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

Quote by 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.

Quote by 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.

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

Quote by symbolipoint

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_f...ano.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).

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davedave	Is this cublc polynomial function solvable? 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.

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

micromass Mentor Blog Entries: 8	Every cubic equation is solvable. You can always use the cubic root algorithm: http://en.wikipedia.org/wiki/Cubic_f...ano.27s_method

symbolipoint Recognitions: Homework Help	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).