Can someone just help me show my work in algebra 1?

Question

alexandervonhumboldt2 · Answer

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henrietepurina · Answer

Here is my question: 

(3x^3)y + 12xy - (3x^2)y - 12y 

Rewrite the expression completely factored. Show the steps of your work. 

And my answer was (3x^3)y−(3x^2)y+12xy−12y

henrietepurina · Answer

(refresh page to remove question marks)

henrietepurina · Answer

@asnaseer

asnaseer · Answer

Didn't we just do exactly this expression?

asnaseer · Answer

http://openstudy.com/users/henrietepurina#/updates/54a03b4ce4b054f0c3b20cbb

henrietepurina · Answer

yeah yeah, but in this one I need to factor it :P

texaschic101 · Answer

3x^3y + 12xy - (3x^2y) - 12y --- factor out common terms...3 and y

3y(x^3 + 4x - x^2 - 4) = 

3y (x^2 + 4)(x - 1)

asnaseer · Answer

Ok - what confused me is your statement:

"And my answer was (3x^3)y−(3x^2)y+12xy−12y"

which is not what we got last time

asnaseer · Answer

last time we got to:$$3y(x^3+4x-x^2-4)$$

henrietepurina · Answer

OMG.... I just messed up, my final answer is not (3x^3)y−(3x^2)y+12xy−12y :P

asnaseer · Answer

Ok, so look at where we got to:$$3y(x^3+4x-x^2-4)$$Now make use of your conclusion in your previous question which said that this will not change if we change the order of the terms inside the brackets. Therefore we can swap the positions of the $4x$ and the $-x^2$ to get:$$3y(x^3+4x-x^2-4)=3y(x^3-x^2+4x-4)$$Next we factor the first two terms in this to get:$$3y(x^3+4x-x^2-4)=3y(x^3-x^2+4x-4)=3y(x^2(x-1)+4x-4)$$Then factor the last two terms to get:$$=3y(x^2(x-1)+4(x-1))$$Now notice that both inner terms have a factor of $(x-1)$

asnaseer · Answer

So we can factor out this $(x-1)$ to finally get:$$3y(x^3+4x-x^2-4)=3y(x-1)(x^2+4)$$

henrietepurina · Answer

Thanks again, you're a big help :)

asnaseer · Answer

yw :)