First, we factor the largest number we can. First I factored a #9# and a #s#. That left me with #28s^2-66+36#, and I noticed that I could factor out another number, a #2#.Now I have #18s(color(red)(14)s^2-color(blue)(33)s+color(green)(18))#. Now we need to factor the rest. To do that, I multiply the leading coefficient (#color(red)(14)#) and the constant (#color(green)(18)#) to get #252#. That tells me the value I must muliply to, and the middle term tells me that those factors must sum to #color(blue)(33)#.
#+33#
#*252#
#..........#
#1*252color(white)(25)|253#
#2*126color(white)(25)|128#
#3*84color(white)(252)|87#
#4*63color(white)(252)|67#
#6*42color(white)(252)|48#
#7*36color(white)(252)|43#
#9*28color(white)(252)|37#
#color(orange)(12*21color(white)(25)|33)#
Now we can factor by grouping. That looks like this #(color(red)(14)s^2-21s)+(-12s+color(green)(18))#. Now we just factor.
#7s(2s-3)+-6(2s-3)#. That becomes #(7s-6)(2s-3)#. Add in the #18s#, and we are done.
#18s(7s-6)(2s-3)#
Just to make sure we're correct, let's copare our solution to the original expression:
graph{y=18x(7x-6)(2x-3)}
vs
graph{y=252x^3-594x^2+324x}
They are the same, so we are correct! Nice work.
