1. **State the problem:** Find all zeros of the polynomial function $$f(x) = x^4 - 2x^3 - 7x^2 + 8x + 12$$. 2. **Recall the problem type:** We want to find values of $x$ such that $$f(x) = 0$$. 3. **Use the Rational Root Theorem:** Possible rational roots are factors of the constant term 12 divided by factors of the leading coefficient 1, so possible roots are $$\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$$. 4. **Test possible roots by substitution or synthetic division:** - Test $x=1$: $$1 - 2 - 7 + 8 + 12 = 12 \neq 0$$ - Test $x=2$: $$16 - 16 - 28 + 16 + 12 = 0$$ so $x=2$ is a root. 5. **Divide $f(x)$ by $(x-2)$:** $$f(x) \div (x-2) = x^3 + 0x^2 - 7x - 6$$ 6. **Find roots of the cubic $x^3 - 7x - 6 = 0$:** - Test $x=1$: $$1 - 7 - 6 = -12 \neq 0$$ - Test $x=3$: $$27 - 21 - 6 = 0$$ so $x=3$ is a root. 7. **Divide the cubic by $(x-3)$:** $$x^3 - 7x - 6 \div (x-3) = x^2 + 3x + 2$$ 8. **Factor the quadratic:** $$x^2 + 3x + 2 = (x+1)(x+2)$$ 9. **Set each factor equal to zero:** $$x+1=0 \Rightarrow x=-1$$ $$x+2=0 \Rightarrow x=-2$$ 10. **List all roots:** $$x = -2, -1, 2, 3$$ 11. **Check for multiplicity:** Each root appears once, so no double roots. **Final answer:** $$x = -2, -1, 2, 3$$