# Factorising Polynomial over the real number field

• May 17th 2011, 05:35 PM
aonin
Factorising Polynomial over the real number field
hey guys I was just wondering whether I'm doing this right
for example the question asks to factor $\displaystyle z^6+8$ over the real number field

Heres what I did
$\displaystyle z^6+8=(z^2)^3+2^3$
$\displaystyle =(z^2+2)(z^4-2z^2+4)$

but when I checked the answers they got
$\displaystyle z^6+8=(z^2+2)(z^2+\sqrt6z+2)(z^2-\sqrt6z+2)$
they got this by pairing up the roots

which is more factored than my expression, the factors I got are not factorised enough right?, I have tried difference of 2 squares, yet they don't give me the same as the answer as they are complex
$\displaystyle =(z^2+2)(z^4-2z^4+4)$
$\displaystyle =(z^2+2)((z^2+1)^2-3i^2)$
$\displaystyle =(z^2+2)(z^2+1-\sqrt3i)(z^2+1+\sqrt3i)$

• Is the method I used originally wrong, since it is not fully factored?
• How would I be able to factorise it to the right answer (eg. provided by the textbook) by just continuing with the factors I originally got, without using the pairing up of roots method the text book used. Is it possible?

Thanks
Aonin (Nod)
• May 17th 2011, 06:53 PM
TKHunny
Difference of squares? Sure they do. Just rewrite a little:

$\displaystyle z^{4} - 2\cdot z^{2} + 4 = z^{4} + 4 - 2\cdot z^{2} = (z^{2} + 2)^{2} - 2\cdot z^{2} - 4\cdot z^{2} = (z^{2} + 2)^{2} - 6\cdot z^{2}$

You're just not used to completing the square on the usually-outside pieces. Add it to your bag of tricks. You might need it again.
• May 17th 2011, 07:41 PM
aonin
so by completing the square on the outside pieces, its possible to factor any polynomial that couldn't be factored otherwise..or is this a generalisation?
• May 17th 2011, 07:59 PM
TKHunny
Didn't you ever demonstarte the development of the Quadratic Formula?

$\displaystyle x^{4} + b^{2}x^{2} + c^{2} =$

$\displaystyle \left(x^{4} + c^{2}\right) + b^{2}x^{2} =$

$\displaystyle (x^{2}+c)^{2} + b^{2}x^{2} - 2cx^{2} =$

$\displaystyle (x^{2}+c)^{2} + \left(b^{2} - 2c\right)x^{2} =$

$\displaystyle (x^{2}+c)^{2} - \left(2c - b^{2}\right)x^{2} =$

That's an important point. $\displaystyle 2c - b^2 > 0$ I'm guessing you'll just get more imaginaries otherwise.

$\displaystyle (x^{2} + c + \sqrt{2c - b^{2}}|x|)\cdot (x^{2} + c - \sqrt{2c - b^{2}}|x|)$

I'm going to stick my neck out and say that you may have a chance of finding this result in a book on quartic equations by Orson Pratt in the middle 19th century. It's not my favorite result. I'm not very happy about the absolute values, either.

I think, if you force a few things into this form, something may occasionally come of it.