Problem
olynomial function $\mathrm{f}(\mathrm{x})$ of least degree having only real coefficients with zeros of $0,2 i$, and $4+i$. ynomial function is $f(x)=$
Solution
Step 1 ：Given the polynomial function $f(x)$ of least degree having only real coefficients with zeros of $0,2 i$, and $4+i$.
Step 2 ：Since the coefficients of the polynomial are real, the complex roots must come in conjugate pairs. Therefore, the polynomial must also have zeros at $-2i$ and $4-i$.
Step 3 ：We can express the polynomial function as $f(x) = a(x - 0)(x - 2i)(x - (-2i))(x - (4+i))(x - (4-i))$ for some real number $a$.
Step 4 ：Simplifying this expression, we find $f(x) = a(x^5 - 8x^4 + 21x^3 - 32x^2 + 68x)$.
Step 5 ：\(\boxed{f(x) = a(x^5 - 8x^4 + 21x^3 - 32x^2 + 68x)}\) is the polynomial function of least degree having only real coefficients with zeros of $0,2 i$, and $4+i$. Here, $a$ is any real number.
