Zeros Polynomial 98Deb3

1. **State the problem:** Find the zeros of the cubic polynomial $$x^3 - 4x^2 + 9x + 5$$ and determine their multiplicities. 2. **Recall the formula and rules:** To find zeros of a polynomial, we solve $$x^3 - 4x^2 + 9x + 5 = 0$$. Zeros can be found by factoring or using the Rational Root Theorem to test possible roots. 3. **Apply the Rational Root Theorem:** Possible rational roots are factors of the constant term 5 divided by factors of the leading coefficient 1, i.e., $$\pm 1, \pm 5$$. 4. **Test $x=1$:** $$1^3 - 4(1)^2 + 9(1) + 5 = 1 - 4 + 9 + 5 = 11 \neq 0$$ 5. **Test $x=-1$:** $$(-1)^3 - 4(-1)^2 + 9(-1) + 5 = -1 - 4 - 9 + 5 = -9 \neq 0$$ 6. **Test $x=5$:** $$5^3 - 4(5)^2 + 9(5) + 5 = 125 - 100 + 45 + 5 = 75 \neq 0$$ 7. **Test $x=-5$:** $$(-5)^3 - 4(-5)^2 + 9(-5) + 5 = -125 - 100 - 45 + 5 = -265 \neq 0$$ No rational roots found, so we use synthetic division or the cubic formula. 8. **Use synthetic division or factor by grouping:** Try to factor or find one root by trial or use the depressed cubic method. 9. **Use the cubic formula:** For $$ax^3 + bx^2 + cx + d = 0$$ with $$a=1, b=-4, c=9, d=5$$. 10. **Calculate discriminant:** $$\Delta = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$$ $$= 18(1)(-4)(9)(5) - 4(-4)^3(5) + (-4)^2(9)^2 - 4(1)(9)^3 - 27(1)^2(5)^2$$ $$= 18(-180) - 4(-64)(5) + 16(81) - 4(729) - 27(25)$$ $$= -3240 + 1280 + 1296 - 2916 - 675 = -4255$$ Since $$\Delta < 0$$, there is one real root and two complex conjugate roots. 11. **Approximate the real root numerically:** Using methods like Newton-Raphson or graphing, the real root is approximately $$x \approx 0.641$$. 12. **Find the quadratic factor:** Divide the cubic by $$x - 0.641$$ to get a quadratic factor. 13. **Divide:** $$x^3 - 4x^2 + 9x + 5 \div (x - 0.641) = x^2 - 3.359x + 7.805$$ (approximate) 14. **Solve quadratic:** $$x = \frac{3.359 \pm \sqrt{(-3.359)^2 - 4(1)(7.805)}}{2}$$ $$= \frac{3.359 \pm \sqrt{11.29 - 31.22}}{2} = \frac{3.359 \pm \sqrt{-19.93}}{2}$$ 15. **Complex roots:** $$x = \frac{3.359 \pm i\sqrt{19.93}}{2}$$ 16. **Summary:** - One real root $$\approx 0.641$$ with multiplicity 1. - Two complex conjugate roots $$\frac{3.359 \pm i\sqrt{19.93}}{2}$$ each with multiplicity 1. **Final answer:** The polynomial $$x^3 - 4x^2 + 9x + 5$$ has one real root approximately $$0.641$$ and two complex conjugate roots approximately $$1.68 \pm 2.23i$$, all with multiplicity 1.