Poly Question (1 Viewer)

[JustAnotherNE]

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'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).

 

[JustAnotherNE]

yea its mean to have im. roots but i dont know how to get them lol but thx any way

 

[tommykins]

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]

i fink i just got an idea, can u divide both side by X^2 and get somefin from there?

 

[JustAnotherNE]

acually it gets really ugly if u do it that way....maybe theres a easier way

 

[Mark576]

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.

 

[JustAnotherNE]

thx heaps

 

[Mark576]

No problem.

 

[tommykins]

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]

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