1. **Stating the problem:** Factorize the polynomial $$x^4 + 2x^3 - 7x^2 - 8x + 12$$. 2. **Look for rational roots:** Using the Rational Root Theorem, possible roots are factors of 12: $$\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$$. 3. **Test roots:** Substitute values into the polynomial to find zeros. - For $$x=1$$: $$1 + 2 - 7 - 8 + 12 = 0$$, so $$x=1$$ is a root. - For $$x=2$$: $$16 + 16 - 28 - 16 + 12 = 0$$, so $$x=2$$ is a root. 4. **Divide polynomial by $$ (x-1)(x-2) $$:** First, divide by $$x-1$$ using polynomial division or synthetic division to get a cubic. Division by $$x-1$$: $$x^4 + 2x^3 - 7x^2 - 8x + 12 \div (x-1) = x^3 + 3x^2 - 4x - 12$$ 5. **Divide $$x^3 + 3x^2 - 4x - 12$$ by $$x-2$$:** $$x^3 + 3x^2 - 4x - 12 \div (x-2) = x^2 + 5x + 6$$ 6. **Factor the quadratic $$x^2 + 5x + 6$$:** Since $$6 = 2 \times 3$$ and $$2 + 3 = 5$$, we factor as: $$(x + 2)(x + 3)$$ 7. **Combine all factors:** $$x^4 + 2x^3 - 7x^2 - 8x + 12 = (x - 1)(x - 2)(x + 2)(x + 3)$$ **Final answer:** $$\boxed{(x - 1)(x - 2)(x + 2)(x + 3)}$$.