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