1Cademy - Factoring $$2x^4

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Strategy to Factor Polynomials Completely

Example

Factoring $2x^4 - 32$

Factor $2x^4 - 32$ completely by extracting the GCF and then applying the difference of squares pattern twice in succession.

Step 1 — Is there a GCF? Yes, the GCF of $2x^4$ and $32$ is $2$ . Factor it out:

$2x^4 - 32 = 2(x^4 - 16)$

Step 2 — Classify the expression inside the parentheses. The expression $x^4 - 16$ is a binomial. Is it a difference of squares? Yes: $x^4 = (x^2)^2$ and $16 = 4^2$ , so $x^4 - 16 = (x^2)^2 - 4^2$ .

Step 3 — Factor as a product of conjugates:

$2((x^2)^2 - 4^2) = 2(x^2 - 4)(x^2 + 4)$

Step 4 — Check each factor for further factoring. The first binomial $x^2 - 4$ is itself a difference of squares: $x^2 = x^2$ and $4 = 2^2$ , so apply the pattern again:

$2(x^2 - 4)(x^2 + 4) = 2(x - 2)(x + 2)(x^2 + 4)$

The second binomial $x^2 + 4$ is a sum of squares, which does not factor.

Step 5 — Check.

Is the expression factored completely? Yes — none of the remaining binomials is a difference of squares.
Verify by multiplying: $2(x - 2)(x + 2)(x^2 + 4) = 2(x^2 - 4)(x^2 + 4) = 2(x^4 - 16) = 2x^4 - 32$ ✓.

The completely factored form is $2(x - 2)(x + 2)(x^2 + 4)$ . This example illustrates that after one application of the difference of squares pattern, the resulting factors must be re-examined — one of them may itself be a difference of squares that factors further. The sum of squares factor $x^2 + 4$ cannot be factored, so the process stops there.

Updated 2026-04-21

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