Some polynomials of degree 4 or 6 are really quadratics in disguise. It is the chunk idea again, combined with the trinomial factoring you learned in Factoring Completely.
Key idea
If the powers are $x^{2n}$, $x^n$ and a constant (like $x^4 - 5x^2 + 4$), let $u = x^n$. Factor the quadratic in $u$, put $x^n$ back, then check each factor again: differences of squares and cubes may split further.
Why it works
$x^4$ is $(x^2)^2$, so $x^4 - 5x^2 + 4$ is exactly $u^2 - 5u + 4$ with $u = x^2$. Any factoring of the quadratic in $u$ is a true identity for every value of $u$, including $u = x^2$. The substitution only changes how it looks.
Worked examples
Worked example: Four linear factors
Factor $x^4 - 5x^2 + 4$ completely.
- Let $u = x^2$: $u^2 - 5u + 4 = (u - 1)(u - 4)$.
- Put back: $(x^2 - 1)(x^2 - 4)$.
- Both are differences of squares: $(x - 1)(x + 1)(x - 2)(x + 2)$.
Answer: $(x - 1)(x + 1)(x - 2)(x + 2)$
Worked example: A factor that stops
Factor $x^4 + x^2 - 12$ completely over the integers.
- $u^2 + u - 12 = (u + 4)(u - 3)$, so $(x^2 + 4)(x^2 - 3)$.
- $x^2 + 4$ is a sum of squares and $3$ is not a perfect square, so neither factor splits over the integers.
Answer: $(x^2 + 4)(x^2 - 3)$
Worked example: Cubes inside
Factor $x^6 - 9x^3 + 8$.
- Let $u = x^3$: $(u - 1)(u - 8)$, so $(x^3 - 1)(x^3 - 8)$.
- Each is a difference of cubes: $x^3 - 1 = (x - 1)(x^2 + x + 1)$ and $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$.
Answer: $$(x - 1)(x^2 + x + 1)(x - 2)(x^2 + 2x + 4)$$
Common mistake
Stopping after the $u$ step. $(x^2 - 1)(x^2 - 4)$ is equal to the original, but it is not factored completely. After putting $x^n$ back, look at every factor again.