- #1
davedave
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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.
x^3 - x - 2 = 0
I am wondering whether it is solvable or not. Please think about it.
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.
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.
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.
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.
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.
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.
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.