Question

select the correct answer.
what is the complete factorization of $36y^2 - 1$?
a. $6(y + 1)(y - 1)$
b. $(6y + 1)(6y - 1)$
c. $(6y - 1)^2$
d. $(y - 6)(y + 6)$

Explanation:

Step1: Recognize difference of squares

The expression $36y^2 - 1$ fits $a^2 - b^2$, where $a=6y$, $b=1$.

Step2: Apply difference of squares rule

Use $a^2 - b^2=(a+b)(a-b)$. Substitute $a=6y$, $b=1$:
$$(6y+1)(6y-1)$$

Step3: Verify other options

Option A expands to $6y^2-6$, Option C expands to $36y^2-12y+1$, Option D expands to $y^2-36$; none match $36y^2-1$.

Answer:

B. $(6y + 1)(6y - 1)$