1. ## 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^7-3x^6-7x^4+21x^3-8x+24 has a factor (x-3)

a) Divide x^7-3x^6-7x^4+21x^3-8x+24 by x-3 and thus:

b) express it in the form (x-3)(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^7-3x^6-7x^4+21x^3-8x+24 as the product of 7 linear factors.

I managed part a) and got x^6-7x^3-8 and part b was easy enough

c) i substituted in Z=x^3 and got z^2-7z-8 I then used the quadratic formula and got x = 8 and x =-1

From there i used (x-8) and (X+1) multipled them together and tried polynomial division on x^6-7x^3-8 but that didnt work

then tried the same on x^7-3x^6-7x^4+21x^3-8x+24 and that didn't work.

I'm at a bit of a loss now, so any help on this would be apprecciated!

Cheers

2. Hi

So you get $\displaystyle x^7-3x^6-7x^4+21x^3-8x+24 = (x-3)(x^3+1)(x^3-8)$

Now you just need to find the 3 cubic roots of -1 and the 3 cubic roots of 8

$\displaystyle x = r e^{i\theta}$

$\displaystyle x^3 = r^3 e^{3i\theta} = -1 = e^{i\pi}$

$\displaystyle r^3 = 1$ 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 $\displaystyle x+1$
$\displaystyle x^3+1 = (x+1)(x^2-x+1)$
Then you need to solve $\displaystyle x^2-x+1=0$

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# factoring 7th degree polynomials example with solutions

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