Find a polynomial function of least degree having only real coefficients, a leading coefficient of 1 , and zeros of $3-i$ and $4-2 i$. The polynomial function is $f(x)=$ (Simplify your answer.)

Step 1 ：The zeros of a polynomial are the values of x for which the polynomial equals zero. If a polynomial has real coefficients, then any complex zeros must occur in conjugate pairs. This means that if \(3-i\) is a zero, then its conjugate \(3+i\) must also be a zero. Similarly, if \(4-2i\) is a zero, then its conjugate \(4+2i\) must also be a zero.

Step 2 ：We can find the polynomial by multiplying the factors associated with each zero. The factor associated with a zero a is \((x-a)\).

Step 3 ：So, the polynomial we are looking for is \((x-(3-i))(x-(3+i))(x-(4-2i))(x-(4+2i))\).

Step 4 ：We can simplify this expression by multiplying out the factors.

Step 5 ：Final Answer: The polynomial function of least degree having only real coefficients, a leading coefficient of 1 , and zeros of \(3-i\) and \(4-2 i\) is \(\boxed{x^4 - 14x^3 + 78x^2 - 200x + 200}\).