How do you factor #3x^2+ x - 14#?

1 Answer
Nghi N.
Sep 26, 2015

Factor #y = 3x^2 + x - 14#

Explanation:

#y = 3x^2 + x - 14 = #3(x + p)(x + q)
Converted trinomial #y' = x^2 + x - 42.# = (x + p')( + q')
p' and q' have opposite sign. Factor pairs of (-42) -->...(-2, 21)( -3, 14)(-6, 7). This sum is 7 - 6 = 1 = b. Then, p' = -6 and q' = 7.
Therefor, #p = -6/3 = -2# and #q = 7/3#.
Factored form: #y = 3(x - 2)(x + 7/3) = (x - 2)(3x + 7)#

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