \cdot x = 3x
- Last: 3 \cdot 5 = 15
Sum: 2x^2 + 10x + 3x + 15 = 2x^2 + 13x + 15.
Why no shortcut: the identities only cover special structures (same factors, conjugate pairs, same base raised to a power). A generic product of two unrelated binomials has no symmetry to exploit. Don't waste 30 seconds searching for a pattern that isn't there — recognise the genericness and FOIL.
Example 3 — High power: Pascal's triangle crushes brute force
Expand (x - 3)^4.
The brute route. Write (x-3)(x-3)(x-3)(x-3). FOIL the first two: x^2 - 6x + 9. FOIL again with (x-3): x^3 - 9x^2 + 27x - 27. FOIL once more with (x-3). That's three rounds of multiplication, each producing more terms to combine. Easily 90 seconds, and the chance of a sign slip is high.
Pascal's triangle route. Row 4 of Pascal's triangle is 1, 4, 6, 4, 1. For (a-b)^4 the signs alternate +, -, +, -, +. The powers of a go 4, 3, 2, 1, 0 and powers of b go 0, 1, 2, 3, 4. With a = x, b = 3:
About 25 seconds, no FOILing, no like-term collection. Why: Pascal's triangle gives all five coefficients in one glance, and the alternating-sign rule handles the minus automatically — you skip the entire multiply-and-combine phase.
For (a + b)^n with n \geq 3, this approach scales linearly while FOIL scales like a small explosion.
A tiny caveat: don't force an identity
Sometimes students see (2x + 5)(2x + 7) and think "close enough to (a+b)^2". It isn't — the second terms differ. Don't bend the expression to fit an identity that almost works. Why: forcing a wrong identity introduces silent errors that are harder to catch than a clean FOIL mistake. If the pattern doesn't match exactly, FOIL.
The exception is small algebraic rewrites that create an identity:
This is a real technique — completing a difference of squares — and it's worth learning, but only after the basic decision tree feels automatic.
Exam strategy
In CBSE Class 9 and 10 boards, the polynomials chapter is identity-heavy on purpose: examiners reward the student who sees the pattern. In JEE Main and Advanced, identities save 30–60 seconds per problem on algebraic simplification, and that compounds across a 90-question paper. Why: JEE rewards both speed and accuracy, and identities improve both at once — a rare double win.
Train the recognition. Before you start writing, look at the expression for two seconds and ask: square? difference of squares? cube? Only if the answer is no should your pen touch the paper for FOIL.
References
- NCERT, Mathematics Textbook for Class IX, Chapter 2: Polynomials — official Indian syllabus introduction to identities.
- Khan Academy, Multiplying binomials by polynomials (FOIL and beyond) — interactive practice.
- Wikipedia, Binomial theorem — generalises Pascal's triangle to any non-negative integer power.
- Art of Problem Solving, Identities and FOIL — competition-math perspective on when to expand vs factor.