How do you factor #12x^2 + 5x - 2= 0#?

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Answer 1 · 2015-05-31T04:41:12

#y = 12x^2 + 5x - 2 #= (x - p)(x - q). Use the new AC Method

Convert y to #y' = x^2 + 5x - 24. #= (x - p')(x - q')

Compose factor pair of (-24) -> (-2, 12)(-3, 8) OK

Then, p' = -3 and q' = 8.

Then #p = -3/12 = -1/4 and q = 8/12 = 2/3#

Factored form:# y = (x - 1/4)(x + 2/3) = (4x - 1)(3x + 2).#

Check by developing: #y = (12x^2 + 8x - 3x - 2). # OK