How do I factorize this?

3y^2 -12
3 is a common factor so
3(y^2-4)
(y^2-4) is sum and difference of squares
3(y^2-4)=3(y+2)(y-2)
3(y+2)(y-2) (answer)

3y² - 12 = 0
3y² = 12
y² = 4
y = +/- 2

Answer: 3(y + 2)(y - 2) are the factors.

Proof:
= 3(y + 2)(y - 2)
= 3(y² - 2y + 2y - 4)
= 3(y² - 4)
= 3y² - 12

a^2 - b^2 = (a + b)(a - b)

3y^2 - 12
= 3(y^2 - 4)
= 3(y^2 - 2^2)
= 3(y + 2)(y - 2)

3 (y^2 - 4) [common factor]
3 (y - 2) (y + 2) [sum and difference]

There are two terms here; 3y² and -12
They can be broken up into 3×y² and 3×(-4)
Both terms contain 3 which is a common factor.
So, taking the 3 out, we get 3(y² - 4).
Now, recall a² - b² = (a - b)(a + b)
So y² - 4 = y² - 2² = (y - 2)(y + 2)
So, the answer is 3(y - 2)(y + 2).

3y² - 12
3(y² - 4)
3 (y - 2) (y + 2)

Answer: 3 (y - 2) (y + 2)

3 (y ² - 4)
3 (y - 2) (y + 2)

Ok,
the ques is 3y^2-12,

so you can take 3 as common in both the terms,

so your equation is simplified to 3(y^2-4)

now according to formula, (x^2-y^2)=(x-2)(x+2),

the final answer is 3(y-2)(y+2).