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

## Recommended Posts

how do i solve this guy?

thanks

##### Share on other sites

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

##### Share on other sites

well do it this way.

$x^{2}(x-1)-\frac{2}{x}=0$

$x^{3}-x^{2}=\frac{2}{x}$

$x^{4}-x^{3}-2=0$

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.

##### Share on other sites

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.

##### Share on other sites

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

##### Share on other sites

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

##### Share on other sites

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

##### Share on other sites

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

##### Share on other sites

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.

##### Share on other sites

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

## Create an account

Register a new account