
7th degree polynomial
Hi there abit stuck with this problem, i've worked through parts and b and some of c but thats were i'm getting stuck.
The 7th degree polynomial x^73x^67x^4+21x^38x+24 has a factor (x3)
a) Divide x^73x^67x^4+21x^38x+24 by x3 and thus:
b) express it in the form (x3)(ax^6+bx^3+c)
c) By putting z=x^3, find all the factors, real or complex of the 6th degree polynomial and thus:
d) express x^73x^67x^4+21x^38x+24 as the product of 7 linear factors.
I managed part a) and got x^67x^38 and part b was easy enough
c) i substituted in Z=x^3 and got z^27z8 I then used the quadratic formula and got x = 8 and x =1
From there i used (x8) and (X+1) multipled them together and tried polynomial division on x^67x^38 but that didnt work
then tried the same on x^73x^67x^4+21x^38x+24 and that didn't work.
I'm at a bit of a loss now, so any help on this would be apprecciated!
Cheers

Hi
So you get $\displaystyle x^73x^67x^4+21x^38x+24 = (x3)(x^3+1)(x^38)$
Now you just need to find the 3 cubic roots of 1 and the 3 cubic roots of 8
For 1 this leads to
$\displaystyle x = r e^{i\theta}$
$\displaystyle x^3 = r^3 e^{3i\theta} = 1 = e^{i\pi}$
$\displaystyle r^3 = 1$ then $\displaystyle r = 1$
$\displaystyle 3\theta = \pi + 2k \pi$ then $\displaystyle \theta = \frac{\pi}{3} + \frac{2k \pi}{3}$ with k integer
Another way is to say that 1 is one cubic root of 1
Therefore $\displaystyle x^3+1$ can be factorized by $\displaystyle x+1$
$\displaystyle x^3+1 = (x+1)(x^2x+1)$
Then you need to solve $\displaystyle x^2x+1=0$