1Cademy - Factoring $$3x^5y

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Strategy to Factor Polynomials Completely

Example

Factoring $3x^5y - 48xy$

Factor $3x^5y - 48xy$ completely by extracting the GCF and repeatedly applying the difference of squares pattern.

Step 1 — Check for a GCF: The two terms $3x^5y$ and $48xy$ share a numerical factor of $3$ and variable factors $x$ and $y$ . The GCF is $3xy$ . Factor it out: $3xy(x^4 - 16)$

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

Step 3 — Factor as a product of conjugates: $3xy((x^2)^2 - 4^2) = 3xy(x^2 - 4)(x^2 + 4)$

Step 4 — Check each binomial factor for further factoring: The first binomial $x^2 - 4$ is itself a difference of squares: $x^2 = x^2$ and $4 = 2^2$ . Factor it again: $3xy(x - 2)(x + 2)(x^2 + 4)$ The second binomial $x^2 + 4$ is a sum of squares, which does not factor further.

Step 5 — Check: The expression is factored completely. Verify by multiplying: $3xy(x - 2)(x + 2)(x^2 + 4) = 3xy(x^2 - 4)(x^2 + 4) = 3xy(x^4 - 16) = 3x^5y - 48xy$ ✓.

The completely factored form is $3xy(x - 2)(x + 2)(x^2 + 4)$ .

Updated 2026-04-30

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OpenStax Intermediate Algebra

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