Lesson · Polynomials and Factoring · Polynomial arithmetic

Special Products

The difference of squares, the two square-of-a-binomial forms and the cube patterns are ordinary products of “Multiplying Polynomials”. They are singled out because each one runs backwards in “Factoring Special Forms”, where recognition is the whole method.

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(a + b)(a − b) = a² − b² (a + b)² = a² + 2ab + b² (a − b)² = a² − 2ab + b² (a + b)(a² − ab + b²) = a³ + b³ (a − b)(a² + ab + b²) = a³ − b³

The patterns, forwards

Core ideas

What this lesson establishes

Nothing here is new mathematics. What is new is reading a product backwards — seeing 9x² − 25 and thinking "difference of squares" rather than "a quadratic".

These are theorems, not conventions

Each identity is proved in one line by distributing. If you ever doubt a sign, expand it — the derivation is shorter than the doubt.

The middle term is the whole difficulty

(a + b)² is a² + 2ab + b², not a² + b². Squaring distributes over multiplication but never over addition, and this is the error that survives longest into later courses.

Recognition is what you are training

a and b stand for whole expressions, not just letters. In (3x + 5)(3x − 5) you have a = 3x and b = 5, and the answer 9x² − 25 needs no working at all.

The material

The patterns and where they come from

Each derived once, then read as a shape you can spot.

Theorem · Difference of squares

(a + b)(a − b) = a² − b² for all a and b.

Distribute the four products: (a + b)(a − b) = a² − ab + ba − b².

The middle terms −ab and +ba are like terms with opposite signs, so they sum to zero, leaving a² − b².

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The cancellation of the middle is the entire content. It happens because the two brackets differ only in one sign, which is why the pattern is worth naming: three terms collapse to two.

Theorem · Square of a binomial

(a + b)² = a² + 2ab + b² and (a − b)² = a² − 2ab + b².

(a + b)² = (a + b)(a + b) = a² + ab + ba + b² = a² + 2ab + b².

For the second, replace b by −b: the b² term is unchanged because (−b)² = b², and the middle term becomes −2ab.

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So the two forms are one result. The last term is always positive — it is a square — and only the middle sign follows the sign in the bracket.

The claim (a + b)² = a² + b² is false, and one number settles it: at a = 1, b = 3 the left side is (1 + 3)² = 16 and the right side is 1 + 9 = 10. Whenever you are unsure of an algebraic identity, test it on small numbers. A single counterexample is a complete disproof.

The cube patterns

Two more, used in “Factoring Special Forms” and rarely anywhere else:

(a + b)(a² − ab + b²) = a³ + b³ (a − b)(a² + ab + b²) = a³ − b³

For the first: distributing gives a³ − a²b + ab² + a²b − ab² + b³. The a²b terms cancel and the ab² terms cancel, leaving a³ + b³. The second is the same computation with b replaced by −b.

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Note the signs, because they are counter-intuitive: the binomial carries the same sign as the answer, and the middle term of the trinomial carries the opposite one. The trinomial factor is not a perfect square — it has ab, not 2ab.

Also worth having: (a + b)³ = a³ + 3a²b + 3ab² + b³, which is what you get by multiplying (a + b)² by (a + b) once more.

Example · Recognition in practice

(3x + 5)(3x − 5): same terms, opposite signs, so it is a difference of squares with a = 3x, b = 5, giving 9x² − 25. (2x − 7)²: a square of a binomial with a = 2x, b = 7, giving 4x² − 28x + 49 — the middle term is 2·(2x)·7 = 28x, subtracted.

Interactive

The special products

An identity and an honest expansion, computed separately

Method

Spotting a special product

Three questions. If all three fail, expand it the ordinary way.

Are the two brackets identical except for one sign?

If so it is a difference of squares: square the first term, square the second, subtract. The middle vanishes.

Is it a bracket squared?

Then it is a² ± 2ab + b². Identify a and b, and remember the middle term is twice their product — that factor of 2 is the part people drop.

Is it a binomial times a trinomial with no doubled middle term?

Check for the cube pattern: a² ∓ ab + b² alongside a ± b. If the trinomial has 2ab rather than ab, it is not this pattern.

Otherwise, distribute

The patterns are shortcuts, not obligations. An ordinary product expanded carefully is always right; a pattern applied to something that does not match never is.

Patterns

Score 0 / 0

Choose the answer you can justify.

Four products, by pattern

Each one names its pattern first, and only then substitutes.

1. (3x + 5)(3x − 5) difference of squares, a = 3x, b = 5 = (3x)² − 5² = 9x² − 25 2. (2x − 7)² square of a binomial, a = 2x, b = 7 = (2x)² − 2(2x)(7) + 7² = 4x² − 28x + 49 3. (x + 3)² a = x, b = 3 = x² + 6x + 9 NOT x² + 9 check at x = 1: (1 + 3)² = 16, and 1 + 6 + 9 = 16 4. (x + 2)(x² − 2x + 4) sum of cubes, a = x, b = 2 = x³ + 8

Line 3 is the one to fix in memory. The two sides of the false identity differ by 6x, and at x = 1 that is the difference between 16 and 10.

Line 4 is worth expanding by hand once, to watch the four middle terms annihilate each other. After that you can use it as a pattern with a clear conscience.

For a faded pattern check, identify a = 3x and b = 2 in (3x + 2)². The outer terms are supplied: 9x² and 4. Recover the middle term, then test x = 1. The expansion is 9x² + 12x + 4, and both forms give 25; 9x² + 6x + 4 uses ab where the identity requires 2ab.

Common mistakes

Writing (a + b)² = a² + b²The missing 2ab is the most persistent error in school algebra. Squaring does not distribute over addition, and one numerical test exposes it every time.

Halving or dropping the 2 in the middle term(2x − 7)² has middle term 2·(2x)·7 = 28x, not 14x. Both a and b go in, and so does the 2.

Expecting a² + ab + b² to be a perfect squareIt is not: the square has 2ab. The trinomial in the cube patterns is deliberately not a perfect square, and treating it as one sends you looking for a perfect-square factorisation that is not there.

Check your understanding

✓

Finish when you produce 9x² − 25 without writing a middle term at all.

Recognition should be immediate in both directions: given (3x + 5)(3x − 5) you write 9x² − 25, and given 9x² − 25 you should already be uneasy that it factors. “Factoring Special Forms” turns that unease into a method.

Note: These five identities are the only things in the course worth committing to memory, and even they can be re-derived in a line. Everything else is either another recognised identity or a finite search you can carry out, not an inspired guess you must somehow recall.