Quartic Equation 50Cd8E

1. **State the problem:** Solve the quartic equation $$x^4 - 4x^3 - 12x^2 + 100x - 325 = 0$$ completely, finding all real and non-real solutions. 2. **Look for rational roots using the Rational Root Theorem:** Possible roots are factors of 325 (±1, ±5, ±13, ±25, ±65, ±325). 3. **Test possible roots:** Try $$x=5$$: $$5^4 - 4(5)^3 - 12(5)^2 + 100(5) - 325 = 625 - 500 - 300 + 500 - 325 = 0$$ So, $$x=5$$ is a root. 4. **Divide the polynomial by $$x-5$$ using synthetic division:** $$\begin{array}{c|ccccc} 5 & 1 & -4 & -12 & 100 & -325 \\ & & 5 & 5 & -35 & 325 \\ \hline & 1 & 1 & -7 & 65 & 0 \\ \end{array}$$ The quotient is $$x^3 + x^2 - 7x + 65$$. 5. **Solve the cubic equation:** $$x^3 + x^2 - 7x + 65 = 0$$. 6. **Try rational roots again:** Test $$x=-5$$: $$(-5)^3 + (-5)^2 - 7(-5) + 65 = -125 + 25 + 35 + 65 = 0$$ So, $$x=-5$$ is a root. 7. **Divide cubic by $$x+5$$:** $$\begin{array}{c|cccc} -5 & 1 & 1 & -7 & 65 \\ & & -5 & 20 & -65 \\ \hline & 1 & -4 & 13 & 0 \\ \end{array}$$ The quotient is $$x^2 - 4x + 13$$. 8. **Solve quadratic:** $$x^2 - 4x + 13 = 0$$. Use quadratic formula: $$x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(13)}}{2(1)} = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i$$. 9. **Summary of solutions:** - Real solutions: $$x=5$$ and $$x=-5$$. - Non-real solutions: $$x=2+3i$$ and $$x=2-3i$$. **Final answer:** Real solutions are $$x=5, -5$$. Non-real solutions are $$x=2+3i, 2-3i$$.