How do you factor completely #20x^2 + 22xy + 6y^2#?

1 Answer
Nov 17, 2015

Answer:

Factor: #F = 20x^2 + 22xy + 6y^2#

Ans: 2(2x + y)(5x + 3y)

Explanation:

I use the new AC Method to factor trinimilas (Socratic Search)
#F = 2f = 2(10x^2 + 11Xy + 3y^2). #
Converted trinomial #f' = x^2 + 11xy + 30 y^2 =# (x + p')(x + q').
p' and q' have same sign. Factor pairs of (30y^2) --> (5y, 6y). This sum is 11y = b. Then p' = 5y and q' = 6y.
Therefor, #p = (p')/a = (5y)/10 = y/2# and #q' = (6y)/10 = (3y)/5#.
Factored form: #10(x + y/2)(x + (3y)/5) = (2x + y)(5x + 3y)#
Finally:
#F = 2f = 2(2x + y)(5x + 3y)#