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

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Answer 1 · 2015-05-24T18:59:24

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

How do you factor 3p^2 + 2p - 16 = 03p^2 + 2p - 16 = 0?