How do you factor $12 c^{2} - 75$ $12c^2-75$ completely?

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Answer 1 · 2014-10-28T02:28:24

$12 c^{2} - 75$ $12c^2-75$

by factoring out 3,

$= 3 (4 c^{2} - 25)$ $=3(4c^2-25)$

by $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ ,

$= 3 (2 c + 5) (2 c - 5)$ $=3(2c+5)(2c-5)$

I hope that this was helpful.

How do you factor 12c2−7512c^2-75 completely?