The special products work backward too. If you recognize the pattern, you can factor in one step instead of searching for numbers, as you did for general trinomials.
Why it works
Because $(a + b)(a - b) = a^2 - b^2$, any difference of two squares factors: $x^2 - 49 = (x + 7)(x - 7)$. Because $(a + b)^2 = a^2 + 2ab + b^2$, a trinomial whose first and last terms are squares and whose middle term is twice their product is a perfect square. A sum of squares, like $x^2 + 9$, does not factor with real numbers: $x^2 + 9$ is at least $9$ for every real $x$, so it is never $0$, while a factor $(x - r)$ would make it $0$ at $x = r$.
| Pattern | Factors as | Example |
|---|---|---|
| $a^2 - b^2$ | $(a + b)(a - b)$ | $9x^2 - 25 = (3x + 5)(3x - 5)$ |
| $a^2 + 2ab + b^2$ | $(a + b)^2$ | $x^2 + 10x + 25 = (x + 5)^2$ |
| $a^2 - 2ab + b^2$ | $(a - b)^2$ | $4x^2 - 12x + 9 = (2x - 3)^2$ |
| $a^2 + b^2$ | does not factor | $x^2 + 9$ |
Key idea
Look for squares. Square minus square: $(a + b)(a - b)$. Square, twice the product, square: $(a \pm b)^2$. A sum of two squares does not factor over the real numbers.
Worked examples
Worked example: Difference of squares
Factor $16x^2 - 81$.
- $16x^2 = (4x)^2$ and $81 = 9^2$, and they are subtracted.
Answer: $(4x + 9)(4x - 9)$
Worked example: Perfect square trinomial
Factor $4x^2 - 12x + 9$.
- $4x^2 = (2x)^2$ and $9 = 3^2$.
- Middle check: $2 \cdot 2x \cdot 3 = 12x$, with a minus sign.
Answer: $(2x - 3)^2$
Common mistake
Factoring $x^2 + 9$ as $(x + 3)(x - 3)$. Multiply it back: $(x + 3)(x - 3) = x^2 - 9$, the wrong sign. A sum of squares stays as it is; it is already factored completely over the real numbers.