Answer:
Hence, the polynomial function in standard form is:

Step-by-step explanation:
We are given a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros 2,5-i.
As we know that the polynomial function has rational coefficients.
so any complex root always appear in pair.
so as 5-i is a root of f(x) then it's complex conjugate 5+i is also an root of f(x).
Hence f(x) could be represented as:
![f(x)= (x-2)[x-(5-i)][x-(5+i)]\\\\f(x) = (x - 2)[(x-5)+ i][(x-5)- i)\\\\f(x)= (x - 2)[(x-5)^2 - i^2]\\\\f(x)=(x - 2)(x^2 - 10x + 25 + 1)\\\\f(x)= (x - 2)(x^2 - 10x + 26)\\\\f(x) = x^3 - 10x^2 + 26x - 2x^2 + 20x-52\\\\f(x)=x^3 - 12x^2 + 46x - 52](https://tex.z-dn.net/?f=f%28x%29%3D%20%28x-2%29%5Bx-%285-i%29%5D%5Bx-%285%2Bi%29%5D%5C%5C%5C%5Cf%28x%29%20%3D%20%28x%20-%202%29%5B%28x-5%29%2B%20i%5D%5B%28x-5%29-%20i%29%5C%5C%5C%5Cf%28x%29%3D%20%28x%20-%202%29%5B%28x-5%29%5E2%20-%20i%5E2%5D%5C%5C%5C%5Cf%28x%29%3D%28x%20-%202%29%28x%5E2%20-%2010x%20%2B%2025%20%2B%201%29%5C%5C%5C%5Cf%28x%29%3D%20%28x%20-%202%29%28x%5E2%20-%2010x%20%2B%2026%29%5C%5C%5C%5Cf%28x%29%20%3D%20x%5E3%20-%2010x%5E2%20%2B%2026x%20-%202x%5E2%20%2B%2020x-52%5C%5C%5C%5Cf%28x%29%3Dx%5E3%20-%2012x%5E2%20%2B%2046x%20-%2052)
Hence, the polynomial function in standard form is:
