The difference between two squares

E

Elohymn

New member
Joined
Feb 18, 2021
Messages
4
Hi guys,

I have a question in regards to 'the difference between two squares'. Everywhere I look it tells me to work backwards to prove that 'a² - b² = (a+b) (a-b)' but what if you didn't know the answer, how would you factorise to get to '(a+b) (a-b)' in the first place?

I've put my thought process below. Maybe someone can tell me where I'm going wrong?

1. So 6xy + 3x²y can be factored to 3xy(2+x).
2. So why when factoring the difference between two squares does it seem different? a² - b² = (a+b) (a-b).
3. If I use the same method (from step 1) I don't get the same answer? I get a² - b² = 1²(a-b)?
4. What am I missing?

Thanks for your time and help
 
Here's how you factor this:
[math]a^2 - b^2 = a^2 + (ab - ab) - b^2 = (a^2 + ab) - (ab + b^2) = a(a + b) - b(a + b) = (a - b)(a + b)[/math]
How is this different for [math]6xy + 3x^2 y[/math]? The terms are not the difference between two squares. It factors differently.

-Dan
 
Elohymn said:
I have a question in regards to 'the difference between two squares'. Everywhere I look it tells me to work backwards to prove that 'a² - b² = (a+b) (a-b)' but what if you didn't know the answer, how would you factorise to get to '(a+b) (a-b)' in the first place?

I've put my thought process below. Maybe someone can tell me where I'm going wrong?

1. So 6xy + 3x²y can be factored to 3xy(2+x).
2. So why when factoring the difference between two squares does it seem different? a² - b² = (a+b) (a-b).
3. If I use the same method (from step 1) I don't get the same answer? I get a² - b² = 1²(a-b)?
4. What am I missing?
Click to expand...

Factoring is an "undoing" process by nature -- finding a way to get a certain result by multiplying. It's inherently difficult, and inherently backward. Not everything can be factored by the same method (or by any method); typically you either recognize an expression as a result you've run across in a multiplication you've done, or you just discover it by hard work. (Most of the factoring methods we teach are the result of someone else's discovery in such a way.)

Most expressions one could write can't be factored at all. When you find a type that can be factored, you grab onto it and hold on for life.

But there are a variety of ways one can "discover" this factoring; topsquark's is a nice trick when it works (trying to insert terms to make something that can be factored by grouping). Another approach is to recognize that x^2 - y^2 = 0 has solutions x=y and x=-y, which implies that it has (x-y) and (x+y) as factors, so you'd try multiplying those ...
 
[MATH](a - b)(a + b) = a(a - b) - b(a - b) = a^2 - ab + ab - b^2 = a^2 - b^2.[/MATH]
Does that look good to you?

[MATH]p = q \iff q = p.[/MATH]
Can you think of an exception?

[MATH]\therefore (a - b)(a + b) = a^2 - b^2 \implies a^2 - b^2 = (a - b)(a + b).[/MATH]
This is a proof.
 
Here is another way. I would not do it this way but that is me and maybe you'll love it.

Consider x2-b2 = 0. Clearly x= b and x=-b are the two solutions. So x2-b2 factors to (x-b)(x+b). Now let x=a so a2 - b2 = (a-b)(a+b)
 
Hi,

So I've read through all the replies and here is my understanding so far:

-------------------------------------------------------------

1. a² - b² = (a + b)(a - b)

2. a² - b² = 0
- In this case we've just set it equal to 0. We could set it equal to anything but 0 is easier to work out?

3. a = b OR a = -b
- For 'a² - b² = 0' one of the above must be true?

4. I'll set 'b = 4'

5. a² - b² = (a x a) - (b x b)

6. (a x a) - (b x b) = (4 x 4) - (4 x 4)
- OR (-4 x -4) - (4 x 4)

7. (4 x 4) - (4 x 4) = 16 - 16
- OR (-4 x -4) - (4 x 4) = 16 - 16?

8. 16 - 16 = 0
- So if that works I can use the same values in the format '(a + b)(a - b)' to figure out if I still get 0?

9. (a + b)(a - b) = (4 + 4)(4 - 4)
- OR (-4 + 4)(-4 + 4)?

10. (4 + 4)(4 - 4) = 8 x 0
- OR (-4 + 4)(-4 + 4) = 0 x 0

11. 8 x 0 = 0
- OR 0 x 0 = 0

12. So if 'a² - b² = 0' and '(a + b)(a - b) = 0' then 'a² - b²' must also equal '(a + b)(a - b)'?
- So all we are saying is ‘0 = 0’ and we could use any equation that also equalled zero in this manner? Anything that equals zero is also equal to anything else that equals zero? And that is how we prove the two expressions are equal?

-------------------------------------------------------------

As for the other method:

1. a² - b² = a² (+ ab - ab) - b² = (a + b)(a - b)

- I’m failing to see how you go forward. In other factorisations you simply move numbers/letters around, but in this it seems as if you are adding them? In ‘a² - b² = a² (+ ab - ab) - b²’ where did the ‘(+ ab - ab)’ come from? I can see how to find them backwards… but not forwards.

- I had '(a x a) - (b x b)' but now I have '(a x a)(+ ab - ab)-(b x b)'. Extra letters have appeared, but from where?

- Is this only possible to understand backwards? People found the answer in other ways (like 0 = 0) and then connected the two? Or am I still missing something?

-------------------------------------------------------------

Thanks very much for all the answers. I’m trying to study at home and have no one to ask these questions.
 
Last edited:
Elohymn said:
Hi,

So I've read through all the replies and here is my understanding so far:

-------------------------------------------------------------

1. a² - b² = (a + b)(a - b)

2. a² - b² = 0
- In this case we've just set it equal to 0. We could set it equal to anything but 0 is easier to work out?



Thanks very much for all the answers. I’m trying to study at home and have no one to ask these questions.
Click to expand...
You need to set it to 0! See if your procedure works if you set it equal to 7.

You do have someone to ask. Actually you have many people to ask! Just ask on this forum.
 
Elohymn said:
1. a² - b² = (a + b)(a - b)
Click to expand...
As I understand it, your goal is not to prove that this is true, which you presumably accept, but how you would figure it out in the first place without being told. So I'm not sure why you state it as the first line of your work.

Elohymn said:
2. a² - b² = 0
- In this case we've just set it equal to 0. We could set it equal to anything but 0 is easier to work out?
Click to expand...
The reason for setting a polynomial to 0 in trying to factor it is the theorem that a polynomial is equal to zero only when one of its factors is zero, so that finding zeros (solutions of that equation) tells you factors. Setting it to something else wouldn't help.

Elohymn said:
3. a = b OR a = -b
- For 'a² - b² = 0' one of the above must be true?
Click to expand...
This is because if a² = b², then a must be one of the square roots of b², namely b or -b.

Elohymn said:
4. I'll set 'b = 4'
Click to expand...
The rest of your work is not useful. Using a specific number is helpful only to make things more concrete.

Suppose you want to factor a² - 4². Setting this to zero, we want to solve a² - 4² = 0. So a² = 4², from which we conclude that a = 4 or -4.

Since the solutions of a² - 4² = 0 are a = 4 and a = -4, it can be factored as (a - 4)(a + 4). Then we check by multiplication and confirm.

The reasoning with b in general is identical, with 4 replaced with b.

Elohymn said:
1. a² - b² = a² (+ ab - ab) - b² = (a + b)(a - b)

- I’m failing to see how you go forward. In other factorisations you simply move numbers/letters around, but in this it seems as if you are adding them? In ‘a² - b² = a² (+ ab - ab) - b²’ where did the ‘(+ ab - ab)’ come from? I can see how to find them backwards… but not forwards.
Click to expand...
Are asking what to do next ("go forward" from here), or how the first step would have been done?

I mentioned that this approach means adding in a term that (you hope) will make it possible to factor by grouping. We want to add and subtract the same thing, so we don't change the value of the expression, so we are hoping for something like

a2 + ____ - ____ - b2​

We want the added terms to have a common factor with both the term to the left and the term to the right, so we try "ab":

a2 + ab - ab - b2​

Trying that out (factoring by grouping), we factor the first pair and the second pair:

a(a + b) - b(a + b)​

Those do have a common factor as we hoped, so we finish:

(a - b)(a + b)​

And there we are. Factoring often works by hope like this: If we don't know a method we are sure will work, we try something and see if it does. If not, we try another idea, and if we run out, we guess that it might be one of the majority of polynomials that we can't factor. For example, a2 + b2 can't be factored.
 
Karl Popper made an important distinction between the psychology of discovery and the logic of demonstration. What I gave was a logical demonstration.

I believe most basic mathematical facts were initially discovered through inductive reasoning. Arithmetic was once a science.

[MATH]7^2 - 3^2 = 49 - 9 = 40 = 4 * 10 = (7 - 3)(7 + 3).[/MATH]
[MATH]10^2 - 6^2 = 100 - 36 = 64 = 4 * 16 = (10 - 6)(10 + 6).[/MATH]
Do a few more. It seems true for every example. Is it always true? Now you try a general proof (which I already gave you).

EDIT: Mathematicians like to proceed from unproven axioms to proven theorems. That is not how the human mind normally works. We see patterns; some represent real regularities; some do not. The practical utility of math is to allow us to determine which patterns we see are meaningful.
 
From what I can see, they can never be equal to '7' if 'a' is equal to both 'b' and '-b'. They can only be equal to zero?

--------------------------------------------------

Ah, sorry. It was my thought process that I wrote down. I was just writing as I thought, really.

1. a² + ab - ab - b²
- So we are simply looking to add terms that don't change the value as they will cancel out?

2. a(a + b) - b(a + b)
- We then factor in the same way as my original example '6xy + 3x²y = 3xy(2+x)'. But this time for 2 brackets, not one.

3. (a - b)(a + b)
- The next step I'm still unsure on. Why did one of the signs flip? Is that to not change the original value?

--------------------------------------------------

I think seeing it written with numbers, not letters, helped a lot.

a = 10
b = 5

10² - 5² = 100 - 25 = 75 = 5 x 15 = (10 - 5)(10 + 5)

When I see it written out like this I can clearly see the path 'forward' and 'backwards'. Perhaps, the letters were throwing me off and next time I should just completely switch them out for numbers if I have this problem again.
 
Elohymn said:
2. a(a + b) - b(a + b)
- We then factor in the same way as my original example '6xy + 3x²y = 3xy(2+x)'. But this time for 2 brackets, not one.
Click to expand...
Have you not learned factoring by grouping yet?

Elohymn said:
3. (a - b)(a + b)
- The next step I'm still unsure on. Why did one of the signs flip? Is that to not change the original value?
Click to expand...
What sign do you think flipped??

Elohymn said:
I think seeing it written with numbers, not letters, helped a lot.

a = 10
b = 5

10² - 5² = 100 - 25 = 75 = 5 x 15 = (10 - 5)(10 + 5)

When I see it written out like this I can clearly see the path 'forward' and 'backwards'. Perhaps, the letters were throwing me off and next time I should just completely switch them out for numbers if I have this problem again.
Click to expand...
The trouble is, nothing you did here applies in general; you've only verified that 10² - 5² and (10 - 5)(10 + 5) equal the same thing, with no reasoning that can be generalized. (Why did you pick 5*15 rather than 3*25, for example? That's the same "working backward" thing you complained about.)

I used a specific number for only one variable, and made sure all my work would be valid in general. I do recommend using numbers for a trial run when you do unfamiliar algebra, because that feels familiar; but be sure you then do the actual algebra.

It will be helpful if you can use the Reply button (either for the whole message, or the one that pops up when you select part of it), so we can know who, and what part, you are answering.
 
You must log in or register to reply here.
Top