# Math Help - Factorisation

1. ## Factorisation

Hi,

I have managed to factorise $A^4 - B^4$. But can't factorise $A^4 + B^4$. Can anyone help me out? Thanks

2. Originally Posted by M.R
Hi,

I have managed to factorise A^4 - B^4. But can't factorise A^4 + B^4. Can anyone help me out? Thanks
You can't without using complex numbers, have you met these?

3. Since any expression $a^{2n}$ with $n \in \mathbb{N}$ is always positive, then $a^{2n} + b^{2n'}$ will always be positive. Thus, there is no possible factorization of such expressions in $\mathbb{R}$. However, there is a factorization in $\mathbb{C}$, which involves complex numbers.

4. I haven't done. We just started the maths course. In Chapter 1 we have so far done:
1A - Terms, Factors and indices
1B - Expanding Brackets
1C - Factorisation
1D - Algebraic Factions
1E - Four Cubic Identities. And this is where we are up to.

The answer for $A^4 + B^4$ is given as $(A^2 - \sqrt{2}AB + B^2) (A^2 + \sqrt{2}AB + B^2)$.

So i dont' think they used complex numbers. I have tired expanding out the answer and it is correct. But i can't figure it out, how to get there?

5. No, it is wrong. You cannot obtain a full factorization of $a^4 + b^4$ in $\mathbb{R}$.

Here is an example : someone asks you to find the solutions of $6^4 + x^4 = 0$. You would apply your factorization to this problem, so it becomes :

$(6^2 - \sqrt{2} \times 6x + x^2) (6^2 + \sqrt{2}\times 6x + x^2) = 0$

Thus we are left with :

$x^2 - \sqrt{2}\times 6x + 36 = 0$

and

$x^2 + \sqrt{2}\times 6x + 36 = 0$

Oops ! None of these quadratic equations admit roots in $\mathbb{R}$. Now you are screwed without complex numbers.

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

I rather think the factorization they come up with is a bit useless, since any use you would eventually make of this factorization would anyway require complex numbers to be successfully accomplished.

6. Hi Bacterius,

The question is from Cambridge Mathematics 3 Unit (year 11). Exercise 1E, question 10.

Furthermore, similar answer are given from question 11. Please have a look at http://www.mathhelpforum.com/math-he...orisation.html

So i don't know how else to figure it out with basic maths. Mind you it is from the extension question section

7. I know what you mean.
The factorization you are given is correct, but you will need to learn complex numbers to effectively use it to solve problems (furthermore, you will learn a much simpler factorization involving complex numbers when you will have learnt the later).

8. Yes, your right. I am still a bit far from complex number. I think it's covered in the 4 unit (extension 2) course. But thanks again for you help

9. Thanks to Soroban, from post (http://www.mathhelpforum.com/math-he...orisation.html) i have figured it out:

$A^4+B^4$
$=A^4+2A^2B^2+B^4 - 2A^2B^2$
$=(A^2+B^2)^2 - (\sqrt{2}AB)^2$
$=(A^2-\sqrt{2}AB+B^2) (A^2+\sqrt{2}AB+B^2)$ [Difference of squares]