How do you factor #6x^2 +5xy -21y^2#?

1 Answer
May 30, 2016

Answer:

(2x - 3y)(3x + 7y)

Explanation:

Use the new AC Method (Socratic Search)
Consider x as a variable, y as a constant, and factor f(x).
#f(x) = 6x^2 + 5yx - 21 y^2 =# 6(x + p)(x + q)
Converted trinomial #f'(x) = x^2 + 5yx - 126y^2 =# (x + p')(x + q').
p' and q' have opposite signs because ac < 0.
Factor pairs of #(ac = 126y^2)# -->...(-6y, 21y)(-9y, 14y). This last sum is (5y = b). Then p' = -9y and q' = 14y.
Back to f(x) --> #p = (p')/a = (-9y)/6 = -(3y)/2#, and
#q = (q')/a = (14y)/6 = (7y)/3#
Factored form : #f(x) = 6(x - (3y)/2)(x + (7y)/3) = (2x - 3y)(3x + 7y)#