Can polynomials have odd roots (like f(x)= cube root of i’ll give branliest!

I think you're asking if it's possible to have a cube root, fifth root, 7th root, etc of a number as a solution to f(x). The answer is yes it's possible.

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Example:

f(x) = x^3 - 29

This function has one real-number root of (cube root of 29) and the other two roots are complex or imaginary roots.

Yes.

Step-by-step explanation:

A simple example is  x^3 - 27 = 0  where  one root is x = ∛27

answer:     true

step-by-step explanation:

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.