How do you factor #12x^3 - 60x^2 + 75x = 0#?

1 Answer
Apr 29, 2016

Answer:

#3x(2x + 5)^2#

Explanation:

f(x) = x(12x^2 - 60x + 75)
Factor the trinomial y in parentheses by the new AC Method (Socratic Search).
#y = 12x^2 - 60x + 75 = #12(x + p)(x + q)
Converted trinomial #y' = x^2 - 60x + 900 =# (x + p')(x + q').
p'and q' have same sign because ac > 0.
#y' = x^2 - 60x + 900 = (x - 30)^2#. Then p' = q' = 30 (double roots)
Back to y, #p = q = p'/(a) = q'/(a) = 30/12 = 5/2#.
#f(x) = 12x(x + 5/2)^2 = 3x(2x + 5)^2#