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Factor completely: 2x^2 + 14x + 12 A. (x + 1)(x + 6) B. 2(x + 2)(x + 3) C. 2(x + 1)(x + 6) D. 2(x + 2)(x + 6)
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@TechnoSoul
Start this by factoring out the 2. It will make the coefficient in front of the x^2 term be 1 instead of 2.\[2(x ^{2}+7x+6)\]Now factor what's inside the parenthesis. That comes to 2(x + 6)(x + 1). If you were to FOIL that back out and then redistribute the 2, you would be back at your original polynomial.
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