Polynomial Equation (1 Viewer)

[wagig]

i) Solve
ii) Write this equation in the form of two quadratic factors

 

[rumbleroar]

What have you tried so far?

 

[HeroicPandas]

Part (i), i can think of 2 methods

Method 1:
It is a series and there are only 2 series u know: arithmetic or geometric

Method 2:
Use this z^n - 1 = (z-1)(1+z+z^2+z^3+...+z^(n-1))

ii) When u solve (i) you know which roots belong to that equation (say A, B, C, D), but to express it in the form of quadratic factors...
u do this z^4 + z^3 + z^2 + z + 1 = (z-A)(z-B)(z-C)(z-D)(z-E) then expand

 

[wagig]

What have you tried so far?

My friend solved both parts using the fifth roots of unity:

I.e.
(z^5-1) = (z-1)(z^4+z^3+z^2+z+1)
Then found the 5 roots using de moivre's theorem and just said that 1 is not a solution.

But to me it just seems like something too unrelated, and there's probably a less obscure method of solving it,
any ideas?

 

[obliviousninja]

My friend solved both parts using the fifth roots of unity:

I.e.
(z^5-1) = (z-1)(z^4+z^3+z^2+z+1)
Then found the 5 roots using de moivre's theorem and just said that 1 is not a solution.

But to me it just seems like something too unrelated, and there's probably a less obscure method of solving it,
any ideas?

looks about right to me.

 

[wagig]

Part (i), i can think of 2 methods

Method 1:
It is a series and there are only 2 series u know: arithmetic or geometric

Method 2:
Use this z^n - 1 = (z-1)(1+z+z^2+z^3+...+z^(n-1))

ii) When u solve (i) you know which roots belong to that equation (say A, B, C, D), but to express it in the form of quadratic factors...
u do this z^4 + z^3 + z^2 + z + 1 = (z-A)(z-B)(z-C)(z-D)(z-E) then expand

Ohhh okay thats cool; i didn't know about the

Spoiler

z^n - 1 = (z-1)(1+z+z^2+z^3+...+z^(n-1)).

Guess that makes what my friend did less of a stretch.
Thanks

 

[Drongoski]

(z² -2 ... + 1) (z² -2 ... +1) = 0

 

[Squar3root]

some nice trolling in this thread

 

[RealiseNothing]

some nice trolling in this thread

wot

 

[Squar3root]

wot

See what I did there?

 

[SpiralFlex]

Its not trolling. It's customary (generally) in math/science forums for people to hide their solutions and give hints to the OP without starving him the opportunity to learn and find the answer themselves.

 

[dunjaaa]

Part (i) can also be done by dividing by z^2 so you get (z^2+1/z^2) + (z+1/z) + 1 = 0, then using the result z^n+1/z^n=2cos(nƟ) solving for Ɵ and interpreting solutions in terms of cis(Ɵ) by taking into account the principle argument -π≤Ɵ≤π when deciding which angles to select. Well since the polynomial is of degree 4 and real, there should be 2 conjugate pairs of roots.
Part (ii), just like HeroicPanda mentioned, express the roots in linear factors as such (z-A)(z-A(conjugate))(z-B)(z-B(conjugate)) and by expanding you should get a product of 2 quadratic factors. Just a quick tip, when you expand (z-A)(z-A(conjugate)), it becomes z^2-2Re(A)+1. The main purpose for why they make you express in the form of 2 quadratic factors is because its the simplest form in which the polynomial can be factorised over the real field.

 

[Dabest]

Easy... if you have been using the terry lee book