This is a trivial example of a cubic polynomial. The x term appears only once, therefore you can solve directly for x:
Good question. The answer is that the numbers will not simply be imaginary: they will be of the form , with form a real and an imaginary component. It is possible for such numbers to be cube roots of a real number. For example, the complex cube roots of 1 are:i also don't understand how a polynomial with odd degree can have a complex root.. if someone can help me please give me a pointer for this. then thank you. what i don't understand is that with the odd degree, the imaginary part will not disappear as it is supposed to.
In addition, to get the complex cube roots of any real number, including the one in the problem, just multiply these numbers by the real cube root. This is because when you cube the product, you get the product of the cubes, one of which will simply be 1.
So the three solutions to your polynomial are: