We are given the cubic equation:

$x^3 - 4x^2 + x + 6 = 0$

Step 1: Find Rational Roots Using the Rational Root Theorem

The possible rational roots are factors of the constant term $6$ divided by factors of the leading coefficient $1$. The possible rational roots are:

$\pm 1, \pm 2, \pm 3, \pm 6$

Let's test $x = 1$:

$(1)^3 - 4(1)^2 + 1 + 6 = 1 - 4 + 1 + 6 = 4 \neq 0$

So, $x = 1$ is not a root.

Let's test $x = -1$:

$(-1)^3 - 4(-1)^2 + (-1) + 6 = -1 - 4 - 1 + 6 = 0$

Since $x = -1$ is a root, we use synthetic division to factor the polynomial.

Step 2: Synthetic Division

We divide $x^3 - 4x^2 + x + 6$ by $(x + 1)$.

-1 & 1 & -4 & 1 & 6 \\ & & -1 & 5 & -6 \\ \hline & 1 & -5 & 6 & 0 \\ \end{array}$$ The quotient is $$x^2 - 5x + 6$$, so we can rewrite the equation as: $$(x + 1)(x^2 - 5x + 6) = 0$$ ### Step 3: Solve the Quadratic Equation Factorizing $$x^2 - 5x + 6$$: $$(x - 2)(x - 3) = 0$$ ### Step 4: Find the Roots Setting each factor to zero: $$x + 1 = 0 \quad \Rightarrow \quad x = -1$$ $$x - 2 = 0 \quad \Rightarrow \quad x = 2$$ $$x - 3 = 0 \quad \Rightarrow \quad x = 3$$ ### Final Answer: $$x = -1, 2, 3$$ Would you like a more detailed explanation or another method, such as graphical verification? --- Here are some related problems to try: 1. Solve $$x^3 + 2x^2 - 5x - 6 = 0$$. 2. Factorize $$x^3 - 3x^2 - 4x + 12$$. 3. Solve $$x^3 + x^2 - 10x + 8 = 0$$ by synthetic division. 4. Find all real roots of $$x^3 - 6x^2 + 11x - 6 = 0$$. 5. Verify if $$x = -2$$ is a root of $$x^3 + 3x^2 - x - 3 = 0$$. **Tip:** Always check for rational roots first using the Rational Root Theorem before applying synthetic division!