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.!
