1Cademy - Factoring $$12y^2

Learn Before

Strategy to Factor Polynomials Completely

Example

Factoring $12y^2 - 75$

Factor $12y^2 - 75$ completely by first extracting the GCF and then applying the difference of squares pattern.

Step 1 — Is there a GCF? Yes. The two terms $12y^2$ and $75$ share a common factor of $3$ . Factor it out:

$12y^2 - 75 = 3(4y^2 - 25)$

Step 2 — Classify the expression inside the parentheses. The expression $4y^2 - 25$ is a binomial. It is not a sum — it is a difference. Is it a difference of squares? Yes: $4y^2 = (2y)^2$ and $25 = 5^2$ , so $4y^2 - 25 = (2y)^2 - 5^2$ .

Step 3 — Factor as a product of conjugates. Apply the difference of squares pattern $a^2 - b^2 = (a - b)(a + b)$ with $a = 2y$ and $b = 5$ :

$3((2y)^2 - 5^2) = 3(2y - 5)(2y + 5)$

Step 4 — Check.

Is the expression factored completely? Yes — neither binomial $(2y - 5)$ nor $(2y + 5)$ is itself a difference of squares, so no further factoring is possible.
Verify by multiplying: $3(2y - 5)(2y + 5) = 3(4y^2 - 25) = 12y^2 - 75$ ✓.

The completely factored form is $3(2y - 5)(2y + 5)$ . This example demonstrates a two-step factoring process: extracting the GCF first reveals a difference of squares hidden inside the parentheses. Without factoring out the GCF of $3$ first, the original binomial $12y^2 - 75$ does not look like a difference of squares — neither $12$ nor $75$ is a perfect square. The GCF step transforms the expression into $3(4y^2 - 25)$ , where the difference of squares structure $(2y)^2 - 5^2$ becomes visible.

Updated 2026-04-21

Contributors are:

Who are from:

References

Learn Before

Related

Learn After