How do you factor completely #9xy^2 - 24xy + 15x#?

1 Answer
Tony B
Nov 26, 2015

#3x(3y-5)(y-1)#

Explanation:

#x# is a factor of everything so we can take that out. Also 3 is likewise a factor of everything so we can take that out as well!

#3x(3y^2-8y+5) #

Both 3 and 5 are prime number so we can only use 1 and themselves as whole number factors. Factors of #5->{1,5}#
Factors of #3->{1,3}#. So it a matter of experimenting with the order to see if we can reproduce #3y^2-8y+5#

Strait off there is no choice other than #(3y+?)(y+?)# to give #3y^2#

#3x(3y-5)(y-1)#

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