How do you factor #9y^2+18y+8#?

2 Answers
Alan P.
May 9, 2015

#9y^2+18y+8#

#=9(y^2+2y+8/9)#

Looking for #axxb =8/9# with #a+b =2#
reveals the solution:

#=9(y+4/3)(y+2/3)#

Nghi N.
May 9, 2015

I use the new AC Method (Google, Yahoo Search) to factor:
f(x) = 9y^2 + 18y + 8. = (y - p)(y - q)
Converted trinomial: f'(y) = y^2 + 18y + 72 = (y -p') (y - q') with
(a.c = 9.(8) = 72.
Compose factor pairs of ac = 72. Proceed: (1, 72)(2, 36)...(6, 12).OK
#p' = 6 -> p = (p')/a# = #6/9 = 2/3 #
#and q' = 12 -> q = (q')/a = 12/9 = 4/3#

f(y) = (y + 2/3)(y + 4/3) = (3y + 2)(3y + 4)

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