How do you factor #12c^2-75# completely?

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Answer 1 · 2014-10-28T02:28:24

#12c^2-75#

by factoring out 3,

#=3(4c^2-25)#

by #a^2-b^2=(a+b)(a-b)#,

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

I hope that this was helpful.