Expert: Paul Klarreich - 9/24/2007

Question
factorise the polynomial x^8 + 16 into polynomial factors with real coefficients, where the factos are either linear or quadratic.

Could you also give me more questions to try and a general method in Layman terms? Thanks

Answer
Questioner:   Amy
Category:  Advanced Math
Private:  No
 
Subject:  Complex number
Question:  factorise the polynomial x^8 + 16 into polynomial factors with real coefficients, where the factos are either linear or quadratic.

Could you also give me more questions to try and a general method in Layman terms? Thanks
.........................................
Hi, Amy,

Since you are studying complex numbers, you have, I assume, learned deMoivre's theorem, so you normally handle this by

1. writing your equation as  z^8 + 16 = 0, then z^8 = - 16

2. writing  z in polar form:  z = r (cos t + i sin t), which I will abbreviate as:

z = r CiS t  [Cosine + i Sine]

['t' is short for 'theta']

3. writing -16 in polar form in EIGHT ways:  Eight, because you have z^8.

- 16 = 16 cis  180
- 16 = 16 cis  540  << add 360 each time
- 16 = 16 cis  900  << add 360 each time
- 16 = 16 cis 1260  << add 360 each time
- 16 = 16 cis 1620  << add 360 each time
- 16 = 16 cis 1980  << add 360 each time
- 16 = 16 cis 1440  << add 360 each time
- 16 = 16 cis 1800  << add 360 each time

Now for the EIGHTH ROOT of each, you take the (real, positive) eighth root of 16, which is  sqrt(2),  then you take 1/8 of each of those angles.

z1 = sqrt(2) cis   22.5
z2 = sqrt(2) cis   67.5  << add 45 each time
z3 = sqrt(2) cis  112.5  
z4 = sqrt(2) cis  157.5
z5 = sqrt(2) cis  202.5
z6 = sqrt(2) cis  247.5
z7 = sqrt(2) cis  292.5
z8 = sqrt(2) cis  337.5

Now you have EIGHT linear factors of the form:

(z - z1)()()()()()()(z - z8)

Now you can observe that z1 and z8 are Complex Conjugates.   [C-C]
So are Z2 and Z7
So are Z3 and Z6
So are Z4 and Z5

.........................
Take the two factors using the first pair of C-C roots:

Note that:
A. If you add two C-C numbers, the sum is real.
B. If you multiply two C-C, the sum is real and positive.

Multiply these factors:
(z - z1)(z - z8)

(z - S cis 22.5)(z - S cis 337.5) =

[S stands for sqrt(2) to save typing]

(z^2 - S[cis 22.5 + cis 337.5]z  + S^2 cis(22.5 + 337.5) )
      -----------------------     --------------------
        sum of C-C numbers        product of C-C numbers


(z^2 - S[cis 22.5 + cis 337.5]z + S^2 cis(22.5 + 337.5) )

OK. That is a quadratic factor with real coefficients.  If you work out the others, you get your four quadratic factors with real coefficients.

That's the best I can think of now.