Maths help please! (finding roots of a degree 5 polynomial) +REP

[Matthew]

Basically, I've got to factorise 16x5 - 20x3 + 5x - 1 = 0

So far I just used the factor theorem to find (x-1) to be a root, and then divided the above polynomial by (x-1).

This left me with 16x4 + 16x3 - 4x2 - 4x + 1

where do I go from here? :(

this is FP3 by the way!
@Kardan (http://www.habboxforum.com/member.php?u=3428); as you normally seem to be able to answer my questions haha

I'm probably missing something super obvious but I've just stared at it for 10mins :P

[Kardan]

First step is correct.

We now have 16x^4 + 16x^3 - 4x^2 - 4x + 1

We can represent this equation as a quadratic equation squared, so we will have to find (ax^2 + bx + c)^2 = 16x^4 + 16x^3 - 4x^2 - 4x + 1

Multiplying out the bracket gives us: (ax^2 + bx + c)(ax^2 + bx + c) = (a^2)x^4 + abx^3 + acx^2 + abx^3 + (b^2)x^2 + bcx + acx^2 + bcx + c^2

Collate coefficients together:

a^2(x^4) + 2ab(x^3) + (2ac + b^2)x^2 + (2bc)x + c^2

Equate this to our degree 4 polynomial:

a^2(x^4) + 2ab(x^3) + (2ac + b^2)x^2 + (2bc)x + c^2 = 16x^4 + 16x^3 - 4x^2 - 4x + 1

Equate coefficients:

a^2(x^4) = 16(x^4), so a = 4.

2ab(x^3) = 16(x^3), so 2ab = 16, sub in a = 4 to give 8b = 16, so b = 2.

2bc(x) = -4x, so 2bc = -4, sub in b = 2 to give 4c = -4, so c = -1.

Check that a, b and c are correct with x^2 coefficient:

(2ac + b^2)x^2 = -4(x^2)

So 2ac + b^2 = -4, sub in a = 4, b = 2 and c = -1 to give 2*4*-1 + 4 = -4 as needed.

So our 4 degree polynomial can be expressed as (4x^2 + 2x - 1)^2

Can you solve it from here? :)

[Matthew]

-snip-

Can you solve it from here? :)

I've never come across that method before so I wouldn't know how to, no :(

Maybe I should have posted the question itself- we're currently using De Moivre's Theorem.

http://i.imgur.com/IA2b7q1.jpg

I've managed to part (i) as it boiled down to a simple hidden quadratic but the '1' being there on part (ii) is what's confusing me!

[Kardan]

With (4x^2 + 2x - 1)^2 = 0, take the square root of both sides, giving you:

4x^2 + 2x - 1 = 0 - Simply solve the quadratic for the roots.

That's what I would do at least, if there's a certain method to it that also involves polynomials, I can't see it at all.

[Matthew]

With (4x^2 + 2x - 1)^2 = 0, take the square root of both sides, giving you:

4x^2 + 2x - 1 = 0 - Simply solve the quadratic for the roots.

That's what I would do at least, if there's a certain method to it that also involves polynomials, I can't see it at all.

Yeah I saw that a minute or so after posting!

Thanks for your help, I've certainly never had to factorize a polynomial like this before :O

I mean the second part of the homework was to do the same sorta thing but with sin6x = 0 (finding it in terms of cosx and sinx yourself), which again is easy as its just a hidden quadratic and DMT is simple anyway. Ah well, thanks again!

[Shockwave.2CC]

So glad I didn't have to do this

This is bloody confusing :s

[Matthew]

So glad I didn't have to do this

This is bloody confusing :s

well i'd imagine it would be if you hadn't learnt/seen it before :P