A polynomial is factored completely when none of its factors (other than numbers and monomials) can be factored again. It usually takes more than one step, using the tools you now have: the GCF, trinomials, special products and grouping.
- GCF first. Take out the greatest common factor, if it is not $1$.
- Count the terms of what is left. Two terms: a difference of squares? Three terms: a perfect square, or a trinomial to factor with two numbers or the ac method. Four terms: grouping.
- Look at every factor again. Can any of them be factored further?
- Check by multiplying back.
Por qué funciona
Taking out the GCF first makes everything smaller and easier: $3x^2 - 12$ hides the pattern, but $3(x^2 - 4)$ shows a difference of squares at once. And a factor like $(x^2 - 4)$ left inside a product is not finished, because it splits into $(x + 2)(x - 2)$. Factoring completely matters later, when the factors give the solutions of equations.
Idea clave
GCF first, then the pattern for the number of terms, then check each factor again until nothing more factors.
Worked examples
Ejemplo resuelto: GCF, then difference of squares
Factor $3x^3 - 12x$ completely.
- GCF $3x$: $3x(x^2 - 4)$.
- $x^2 - 4$ is a difference of squares: $(x + 2)(x - 2)$.
Respuesta: $3x(x + 2)(x - 2)$
Ejemplo resuelto: GCF, then a trinomial
Factor $2x^2 - 8x - 42$ completely.
- GCF $2$: $2(x^2 - 4x - 21)$.
- Numbers with product $-21$ and sum $-4$: $-7$ and $3$.
Respuesta: $2(x - 7)(x + 3)$
Ejemplo resuelto: Twice a difference of squares
Factor $x^4 - 16$ completely.
- $x^4 - 16 = (x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4)$.
- $x^2 - 4 = (x + 2)(x - 2)$; $x^2 + 4$ is a sum of squares and stays.
Respuesta: $(x^2 + 4)(x + 2)(x - 2)$
Error común
Stopping after one step, as in $3x^3 - 12x = 3x(x^2 - 4)$. That is correct but not complete: $x^2 - 4$ still factors. After every step, look at each factor and ask whether it matches a pattern.