how to factor this?

[lovex24]

So I was watching anime, and saw this on it. And I felt ashamed for not being able to solve it as a Calc 2 student. :sad::sad:

Well obviously I know the answer is just there, but I can't figure out what technique I should use to reach that solution. I tried distributing everything out but that doesn't really help. Well normally I probably wouldn't pay so much attention to a problem in anime, but my friend in Pre-cal the other day asked me to help him solve a similar problem possibly involving factoring similar expression like that and I was unable to solve the problem for him, and that's when I realized there's some very basic fundamental algebra that I need to work on. Problem is on the attached pic. Help is very much appreciated.

 

[lovex24]

ok nvm I tried a few times and figured it out. Now I felt silly for posting this on here.

 

[lookagain]

ok nvm I tried a few times and figured it out. Now I felt silly for posting this on here.


\(\displaystyle \ \ \ \) No, do not feel silly for posting it on here. \(\displaystyle \ \ \)I am able to see your thought process.

Be consistent in the expression you posted in the thread and the one you uploaded. For instance,
decide whether that last term is going to be "ca(c - a)" or ac(c - a)."

On your second line, don't distribute ac to make line \(\displaystyle \ ac^2 - a^2c, \ \ \) only to undo
that by factoring ac back out on the third line.

You should keep your fourth line to just rearranging the terms inside the parentheses. Rearranging
on part, and factoring on par,t are two different stages being mixed. \(\displaystyle \ \ \)Hence:

\(\displaystyle b(a^2 - c^2 \ + \ bc - ab) \ + \ ac(c - a)\)


Look at your last line of work that you uploaded:


(i) Your binominal factors need to match the order given at the top of your original post.
You need to show the work for that.
(There is a particular symmetry, or asymmetry, regarding the order of the variables there.)


(ii) As part of that order, you need to show the (relative little amount of) work that so has
the leading negative sign as part of the required end result.

 

[lovex24]

Be consistent in the expression you posted in the thread and the one you uploaded. For instance,
decide whether that last term is going to be "ca(c - a)" or ac(c - a)."

On your second line, don't distribute ac to make line \(\displaystyle \ ac^2 - a^2c, \ \ \) only to undo
that by factoring ac back out on the third line.

You should keep your fourth line to just rearranging the terms inside the parentheses. Rearranging
on part, and factoring on par,t are two different stages being mixed. \(\displaystyle \ \ \)Hence:

\(\displaystyle b(a^2 - c^2 \ + \ bc - ab) \ + \ ac(c - a)\)


Look at your last line of work that you uploaded:


(i) Your binominal factors need to match the order given at the top of your original post.
You need to show the work for that.
(There is a particular symmetry, or asymmetry, regarding the order of the variables there.)


(ii) As part of that order, you need to show the (relative little amount of) work that so has
the leading negative sign as part of the required end result.

Yea thanks for the input. You see this is just on scratch paper, so if I were to copy neatly onto real paper I would have been more careful. (Notice I was using pen as opposed to pencil and crossed out lines, sort of made a mess there) The reason I distributed everything first was because I didn't really have a clue how exactly it should be solved, therefore I simply just tried to distributed everything and see if I could mess with it. Although on the picture from the anime, it does say (c-a) with a negative sign in the front, I've always preferred (And I think most people do as well) to let the letters(variables) go by alphabetical order, not exactly sure why in the anime they would pull out a negative. But yea at the end, I knew I was getting the right answer so I simply stopped since this was mainly for myself not for turning in as an assignment.

Just a little bit of the thoughts I was having. I don't mean to start an argument or anything. I think your suggestions were great, once again I appreciate you looking through my work very thoroughly.

 

[Bob Brown MSEE]

Expanding both ends toward the middle

ab(a-b)+bc(b-c)+ca(c-a)	(a-b)(b-c)(a-c)
aab-abb+bbc-bcc+acc-aac	(a(b-c)-b(b-c))(a-c)
	a(ab-ac-bb+bc)(a-c)
	a(ab-ac-bb+bc)-c(ab-ac-bb+bc)
	aab-aac-abb+abc-abc+acc+bbc-bcc
	aab-aac-abb+acc+bbc-bcc

This is often the quickest path to a proof.
When you construct the final sequence, justify each step.