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

Sagot :

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Hello oddworld7836!

[tex] \huge \boxed{\mathbb{QUESTION} \downarrow}[/tex]

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

[tex] \large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}[/tex]

[tex]12 y ^ { 2 } - 75[/tex]

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.

[tex]12 y ^ { 2 } - 75 \\ = 3\left(4y^{2}-25\right) [/tex]

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...

[tex](4 {y}^{2} - 25) \\ = \left(2y-5\right)\left(2y+5\right) [/tex]

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

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

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

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