Adding "annihilating terms" to factor

[DS2C]

Not a homework problem, just something I came across in my book that makes sense but not fully.

Homework Statement

Factor: $a^4-b^4$

Homework Equations

The Attempt at a Solution

[/B]
The book's solution:

$a^4-b^4$
$=a^4-a^3b+a^3b-a^2b^2+a^2b^2-ab^3+ab^3-b^4$
$=a^3\left(a-b\right)+a^2b\left(a-b\right)+ab^2\left(a-b\right)+b^3\left(a-b\right)$
$=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)$

Just what is going on here? Such a tiny binomial resulting in something way larger, with what seem like arbitrary terms added in.

Additionally, it gives another factorization of the given problem $a^4-b^4$ which is also factored into $\left(a^2-b^2\right)\left(a^2+b^2\right)$
You can have more than one correct solution when factoring a polynomial? In what situations would one be the "correct vs incorrect" way? Or is it synonymous with something like factoring 12 into 4*3, 6*2, 12*1 where they are all correct?

 

[fresh_42]

You could still perform another factorization on $a^2-b^2$. Since it didn't say until only irreducible factors are left, it has indeed more than one solution.

 

[DS2C]

Man this factoring stuff goes deep. Feels like there should be an entire semester focusing on just this. Must be pretty important!

 

[fresh_42]

Man this factoring stuff goes deep. Feels like there should be an entire semester focusing on just this. Must be pretty important!

It is, and of course it's always only up to units, as one can always factor $f(x) = c \cdot (c^{-1} \cdot f(x))$ and which is of no use.

 

[DS2C]

Not at all sure what that means unfortunately. I'm only at factoring polynomials in the form of $ax^2+bx+c$ currently. There's obviously a lot more to learn.

 

[fresh_42]

Well, just apply what you did once again on: $a^4-b^4=(a^2-b^2)(a^2+b^2)$ and you get three factors, which cannot be split any further, so we call them irreducible. The terms $a^4-b^4$ and $a^2-b^2$ are reducible.

 

[epenguin]

A very similar question is already discussed, includinrg what counts as factorization a few threads down.https://www.physicsforums.com/threa...ifference-of-two-squares.925946/#post-5848006

 

[DS2C]

Yeah that's my thread too lol. The reason for a new thread is a new aspect of the factoring. The problem posted, which was out of my book, mentions "inserting annihilating terms" into the expression to factor. This is what I am having trouble wrapping my head around. Why do this and how do you determine what "annihilating" terms to input?

 

[fresh_42]

Yeah that's my thread too lol. The reason for a new thread is a new aspect of the factoring. The problem posted, which was out of my book, mentions "inserting annihilating terms" into the expression to factor. This is what I am having trouble wrapping my head around. Why do this and how do you determine what "annihilating" terms to input?

It is simply another way to describe a division. Instead of calculating $(a^4-b^4):(a-b)$ the entire division is somehow written backwards. If we know the result, then we know which "annihilating terms" to be added. Otherwise it's probably a try and error. I still prefer the division. I doubt there is a system behind it.

 

[DS2C]

That makes more sense. Here's the text in question for reference.

 

[qspeechc]

Factor again $\left(a^2-b^2\right)$ in $\left(a^2-b^2\right)\left(a^2+b^2\right)$

To see that the two factorisations are actually the same, try to re-write the above factorisation as $\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)$. You should see what to do after the first step above.

Normally in factorising we try to factorise each factor as far as possible, so in this case, the first part of this reply is what we normally do.

 

[fresh_42]

Factor again $\left(a^2-b^2\right)$ in $\left(a^2-b^2\right)\left(a^2+b^2\right)$

I assume you've meant $\left(a^2-b^2\right)$ in $\left(a-b\right)\left(a+b\right)$.

 

[qspeechc]

I assume you've meant $\left(a^2-b^2\right)$ in $\left(a-b\right)\left(a+b\right)$.

No. "In" not "into", as in, factor again the part $\left(a^2-b^2\right)$ in the expression $\left(a-b\right)\left(a+b\right)$