1. State the problem. Use the Factor Theorem to determine whether $x-2$ is a factor of $f(x)=x^3-4x^2+x+6$, and to factor $f(x)$ completely. 2. Formula and important rules. The Factor Theorem states that $f(a)=0$ if and only if $x-a$ is a factor of $f(x)$. The Remainder Theorem says the remainder when dividing $f(x)$ by $x-a$ equals $f(a)$. Synthetic division is a quick method to divide a polynomial by a linear factor and obtain the quotient and remainder. 3. Evaluate $f(2)$ to test the candidate factor. Compute $f(2)=8-16+2+6$. Simplify to $f(2)=0$. Therefore $x-2$ is a factor of $f(x)$ by the Factor Theorem. 4. Divide $f(x)$ by $x-2$ using synthetic division to find the quotient. Coefficients: $1, -4, 1, 6$. Synthetic division steps: bring down $1$; multiply by $2$ gives $2$; add to $-4$ gives $-2$. Multiply $-2$ by $2$ gives $-4$; add to $1$ gives $-3$. Multiply $-3$ by $2$ gives $-6$; add to $6$ gives remainder $0$. So the quotient polynomial is $x^2-2x-3$. 5. Factor the quadratic quotient. Factor $x^2-2x-3$ as $(x-3)(x+1)$ because $-3+1=-2$ and $(-3)(1)=-3$. 6. Final factorization and check. Combine the factors to get $f(x)=(x-2)(x-3)(x+1)$. Optionally expand to check: $(x-2)(x-3)(x+1)=x^3-4x^2+x+6$ which matches the original polynomial. Final answer: $f(x)=(x-2)(x-3)(x+1)$.