Question

Factor the expression into an equivalent form 12y^2-75.

Answer 1

Hello oddworld7836!

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

Factor the expression into an equivalent form 12y² - 75.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$12 y ^ { 2 } - 75$

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.

$12 y ^ { 2 } - 75 \\ = 3\left(4y^{2}-25\right)$

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...

$(4 {y}^{2} - 25) \\ = \left(2y-5\right)\left(2y+5\right)$

Now, bring the 3 (common factor) & rewrite the complete expression.

$12 y ^ { 2 } - 75 \\ = \boxed{ \boxed{ \bf \: 3\left(2y-5\right)\left(2y+5\right) }}$

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.

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Hope it'll help you!

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