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Special Patterns: Squares and Cubes

Some polynomials factor instantly once you recognize the pattern: difference of two squares, perfect-square trinomials, and the sum and difference of cubes. Learn the four formulas and the tells that trigger them.

Difference of two squares

These patterns are special products read backwards. The cleanest is the [[difference-of-two-squares|difference of two squares]]: a^2 - b^2 = (a + b)(a - b). Two perfect squares with a minus between them split instantly. Note a *sum* of two squares, a^2 + b^2, does not factor over the real numbers — it is a prime polynomial.

9x^2 - 25 = (3x)^2 - 5^2 = (3x + 5)(3x - 5)

Apply it twice when squares hide inside squares:
x^4 - 16 = (x^2)^2 - 4^2
        = (x^2 + 4)(x^2 - 4)
        = (x^2 + 4)(x + 2)(x - 2)
     (x^2 + 4 is a sum of squares -> stops here)
x^4 − 16 needs the pattern applied twice to be completely factored.

Perfect-square trinomials

A [[perfect-square-trinomial|perfect-square trinomial]] is the square of a binomial expanded: a^2 + 2ab + b^2 = (a + b)^2, and a^2 - 2ab + b^2 = (a - b)^2. The tell: the first and last terms are perfect squares, and the middle term is exactly twice the product of their square roots. You could always grind it out with the AC method, but recognizing the pattern is faster and less error-prone.

Recognize x^2 + 12x + 36
  first term:  x^2  -> sqrt is x
  last term:   36   -> sqrt is 6
  middle test: 2 * x * 6 = 12x  ✓ matches

x^2 + 12x + 36 = (x + 6)^2

4x^2 - 20x + 25 = (2x)^2 - 2(2x)(5) + 5^2 = (2x - 5)^2
Check the middle term equals 2·√first·√last before claiming a perfect square.

Sum and difference of cubes

Unlike squares, cubes factor in both the sum and difference cases. [[sum-of-cubes|Sum of cubes]]: a^3 + b^3 = (a + b)(a^2 - ab + b^2). [[difference-of-cubes|Difference of cubes]]: a^3 - b^3 = (a - b)(a^2 + ab + b^2). A memory aid many use is “SOAP”: the binomial keeps the Same sign, the middle term of the trinomial is Opposite, and the last term is Always Positive.

x^3 + 8  = x^3 + 2^3
         = (x + 2)(x^2 - 2x + 4)

27x^3 - 1 = (3x)^3 - 1^3
         = (3x - 1)(9x^2 + 3x + 1)

The trinomial factor (x^2 - ab + b^2) rarely factors further -> usually prime.
Identify a and b as cube roots, then drop them into the formula.