How do you factor #18x^2-31xy+6y^2#?

1 Answer
Nghi N.
May 28, 2015

#z = 18x^2 - 31yx + 6y^2 #= (x - p)(x - q). Use the new AC Method.
Convert z to #z' = x^2 - 31yx + 108y^2# = (x - p')(x - q').
Compose factor pairs of# (108y^2)# -> (2, 54)(3, 36)(4, 27) = -b.
Then, p' = -4y and q' = -27y.

Next, #p = (-4y)/18 = (-2y)/9# and #q = (q')/18 = -(3y)/2.#

Factored form: #z = (x - (2y)/9)(x - (3y)/2) = (9x - 2y)(2x - 3y)#

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