1. **State the problem:** Factor the polynomial $$P(x) = 3x^4 + x^3 - 14x^2 - 4x + 8$$ completely. 2. **List possible rational zeros:** By the Rational Root Theorem, possible zeros are of the form $$\pm \frac{p}{q}$$ where $$p$$ divides the constant term 8 and $$q$$ divides the leading coefficient 3. Possible values of $$p$$: 1, 2, 4, 8 Possible values of $$q$$: 1, 3 Thus, possible zeros are: $$\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{8}{3}$$ 3. **Test zeros using synthetic division:** Start testing with integer candidates. Test $$x=1$$: $$P(1) = 3(1)^4 + 1^3 - 14(1)^2 - 4(1) + 8 = 3 + 1 - 14 - 4 + 8 = -6 \neq 0$$ Test $$x=-1$$: $$P(-1) = 3(-1)^4 + (-1)^3 - 14(-1)^2 - 4(-1) + 8 = 3 - 1 - 14 + 4 + 8 = 0$$ So, $$x = -1$$ is a root. 4. **Divide $$P(x)$$ by $$x + 1$$ using synthetic division:** Coefficients: 3, 1, -14, -4, 8 Synthetic division setup: -1 | 3 1 -14 -4 8 | -3 2 12 -8 ------------------- 3 -2 -12 8 0 Quotient polynomial: $$3x^3 - 2x^2 - 12x + 8$$ 5. **Factor the cubic $$3x^3 - 2x^2 - 12x + 8$$:** Try rational roots again for the cubic. Possible zeros: $$\pm 1, \pm 2, \pm 4, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}$$ Test $$x=2$$: $$3(2)^3 - 2(2)^2 - 12(2) + 8 = 24 - 8 - 24 + 8 = 0$$ So, $$x=2$$ is a root. 6. **Divide cubic by $$x - 2$$:** Coefficients: 3, -2, -12, 8 Synthetic division: 2 | 3 -2 -12 8 | 6 8 -8 ---------------- 3 4 -4 0 Quotient polynomial: $$3x^2 + 4x - 4$$ 7. **Factor quadratic $$3x^2 + 4x - 4$$:** Use quadratic formula: $$x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} = \frac{-4 \pm \sqrt{16 + 48}}{6} = \frac{-4 \pm \sqrt{64}}{6} = \frac{-4 \pm 8}{6}$$ Roots: $$x = \frac{-4 + 8}{6} = \frac{4}{6} = \frac{2}{3}$$ $$x = \frac{-4 - 8}{6} = \frac{-12}{6} = -2$$ 8. **Write full factorization:** $$P(x) = (x + 1)(x - 2)(3x^2 + 4x - 4) = (x + 1)(x - 2)(3x - 2)(x + 2)$$ **Final answer:** $$\boxed{(x + 1)(x - 2)(3x - 2)(x + 2)}$$