How do you factor completely #3x^2-x-10 #?

1 Answer
Apr 29, 2016

Answer:

(3x + 5)(x - 2)

Explanation:

Use the systematic new AC method (Socratic Search)
#y = 3x^2 - x - 10 = 3(x + p)(x + q)#
Converted trinomial #y' = x^2 - x - 30 =# (x + p')(x + q')
p' and q' have opposite signs, because ac < 0.
Factor pairs of (ac = -30) --> (-5, 6)(5, -6). This sum is -1 = b. Therefor, p' = 5 and q' = -6.
Back to original y, #p = p'/(a) = 5/3# and #q = (q')/a = -6/3 = -2#
Factored form:
#y = 3(x + 5/3)(x - 2) = (3x + 5)(x - 2)#