Final answer:
To determine the fourth-degree polynomial with zeros 2, 1-4i, and f(0) = 17, we use the conjugate root theorem and plug in f(0) to solve for the coefficients. The resulting equation is f(x) = (1/17)(x - 2)(x^2 - 2x + 17)(x + 1).
Step-by-step explanation:
The equation for a fourth-degree polynomial with the given zeros (2, 1-4i) and the value of the polynomial at zero (f(0) = 17) can be found by using the fact that complex roots of polynomials with real coefficients come in conjugate pairs. Here, if 1-4i is a zero, then its conjugate, 1+4i, must also be a zero. To find the fourth zero, we can use the fact that f(0) = 17.
Our polynomial can be expressed as:
f(x) = a(x - 2)(x - (1 - 4i))(x - (1 + 4i))(x - r)
Since 1-4i and 1+4i are complex conjugates, their product will be (x - 1 + 4i)(x - 1 - 4i) = (x - 1)^2 + (4i)^2 which simplifies to x^2 - 2x + 17.
Now, we have:
f(x) = a(x - 2)(x^2 - 2x + 17)(x - r)
To find the values of a and r, we plug in x = 0: a(-2)(17)(-r) = 17. Solving for a and r, we get a = 1/17 and r = -1.
The final polynomial equation is:
f(x) = (1/17)(x - 2)(x^2 - 2x + 17)(x + 1).
To obtain the polynomial in standard form, we would need to expand the product of these terms.