# Simplifying an expression

1. Feb 4, 2010

### Cornraker

1. The problem statement, all variables and given/known data

I have a quiz tomorrow and i know a problem like this is gonna be on it and i need to figure out the process

2. Relevant equations

the expression is: $$(2x^2-3x+1)(4)(3x+2)^3(3)+(3x+2)^4(4x-3)$$

3. The attempt at a solution

i know the final answer$$(3x+2)^3(36x^2-37x+6)$$

Ive tried to work the problem several times and i can't figure out how it turns out to be this. can somebody please do a step by step. it would be greatly appreciated.

2. Feb 4, 2010

### Mentallic

Sure thing. (by the way you should have called it factorizing an expression :tongue:)

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 lets 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?

3. Feb 4, 2010

### Cornraker

well i'll sure try to do it. thank you very much for your time and this lengthy explanation!

4. Feb 4, 2010

### Cornraker

i think i got it! if i'm correct i can factor out a (3x+2)^3 and that makes thing a whole lot simpler

5. Feb 4, 2010

No problem!

Yep