Question

Find the zeros of the function below. Show that the Fundamental Theorem of Algebra is true for the function by counting the number of complex zeros.

j(x) = 4x³ – 4x² – x + 1

Answer 1

The zeros of the function are -1, 1, and 1/2. The Fundamental Theorem of Algebra is true for this function.

Step-by-step explanation:

Find the zeros of the function:

j(x) = 4x³ – 4x² – x + 1

To find the zeros of the function, we set j(x) equal to zero and solve for x:

4x³ – 4x² – x + 1 = 0

Using synthetic division or factoring, we find that the zeros of the function are x = -1, x = 1, and x = 1/2.

Using the Fundamental Theorem of Algebra:

The Fundamental Theorem of Algebra states that every non-constant polynomial function has at least one complex zero.

Since the given function is a cubic polynomial, it can have up to three complex zeros. In this case, we have found all three zeros of the function, and they are all real numbers. Therefore, the Fundamental Theorem of Algebra is true for this function.