To express the cubic polynomial $y = x^3 - x^2 - 8x + 12$ in factored form, we perform the following steps:

Step 1: Identify potential roots using the Rational Root Theorem.

The possible rational roots are factors of the constant term (12) divided by factors of the leading coefficient (1). This gives: $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$

Step 2: Test roots using synthetic or direct substitution.

Substituting $x = 2$: $y = (2)^3 - (2)^2 - 8(2) + 12 = 8 - 4 - 16 + 12 = 0$ Thus, $x = 2$ is a root.

Step 3: Perform synthetic or polynomial division.

Divide $x^3 - x^2 - 8x + 12$ by $(x - 2)$:

2 & 1 & -1 & -8 & 12 \\ & & 2 & 2 & -12 \\ \hline & 1 & 1 & -6 & 0 \\ \end{array}$$ The quotient is $$x^2 + x - 6$$, so: $$x^3 - x^2 - 8x + 12 = (x - 2)(x^2 + x - 6)$$ ### Step 4: Factor the quadratic $$x^2 + x - 6$$. Factorizing $$x^2 + x - 6$$: $$x^2 + x - 6 = (x + 3)(x - 2)$$ ### Final Factored Form: $$y = (x - 2)(x + 3)(x - 2)$$ Or equivalently: $$y = (x - 2)^2(x + 3)$$ Would you like further details or examples? Here are five questions for further practice: 1. How do you factorize higher-degree polynomials like $$x^4 - 3x^3 + x^2 - x + 2$$? 2. What is the significance of the Rational Root Theorem in factorization? 3. Can you verify roots graphically for polynomials? 4. How can synthetic division simplify polynomial division? 5. How would this factored form help solve $$(x^3 - x^2 - 8x + 12) = 0$$? **Tip**: Always check for simple roots (like $$x = 1, -1, 0$$) first when factoring polynomials!