Is this cublc polynomial function solvable?

In summary, the conversation is about two difficult cubic polynomials, one being x^3 - x - 2 = 0 and the other being x^3 + 9x - 1 = 0. The question is whether these polynomials are solvable or not. It is suggested to test for divisibility using synthetic division, but it is also possible to simply evaluate the polynomial to see if it satisfies the equation. It is mentioned that every cubic equation is solvable using the cubic root algorithm. The first polynomial can be factored using (x-2) as a binomial that gives a zero remainder, while the second polynomial can be factored using (x+1)^2.
  • #1
davedave
50
0
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.
 
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  • #2
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.
 
  • #3
EDIT: Deleted totally misleading answer. Ignore if you read it.
 
Last edited:
  • #4
davedave said:
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 said:
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.
 
  • #5
symbolipoint said:
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!
 
  • #6
You can use synthetic division, but isn't it simpler to just evaluate the polynomial and see if the equation is satisfied?
 
  • #8
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).
 

What is a cubic polynomial function?

A cubic polynomial function is a type of mathematical function that can be written in the form of f(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants and x is the independent variable. It is called a "cubic" function because the highest degree of the variable x is 3.

What does it mean for a cubic polynomial function to be solvable?

A cubic polynomial function is considered solvable if there exists a way to find the value(s) of x that make the function equal to a specific number. In other words, it is possible to find the roots or solutions of the function.

Is every cubic polynomial function solvable?

Yes, every cubic polynomial function is solvable. This is because of the Fundamental Theorem of Algebra, which states that every polynomial function of degree n has exactly n complex roots. Since a cubic polynomial function has a degree of 3, it will always have 3 complex roots, which means it is always solvable.

What methods can be used to solve a cubic polynomial function?

There are multiple methods that can be used to solve a cubic polynomial function, including factoring, the rational root theorem, synthetic division, and the cubic formula. The method used will depend on the specific function and the given information.

Can a cubic polynomial function have imaginary solutions?

Yes, a cubic polynomial function can have imaginary solutions. This is because of the Fundamental Theorem of Algebra, which states that every polynomial function has complex roots. Imaginary solutions will occur when the function has a discriminant that is less than 0.

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