How do you factor #2x^2-11xy+5y^2#?

1 Answer
Jul 14, 2016

(2x - y)(x - 5y)

Explanation:

Use the new AC Method to factor trinomials (Socratic Search).
#y = 2x^2 - 11xy + 5y^2 =# 2(x + p)(x + q)
Converted trinomial: #y' = x^2 - 11xy + 10y^2 =# (x + p')(x + q')
p' and q' have same sign because ac > 0
Factor pairs of #(ac = 10y^2)# --> (-y, -10y). This sum is (-11y = b). Consequently, p' = -y and q' = -10y.
Back to original trinomial y --> #p = (p')/a = -y/2#, and #q = (q')/a = -10y/2 = -5y.#
Factored form of y:
#y = 2(x - y/2)(x - 5y) = (2x - y)(x - 5y)#