1. **State the problem:** Solve the equation $$0=36x^4+60x^3-696x^2-480x$$ algebraically using the factor theorem and polynomial long division. 2. **Rewrite the equation:** $$36x^4+60x^3-696x^2-480x=0$$ 3. **Factor out the greatest common factor (GCF):** $$12x(3x^3+5x^2-58x-40)=0$$ This is factoring by taking out the GCF, which simplifies the polynomial. 4. **Set each factor equal to zero:** $$12x=0 \implies x=0$$ and $$3x^3+5x^2-58x-40=0$$ 5. **Use the factor theorem to find a root of the cubic:** Try possible rational roots using factors of constant term (-40) over factors of leading coefficient (3): $$\pm1, \pm2, \pm4, \pm5, \pm8, \pm10, \pm20, \pm40, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{4}{3}, \pm\frac{5}{3}, \pm\frac{8}{3}, \pm\frac{10}{3}, \pm\frac{20}{3}, \pm\frac{40}{3}$$ 6. **Test $x=2$:** $$3(2)^3 + 5(2)^2 - 58(2) - 40 = 3(8) + 5(4) - 116 - 40 = 24 + 20 - 116 - 40 = -112 \neq 0$$ 7. **Test $x=4$:** $$3(4)^3 + 5(4)^2 - 58(4) - 40 = 3(64) + 5(16) - 232 - 40 = 192 + 80 - 232 - 40 = 0$$ So, $x=4$ is a root. 8. **Divide the cubic by $(x-4)$ using polynomial long division:** Divide $$3x^3 + 5x^2 - 58x - 40$$ by $$x - 4$$. - Divide leading terms: $$\frac{3x^3}{x} = 3x^2$$ (quotient term) - Multiply back: $$3x^2(x - 4) = 3x^3 - 12x^2$$ - Subtract: $$(3x^3 + 5x^2) - (3x^3 - 12x^2) = 17x^2$$ - Bring down $$-58x$$ - Divide leading terms: $$\frac{17x^2}{x} = 17x$$ - Multiply back: $$17x(x - 4) = 17x^2 - 68x$$ - Subtract: $$(17x^2 - 58x) - (17x^2 - 68x) = 10x$$ - Bring down $$-40$$ - Divide leading terms: $$\frac{10x}{x} = 10$$ - Multiply back: $$10(x - 4) = 10x - 40$$ - Subtract: $$(10x - 40) - (10x - 40) = 0$$ Quotient is $$3x^2 + 17x + 10$$ and remainder is 0, confirming $(x-4)$ is a factor. 9. **Solve the quadratic $$3x^2 + 17x + 10 = 0$$ using the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-17 \pm \sqrt{17^2 - 4(3)(10)}}{2(3)} = \frac{-17 \pm \sqrt{289 - 120}}{6} = \frac{-17 \pm \sqrt{169}}{6}$$ 10. **Simplify the square root:** $$\sqrt{169} = 13$$ 11. **Find the two roots:** $$x = \frac{-17 + 13}{6} = \frac{-4}{6} = -\frac{2}{3}$$ $$x = \frac{-17 - 13}{6} = \frac{-30}{6} = -5$$ 12. **List all solutions:** $$x = 0, 4, -\frac{2}{3}, -5$$ **Final answer:** $$\boxed{\left\{0, 4, -\frac{2}{3}, -5\right\}}$$