Find all possible roots/zeros of the function $f(x)=x^3-4x^2+x+6$

Step 1 ：First, let's use the Rational Root Theorem (RRT). The RRT states that any rational root, say $\frac{p}{q}$, of the polynomial $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_{1}x+a_0$ must have $p$ as a factor of the constant term $a_0$ and $q$ as a factor of the leading coefficient $a_n$. In this case, the constant term is 6 and the leading coefficient is 1.

Step 2 ：The factors of 6 are $\pm 1$, $\pm 2$, $\pm 3$, and $\pm 6$. Since the leading coefficient is 1, the possible rational roots of the polynomial are $\pm 1$, $\pm 2$, $\pm 3$, and $\pm 6$.

Step 3 ：We substitute each possible rational root into the function until we find one that makes the function equal to zero. By doing this, we find that $-1$, $2$, and $3$ are the roots of the function.