How do you factor #3p^2 + 2p - 16 = 0#?

1 Answer
Nghi N.
May 24, 2015

Factor f(x) = 3x^2 + 2x - 16 = (x - p)(x - q).
I use the new AC Method.
Converted f(x) --> #f'(x) = x^2 + 2x _ 48 #= (x - p')(x - q').

Find p' and q' by composing factor pairs of a.c = -48 -> (-4, 12)(-6, 8). This sum is 2 = b. Then p' = -6 and q' = 8.

Then# p = (p')/a = -6/3 = -2# and #q = (q')/a = 8/3#

Factored form: #f(x) = (x - 2)(x + 8/3) = (x - 2)(3x + 8).#

Check by developing: #f(x) = 3x^2 + 8x - 6x - 16#. OK

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