Let #g(x)=y=2x^3-5x^2+4#.
We observe that sum of the co-effs.#=2-5+4=1!=0, so, (x-1)# is not a factor.
Also, the sum co-effs. of odd powered terms #=2!=-1=-5+4# = the sum of co-effs. of even powered terms, so #(x+1) can not be a factor.
Now to guess the probable factors of #g(x)#, we look at its leading co-eff. #=2,# having factors #1,2# and the const. term#=4,# having factors #1,2,4.#
The probable factors can be guessed by multiplying #x# with the factors of leading co-eff. and taking with this, #+-# the factors of the const. term. This leads us to the guess of probable factors of #g(x)# as #x+-1,x+-2,x+-4, 2x+-1,2x+-2,2x+-4.#
Among these, we have already verified the factors #x+-1, 2x+-2=2(x+-1).#
Next we verify whether #(x-2)# is a factor of #g(x)# by checking #g(2)=2*8-5*4+4=16-20+4=0,#, so, (x-2) is a factor.
Now we arrange the terms of #g(x)# in such a way (shown below) that #(x-2)# turns out as a common factor :-
#g(x)=2x^3 -5x^2+4=ul(2x^3-4x^2)-ul(x^2+2x)-ul(2x+4)=2x^2(x-2)-x(x-2)-2(x-2)=(x-2)(2x^2-x-2)#.
Enjoy maths.!
