Is this cublc polynomial function solvable?

[davedave]

Here is a very difficult cubic polynomial.

x^3 - x - 2 = 0

I am wondering whether it is solvable or not. Please think about it.

 

[aew782]

I need help with this too I posted a similar one and haven't got a response... mine was x^3 + 9x -1=0...They are solvable, but I don't know how to get an answer algebraically or graphically.

 

[rasmhop]

EDIT: Deleted totally misleading answer. Ignore if you read it.

 

[symbolipoint]

Here is a very difficult cubic polynomial.

x^3 - x - 2 = 0

I am wondering whether it is solvable or not. Please think about it.

I need help with this too I posted a similar one and haven't got a response... mine was x^3 + 9x -1=0...They are solvable, but I don't know how to get an answer algebraically or graphically.

Are they solvable? Let's have a guess: The first one can be tested for divisibility by x+1, x-1, x+2, and x-2. The second one can be tested for divisibility by x+1 and x-1. The results may be faster if you know synthetic division.

 

[aew782]

Are they solvable? Let's have a guess: The first one can be tested for divisibility by x+1, x-1, x+2, and x-2. The second one can be tested for divisibility by x+1 and x-1. The results may be faster if you know synthetic division.

Thank you that's all I needed!

 

[SteamKing]

You can use synthetic division, but isn't it simpler to just evaluate the polynomial and see if the equation is satisfied?

 

[micromass]

Every cubic equation is solvable. You can always use the cubic root algorithm: http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

 

[symbolipoint]

The first binomial rendering zero remainder is (x-2). The quotient is x^2+2x+1 which is (x+1)^2. No need to use Cardano's or Vietas substitution for this cubic (micromass gave a good reference Wikipedia article).