Question

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

Answer 1

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)`