According to the fundamental theorem of algebra, how many roots exist for the polynomial function? f(x) = (x3-3x+1)2

The Fundamental Theorem of Algebra says that every non-constant single-variable polynomial with complex coefficients has at least one complex root.

We will see that the polynomial has 6 roots.

Let's see how we get that

The theorem means that even if we do not see the graph crossing the x-axis, there is still a root (a complex root).

Then the number of roots is just the same as the degree of the polynomial (such that there may be multiple roots in one given point).

The degree of the polynomial function:

p(x) = (x^3 - 3*x + 1)^2

is 6, so we have a total of 6 roots. (We only look at the larger exponent to find the degree of the polynomial).

If you want to learn more, you can read:

link

The fundamental theorem of the algebra, establishes that for any polynomial of degree n, the roots of the polynomial are also n roots.

Where n is an integer greater than zero.

We then have the following function:

We observe that the term of greatest exponent for the polynomial is:

Therefore, the polynomial has 6 roots.

exist 6 roots for the polynomial function

the answer is d

step-by-step explanation:

you can also use mathway to get you answer + steps