To factorize $x^3 + 6x^2 + 11x - 6 = 0$ using synthetic division, follow these steps:

Step 1: Find a possible root using the Rational Root Theorem.

The possible rational roots are factors of the constant term (-6) divided by factors of the leading coefficient (1). So the possible roots are: $\pm 1, \pm 2, \pm 3, \pm 6$

Step 2: Test for roots using synthetic division.

Let's start by testing $x = 1$.

Set up synthetic division:

1 & 1 & 6 & 11 & -6 \\ & & 1 & 7 & 18 \\ \hline & 1 & 7 & 18 & 12 \end{array}$$ Since the remainder is not zero, $$x = 1$$ is not a root. Now let's try $$x = -1$$. ### Step 3: Try $$x = -1$$ using synthetic division. $$\begin{array}{r|rrrr} -1 & 1 & 6 & 11 & -6 \\ & & -1 & -5 & -6 \\ \hline & 1 & 5 & 6 & 0 \end{array}$$ The remainder is zero, which means $$x = -1$$ is a root. Therefore, $$x^3 + 6x^2 + 11x - 6$$ can be factorized as: $$(x + 1)(x^2 + 5x + 6)$$ ### Step 4: Factorize the quadratic $$x^2 + 5x + 6$$. Now factor the quadratic: $$x^2 + 5x + 6 = (x + 2)(x + 3)$$ ### Step 5: Write the fully factorized form. Thus, the full factorization of $$x^3 + 6x^2 + 11x - 6 = 0$$ is: $$(x + 1)(x + 2)(x + 3) = 0$$ The roots of the equation are $$x = -1$$, $$x = -2$$, and $$x = -3$$. Would you like further details on any step or have any questions? ### Related Questions: 1. How is the Rational Root Theorem applied in synthetic division? 2. Can synthetic division be used for higher-degree polynomials? 3. How do we factor quadratics like $$x^2 + 5x + 6$$? 4. What are the common mistakes in synthetic division? 5. How does synthetic division compare to long division? **Tip:** Always test simpler potential roots first, like $$\pm 1$$, to minimize calculations.