1Cademy - Factoring $$x^4

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Difference of Squares Pattern

Example

Factoring $x^4 - y^4$

Factor $x^4 - y^4$ completely by applying the difference of squares pattern twice in succession — once to the original expression and again to one of the resulting factors.

Step 1 — Is this a difference of squares? Yes. Rewrite each fourth power as a squared quantity: $x^4 = (x^2)^2$ and $y^4 = (y^2)^2$ , so the expression becomes $(x^2)^2 - (y^2)^2$ .

Step 2 — Factor as the product of conjugates: Apply the pattern $a^2 - b^2 = (a - b)(a + b)$ with $a = x^2$ and $b = y^2$ :

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

Step 3 — Check each factor for further factoring. The first binomial $x^2 - y^2$ is itself a difference of squares: $(x)^2 - (y)^2$ . Apply the pattern again:

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

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

Step 4 — Check by multiplying:

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

The completely factored form is $(x - y)(x + y)(x^2 + y^2)$ . This example demonstrates that a fourth-power difference can be treated as a difference of squares by viewing each fourth power as the square of a second power. After the first application of the pattern, one of the resulting factors may itself be a difference of squares that factors further, while the sum of squares factor cannot be factored and remains as is.

Updated 2026-04-21

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OpenStax Elementary Algebra

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