Sure thing. (by the way you should have called it factorizing an expression
)
You need to have the basic idea of factorizing deeply embedded into your head. Mainly, [tex]ab+ac=a(b+c)[/tex]
(1). a,b and c could be anything much more complicated.
Lets take [tex]a=x^2(x+1)^2[/tex]
Then we would need to factorize [tex]x^2(x+1)^2b+x^2(x+1)^2c[/tex]
Can you now see how we can factorize out the a (or in this case the [tex]x^2(x+1)^2[/tex]) ? We now get the same thing as in (1): [tex]a(b+c)=x^2(x+1)^2(b+c)[/tex]
At the same time, b and c can be something more complicated as well. If we let [tex]b=x(x+1)[/tex] and [tex]c=x+1[/tex] then we now have:
[tex]a\left(x(x+1)+(x+1)\right)[/tex] but this time we aren't completely done because b and c have a common factor also. [tex]x(x+1)+(x+1)=x(x+1)+1(x+1)=(x+1)(x+1)=(x+1)^2[/tex]
So let's put it all together now in [tex]ab+ac=a(b+c)[/tex] where [tex]a=x^2(x+1)^2, b=x(x+1), c=x+1[/tex]
[tex]x^2(x+1)^2(x(x+1)+(x+1))=x^2(x+1)^2(x+1)^2=x^2(x+1)^4[/tex] This last form is completely factorized.
Now looking at your expression: let some other variable such as [tex]y=(3x+2)^3[/tex] and see if that makes things easier to factorize. Also you'll need to factorize [tex]2x^2-3x+1[/tex], can you do this?
