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how do you factor 16x^2-4x-6?
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you try to find two numbers that multiply to c (6) and two that add to b (4)
take 2 out as a common factor leaving \[2(8x^2 - 2x - 3)\] multiply 8 and -3 then find the factors that add to -2 then you can substitute into binomials \[\frac{2(8x + factor1)(8x + factor 2)}{8}\] cancel common factors from the numerator and denominator for the answer
yep that's wat i was going to say
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