x^2(x-1)-2/x

[kingjewel1]

how do i solve this guy?

 

thanks

[bloodhound]

solve what?

 

i assume u mean finding the root of that function

 

it looks like u will need to find the roots of a quartic of a form ax^4 + bx^3 + c = 0

there should be a general formula for that, i dont know it. lemme just try to look for it on the net

[bloodhound]

well do it this way.

[math]x^{2}(x-1)-\frac{2}{x}=0[/math]

[math]x^{3}-x^{2}=\frac{2}{x}[/math]

[math]x^{4}-x^{3}-2=0[/math]

 

then -1 iby inspection is a root, therefore (x+1) is a factor of f(x). then you can reduce the polynomial into a quintic

 

information on solving quartic

http://mathworld.wolfram.com/QuarticEquation.html

there is also a link at the bottom to solutions to quintic

 

I banged the equation into Maple, and the solutions are horrible. the only nice solutions is -1, the others are massive strings of numbers.

[JaKiri]

That should have been fairly obvious, as you can see that x +- 2 can't be a factor, so you've straight ruled out all the other integrals.

[bloodhound]

jakiri i dont understand your last sentence, could u explain in more detail.

[Dave]

I think he meant intervals instead of integrals, but I could be wrong.

[kingjewel1]

thanks.

i could see the -1 as being factor but i thought there might be a way to figure out the other one(s): relatively easily i've looked on the net but i don't seem to be coming up with any answer. :S

[bloodhound]

well u know there is 1 real root, so there must be another real root, and then two complex roots which are conjugates.

 

here is my screenshot from maple

[JaKiri]

I think he meant intervals instead of integrals, but I could be wrong.

 

All the other integral roots.

 

You have an equation which ends in a prime. To factorise it into an (x+a)(x+b)(x+....) form, a*b*c*whatever must equal the prime. To get integral values, one of those numbers must be the prime, and the rest must be one, by the definition of a prime number.

 

As it's fairly clear that 2 is not a factor, the solutions therefore can't be nice integrals.

[bloodhound]

Another useful test in factorising polynomials over rationals, or to test if it has any rational roots would be the Rational Root Test

[kingjewel1]

thanks for your help