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

6

Step-by-step explanation:

First, we can expand the function to get its expanded form and to figure out what degree it is. For a polynomial function with one variable, the degree is the largest exponent value (once fully expanded/simplified) of the entire function that is connected to a variable. For example, x²+1 has a degree of 2, as 2 is the largest exponent value connected to a variable. Similarly, x³+2^5 has a degree of 2 as 5 is not an exponent value connected to a variable.

Expanding, we get

(x³-3x+1)²  = (x³-3x+1)(x³-3x+1)

= x^6 - 3x^4 +x³ - 3x^4 +9x²-3x + x³-3x+1

= x^6 - 6x^4 + 2x³ +9x²-6x + 1

In this function, the largest exponential value connected to the variable, x, is 6. Therefore, this is to the 6th degree. The fundamental theorem of algebra states that a polynomial of degree n has n roots, and as this is of degree 6, this has 6 roots

it should be a.

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