Hello oddworld7836!
![\huge \boxed{\mathbb{QUESTION} \downarrow}](https://tex.z-dn.net/?f=%20%5Chuge%20%5Cboxed%7B%5Cmathbb%7BQUESTION%7D%20%5Cdownarrow%7D)
Factor the expression into an equivalent form 12y² - 75.
![\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}](https://tex.z-dn.net/?f=%20%5Clarge%20%5Cboxed%7B%5Cmathfrak%7BAnswer%20%5C%3A%20with%20%5C%3A%20Explanation%7D%20%5Cdownarrow%7D)
![12 y ^ { 2 } - 75](https://tex.z-dn.net/?f=12%20y%20%5E%20%7B%202%20%7D%20-%2075)
By observing the expression, we can see that, 3 is the only common factor in both the terms of the expression. So, take the common factor 3 out.
![12 y ^ { 2 } - 75 \\ = 3\left(4y^{2}-25\right)](https://tex.z-dn.net/?f=12%20y%20%5E%20%7B%202%20%7D%20-%2075%20%5C%5C%20%20%3D%203%5Cleft%284y%5E%7B2%7D-25%5Cright%29%20)
Now, look at (4y² - 25). They don't have any common factors but they appear in the form of the algebraic identity ⇨ a² - b² = (a + b) (a - b). Here,
- a² = 4, a = 2 (√a² = ✓4 = 2)
- b² = 25, b = 5 (√b² = ✓25 = 5)
So, the (4y² + 25) becomes...
![(4 {y}^{2} - 25) \\ = \left(2y-5\right)\left(2y+5\right)](https://tex.z-dn.net/?f=%284%20%7By%7D%5E%7B2%7D%20%20-%2025%29%20%5C%5C%20%20%3D%20%5Cleft%282y-5%5Cright%29%5Cleft%282y%2B5%5Cright%29%20)
Now, bring the 3 (common factor) & rewrite the complete expression.
![12 y ^ { 2 } - 75 \\ = \boxed{ \boxed{ \bf \: 3\left(2y-5\right)\left(2y+5\right) }}](https://tex.z-dn.net/?f=12%20y%20%5E%20%7B%202%20%7D%20-%2075%20%5C%5C%20%20%3D%20%20%5Cboxed%7B%20%5Cboxed%7B%20%5Cbf%20%5C%3A%203%5Cleft%282y-5%5Cright%29%5Cleft%282y%2B5%5Cright%29%20%7D%7D)
We can't further simplify it. Also, remember that the simplified form of an expression is equivalent to the expression. So, 3 (2y - 5) (2y + 5) is equivalent to 12y² - 75.
__________________
Hope it'll help you!
ℓu¢αzz ッ
Answer:
im sorry
Explanation:
(not really we will never meet in person)