The given expression is:

$x^3 - 7x^2 + 14x - 6$

Let’s find its factors using the factor theorem or synthetic division. First, let's check for possible rational roots by applying the rational root theorem. The possible rational roots would be the factors of the constant term $(-6)$ divided by the factors of the leading coefficient $1$.

Possible rational roots: $\pm 1, \pm 2, \pm 3, \pm 6$

Testing $x = 1$: $1^3 - 7(1)^2 + 14(1) - 6 = 1 - 7 + 14 - 6 = 2 \quad (\text{not a root})$

Testing $x = 2$: $2^3 - 7(2)^2 + 14(2) - 6 = 8 - 28 + 28 - 6 = 2 \quad (\text{not a root})$

Testing $x = 3$: $3^3 - 7(3)^2 + 14(3) - 6 = 27 - 63 + 42 - 6 = 0 \quad (\text{root})$

Since $x = 3$ is a root, we can factor the polynomial as:

$x^3 - 7x^2 + 14x - 6 = (x - 3)(\text{quadratic})$

Now, we perform synthetic division to divide the cubic expression by $(x - 3)$:

3 & 1 & -7 & 14 & -6 \\ & & 3 & -12 & 6 \\ \hline & 1 & -4 & 2 & 0 \\ \end{array}$$ This gives the quotient $$x^2 - 4x + 2$$. Thus, the factored form of the polynomial is: $$(x - 3)(x^2 - 4x + 2)$$ The quadratic $$x^2 - 4x + 2$$ can be solved using the quadratic formula: $$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)} = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}$$ So, the complete factorization of the polynomial is: $$(x - 3)(x - (2 + \sqrt{2}))(x - (2 - \sqrt{2}))$$ ### Would you like more details or clarifications? Here are five related questions you might explore next: 1. What is the discriminant and how does it help with quadratic equations? 2. How do synthetic division and long division of polynomials differ? 3. Can the Rational Root Theorem be applied to any polynomial? 4. How do you solve cubic equations when no rational roots exist? 5. How is the quadratic formula derived from completing the square? **Tip:** Always test small integers first when applying the Rational Root Theorem, as they are often simpler roots.