What should be my sequence of operations here? (factoring)

[ProjectTruth]

Hey Guys,

So I'm currently factoring polynomials.

As you all know,when trying to factor a polynomial with 4 terms,the answer can be the difference of a trinomial and monomial squares.

So I had the problem 4a^2-4ab+b^2-c^2

I thought(book doesn't explain it well) that your first step was to group like terms to create your trinomial which we then end up with:

(4a^2-4ab+b^2)-c^2

From there we solve and get:

(2a-b)^2-c^2

Thus being solved:

=(2a-b-c)(2a-b+c)

Which is the correct answer.

But then they had this problem:

a^2+b^2-c^2-2ab

My logic behind grouping like terms then falls short in this one,in which I now assume we just randomly select 3 terms to isolate. This is where things start getting sticky...

The correct answer is (a-b+c)(a-b-c)

But I can't seem to find my order of operations...

I find that I group (a^2-b^2+c^2)-2ab

But that's as far as I get...

Any insight?

Thanks in advance.

 

[RUber]

Your trinomial term should be made up of only two variables. In this case, use a^2 - 2ab + b^2. It is not a random process.

 

[krebs]

If you use latex or superscripts, you're more likely to get help.
$$4a^2-4ab+b^2-c^2$$
$$(4a^2-4ab+b^2)-c^2$$
$$(2a-b)^2-c^2$$
$$(2a-b-c)(2a-b+c)$$---------------------------------------------

$$a^2+b^2-c^2-2ab$$
Why are you selecting three random terms?

you have ## a\space b\space c\space ab## which three are like terms?

 

[ProjectTruth]

Yes,thank you both,it seems I was hasty in my assumption.

Why I asked,is it seemed like a problem I had with a previous problem,namely:

r^2-s^2-25+10s

Which I could simply not solve if I grouped like terms...

The answer is (r+s-5)(r-s+5)

I could simply not even start the process for this one...

Any time I grouped like terms,it was clearly wrong. So I then just grouped the first 3 in which I get

(r^2-s^2-25)+10s

Which then seems to be correct at first,but I then get no farther with it... any insight on this one?

 

[RUber]

Your goal is to simplify one part into a square of binomials.
In your example,
##r^2 - s^2 - 25 + 10s##
Which might be written as:
## r^2 - s^2 - 5^2 + 2(5s) ##
The terms that should stand out to you are s and 5.
Then once you have one piece written as a square: ##(s-5)^2##
You look again at the whole...## r^2 - (s-5)^2##
Which follows the difference of squares rule:
( r+ (s-5) ) ( r - (s - 5))

 

[ProjectTruth]

Your goal is to simplify one part into a square of binomials.
In your example,
##r^2 - s^2 - 25 + 10s##
Which might be written as:
## r^2 - s^2 - 5^2 + 2(5s) ##
The terms that should stand out to you are s and 5.
Then once you have one piece written as a square: ##(s-5)^2##
You look again at the whole...## r^2 - (s-5)^2##
Which follows the difference of squares rule:
( r+ (s-5) ) ( r - (s - 5))

Hmm,well I wish my book would've told me that...

My book never said seperating the 4 terms into a group of 2 binomials was an option...

It's only ever said seperating them into a monomial and trinomial was my only option.

Guess I need to look into this more...

 

[HallsofIvy]

When you look at [itex]r^2- s^2- 25+ 10s[/itex] you should immediately see that there are two terms involving s, [itex]-s^2+ 10s[/itex], one involving [itex]s^2[/itex] and [itex]10s[/itex]. That should be enough for you to decide that the [itex]r^2[/itex] term should be separated: [itex]r^2- (s^2- 10s+ 25)[/itex]. Now you should recognize that [itex]s^2- 10s+ 25= (s- 5)^2[/itex] so your original polynomial, [itex]r^2- s^2- 25+ 10x[/itex] is equal to [itex]r^2- (s- 5)^2[/itex].

 

[RUber]

Hmm,well I wish my book would've told me that...

My book never said seperating the 4 terms into a group of 2 binomials was an option...

It's only ever said seperating them into a monomial and trinomial was my only option.

Guess I need to look into this more...

I was not saying to break the 4 terms into two binomials, I was saying that the trinomial you are looking for in this sort of problem is the square of a binomial.
All of your examples have been of the form: ## a^2 + 2ab + b^2 - c^2 ## or some variation of that.
Your examples demonstrate that you first find a square of the form ## a^2 + 2ab + b^2 = (a+b)^2 ##. Your odd term out is a standalone square, like ##c^2##.
Then, when you put it back together you have a difference of squares.
##( (a+b)^2 - c^2 ) = ( (a+b) + c ) ( (a+b) - c) ##

 

[ProjectTruth]

When you look at [itex]r^2- s^2- 25+ 10s[/itex] you should immediately see that there are two terms involving s, [itex]-s^2+ 10s[/itex], one involving [itex]s^2[/itex] and [itex]10s[/itex]. That should be enough for you to decide that the [itex]r^2[/itex] term should be separated: [itex]r^2- (s^2- 10s+ 25)[/itex]. Now you should recognize that [itex]s^2- 10s+ 25= (s- 5)^2[/itex] so your original polynomial, [itex]r^2- s^2- 25+ 10x[/itex] is equal to [itex]r^2- (s- 5)^2[/itex].

Hmm,It had not crossed my mind to isolate r^2...

But besides that,it turns out my book had failed to mention that I might also be able to solve polynomials with 4 terms,by grouping them into two binomials consisting of similar terms...

Blast it all... More work to be done yet... but you've got to learn one way or another eh?

Thanks fellas.

Cheers.

 

[ProjectTruth]

I was not saying to break the 4 terms into two binomials, I was saying that the trinomial you are looking for in this sort of problem is the square of a binomial.
All of your examples have been of the form: ## a^2 + 2ab + b^2 - c^2 ## or some variation of that.
Your examples demonstrate that you first find a square of the form ## a^2 + 2ab + b^2 = (a+b)^2 ##. Your odd term out is a standalone square, like ##c^2##.
Then, when you put it back together you have a difference of squares.
##( (a+b)^2 - c^2 ) = ( (a+b) + c ) ( (a+b) - c) ##

I see that now.

But what I mean,is that after a search through the Internet through frustration netted me the knowledge that you may break up polynomials into two binomials consisting of similar terms if you cannot group them into a monomial and trinomial...

Funny thing is though,I literally grouped them up every way imaginable EXCEPT the correct way...geez...

Ha,thanks again.