Poly Question (1 Viewer)

JustAnotherNE

New Member
Joined
Oct 13, 2007
Messages
16
Gender
Male
HSC
2008
Can some one please teach me how to solve this question?

If P(x) = 5X^4 - 11X^3 + 16X^2 - 11X + 5, solve P(x) = 0 over Complex field and factorise P(x) fully over Real field.
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
I've graphed that on Graphmatica, I don't think theres any real solutions for it. (hence why i couldn't factorise it over the real field).
 
Last edited:

JustAnotherNE

New Member
Joined
Oct 13, 2007
Messages
16
Gender
Male
HSC
2008
yea its mean to have im. roots but i dont know how to get them lol but thx any way
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
Neither, normally the equation has real and imaginary roots, so it's possible to find them but for this I'm pretty lost on.

I'm thinking of letting z = x + iy and subbing it in/expand/collect like terms (real/imaginary), but I can't be bothered doing it and I'm trying to figure out how it'll help me.
 

JustAnotherNE

New Member
Joined
Oct 13, 2007
Messages
16
Gender
Male
HSC
2008
i fink i just got an idea, can u divide both side by X^2 and get somefin from there?
 

JustAnotherNE

New Member
Joined
Oct 13, 2007
Messages
16
Gender
Male
HSC
2008
acually it gets really ugly if u do it that way....maybe theres a easier way
 

Mark576

Feel good ...
Joined
Jul 6, 2006
Messages
515
Gender
Male
HSC
2008
P(x) = 5x4 - 11x3 + 16x2 - 11x + 5 = 5x4 + 5 - 11x3 - 11x + 16x2 = 0
[Divide through by x2]
5x2 + 5/x2 - 11x - 11/x + 16 = 0
5(x2 + 1/x2) - 11(x + 1/x) + 16 = 0
5(x + 1/x)2 - 10 - 11(x + 1/x) + 16 = 0
5(x + 1/x)2 - 11(x + 1/x) + 6 = 0
Let z = x + 1/x;
5z2 - 11z + 6 = 0
(5z - 6)(z - 1) = 0
z = 6/5, 1
Solve the remaining quadratic equations in x to find four solutions.

EDIT: Notice this works because the coefficients of P(x) are symmetrical. So look out for this method.
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
Mark576 said:
P(x) = 5x4 - 11x3 + 16x2 - 11x + 5 = 5x4 + 5 - 11x3 - 11x + 16x2 = 0
[Divide through by x2]
5x2 + 5/x2 - 11x - 11/x + 16 = 0
5(x2 + 1/x2) - 11(x + 1/x) + 16 = 0
5(x + 1/x)2 - 10 - 11(x + 1/x) + 16 = 0
5(x + 1/x)2 - 11(x + 1/x) + 6 = 0
Let z = x + 1/x;
5z2 - 11z + 6 = 0
(5z - 6)(z - 1) = 0
z = 6/5, 1
Solve the remaining quadratic equations in x to find four solutions.

EDIT: Notice this works because the coefficients of P(x) are symmetrical. So look out for this method.
I thought of doing a dummy variable, but had no idea how to obtain it

awesome :)
 

paddy@2011

New Member
Joined
Jun 28, 2004
Messages
7
Gender
Undisclosed
HSC
N/A
P(x) = 5X^4 - 11X^3 + 16X^2 - 11X + 5 = 0

Divide both sides by x^2 and we get

5x^2 - 11x + 16 - 11/x + 5/x^2 = 0

5(x^2 + 1/x^2) -11(x+1/x) +16 =0

[Now if u = x + 1/x, then x^2 + 1/x^2 = u^2 - 2]

so the above equation becomes 5(u^2 - 2) -11u + 16 = 0

5u^2 - 11u + 6 = 0
(5u - 6) (u - 1) = 0

so u = 5/6 or 1

and now solve x + 1/x = 5/6 and then x + 1/x = 1
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top