Final answer:
To find all the zeros of f(x), we divide the polynomial by (x-1) to get a quadratic equation, then we apply the quadratic formula, resulting in one real zero at x=1 and two complex zeros 4+i and 4-i.
Step-by-step explanation:
Given that f(1)=0 for the third-degree polynomial f(x)=x^3-9x^2+33x-25, we can confirm that x=1 is a zero of the function. To find the other zeros, we must divide the polynomial by (x-1) to reduce it to a quadratic equation. The polynomial division yields x^2-8x+25. The resulting quadratic equation can be written as ax^2+bx+c=0, where a=1, b=-8, and c=25.
To find the remaining zeros, we apply the quadratic formula:
x = [-b ± sqrt(b^2-4ac)] / (2a). Substituting in the values, we get:
x=(-(-8) ± sqrt((-8)^2-4(1)(25)))/(2(1)), which simplifies to x = (8 ± sqrt(64-100))/2. Since sqrt(64-100) results in an imaginary number, the remaining zeros are complex numbers.
The roots of the quadratic are thus x=4 ± i, where i is the imaginary unit. Therefore, the zeros of the polynomial f(x) are 1, 4+i, and 4-i.