# Adding "annihilating terms" to factor

1. Sep 27, 2017

### DS2C

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

1. The problem statement, all variables and given/known data
Factor: $a^4-b^4$

2. Relevant equations

3. The attempt at a solution

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?

2. Sep 27, 2017

### Staff: Mentor

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.

3. Sep 27, 2017

### DS2C

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

4. Sep 27, 2017

### Staff: Mentor

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.

5. Sep 27, 2017

### 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.

6. Sep 27, 2017

### Staff: Mentor

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.

7. Sep 27, 2017

### epenguin

Last edited: Oct 13, 2017
8. Sep 27, 2017

### DS2C

Yeah thats 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 Im having trouble wrapping my head around. Why do this and how do you determine what "annihilating" terms to input?

9. Sep 27, 2017

### Staff: Mentor

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.

Last edited: Sep 27, 2017
10. Sep 27, 2017

### DS2C

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

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11. Oct 13, 2017

### 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.

12. Oct 13, 2017

### Staff: Mentor

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

13. Oct 13, 2017

### qspeechc

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