Solve Polynomial 03B933

1. **State the problem:** We need to solve the polynomial equation $$6x^5 + x^2 - 43x^3 - 43x^2 + x + 6 = 0.$$ 2. **Rewrite and group terms:** Group like terms to simplify: $$6x^5 - 43x^3 + x^2 - 43x^2 + x + 6 = 0$$ Combine like terms for $x^2$: $$6x^5 - 43x^3 - 42x^2 + x + 6 = 0.$$ 3. **Look for rational roots using Rational Root Theorem:** Possible rational roots are factors of constant term 6 over factors of leading coefficient 6, i.e., $\pm1, \pm2, \pm3, \pm6, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{1}{6}$. 4. **Test $x=1$:** $$6(1)^5 - 43(1)^3 - 42(1)^2 + 1 + 6 = 6 - 43 - 42 + 1 + 6 = -72 \neq 0.$$ 5. **Test $x=2$:** $$6(2)^5 - 43(2)^3 - 42(2)^2 + 2 + 6 = 6(32) - 43(8) - 42(4) + 2 + 6 = 192 - 344 - 168 + 8 = -312 \neq 0.$$ 6. **Test $x=3$:** $$6(3)^5 - 43(3)^3 - 42(3)^2 + 3 + 6 = 6(243) - 43(27) - 42(9) + 9 = 1458 - 1161 - 378 + 9 = -72 \neq 0.$$ 7. **Test $x=-1$:** $$6(-1)^5 - 43(-1)^3 - 42(-1)^2 - 1 + 6 = -6 + 43 - 42 - 1 + 6 = 0.$$ So, $x = -1$ is a root. 8. **Divide polynomial by $(x + 1)$:** Use polynomial division or synthetic division: $$\frac{6x^5 - 43x^3 - 42x^2 + x + 6}{x + 1} = 6x^4 - 6x^3 - 37x^2 - 5x + 6.$$ 9. **Solve the quartic:** $$6x^4 - 6x^3 - 37x^2 - 5x + 6 = 0.$$ 10. **Try rational roots again for quartic:** Possible roots are factors of 6 over factors of 6, same as before. 11. **Test $x=1$:** $$6 - 6 - 37 - 5 + 6 = -36 \neq 0.$$ 12. **Test $x=2$:** $$6(16) - 6(8) - 37(4) - 5(2) + 6 = 96 - 48 - 148 - 10 + 6 = -104 \neq 0.$$ 13. **Test $x=3$:** $$6(81) - 6(27) - 37(9) - 5(3) + 6 = 486 - 162 - 333 - 15 + 6 = -18 \neq 0.$$ 14. **Test $x=-1$:** $$6 - (-6) - 37 - (-5) + 6 = 6 + 6 - 37 + 5 + 6 = -14 \neq 0.$$ 15. **Test $x=-2$:** $$6(16) - 6(-8) - 37(4) - 5(-2) + 6 = 96 + 48 - 148 + 10 + 6 = 12 \neq 0.$$ 16. **Try factoring quartic by grouping or quadratic form:** Try to factor as $(ax^2 + bx + c)(dx^2 + ex + f) = 6x^4 - 6x^3 - 37x^2 - 5x + 6.$ 17. **Find factors:** Possible factor pairs for 6 are (6,1), (3,2), (2,3), (1,6). 18. **Try $(3x^2 + mx + n)(2x^2 + px + q)$:** Expanding: $$6x^4 + (3p + 2m)x^3 + (3q + 2n + mp)x^2 + (mq + np)x + nq.$$ 19. **Match coefficients:** $$6x^4 - 6x^3 - 37x^2 - 5x + 6$$ So, system: $3p + 2m = -6$ $3q + 2n + mp = -37$ $mq + np = -5$ $nq = 6$ 20. **Try $n=2$, $q=3$ (since $nq=6$):** Then: $3(3) + 2n + mp = 9 + 2n + mp = -37$ $mq + np = m(3) + n p = 3m + 2p = -5$ $3p + 2m = -6$ 21. **Solve system:** From $3p + 2m = -6$, express $m = \frac{-6 - 3p}{2}$. Substitute into $3m + 2p = -5$: $$3\left(\frac{-6 - 3p}{2}\right) + 2p = -5$$ $$\frac{-18 - 9p}{2} + 2p = -5$$ Multiply both sides by 2: $$-18 - 9p + 4p = -10$$ $$-18 - 5p = -10$$ $$-5p = 8$$ $$p = -\frac{8}{5}.$$ 22. **Find $m$:** $$m = \frac{-6 - 3(-8/5)}{2} = \frac{-6 + 24/5}{2} = \frac{-30/5 + 24/5}{2} = \frac{-6/5}{2} = -\frac{3}{5}.$$ 23. **Check $9 + 2n + mp = -37$:** $$9 + 2n + m p = 9 + 2n + \left(-\frac{3}{5}\right)\left(-\frac{8}{5}\right) = 9 + 2n + \frac{24}{25} = -37.$$ $$2n = -37 - 9 - \frac{24}{25} = -46 - \frac{24}{25} = -\frac{1150}{25} - \frac{24}{25} = -\frac{1174}{25}.$$ $$n = -\frac{1174}{50} = -\frac{587}{25}.$$ 24. **Contradiction with $n=2$ assumption:** So $n=2$ is invalid. Try $n=3$, $q=2$. 25. **Try $n=3$, $q=2$:** $3q + 2n + mp = 3(2) + 2(3) + m p = 6 + 6 + m p = 12 + m p = -37$ $m q + n p = 2 m + 3 p = -5$ $3 p + 2 m = -6$ 26. **From $3 p + 2 m = -6$, express $m = \frac{-6 - 3 p}{2}$.** Substitute into $2 m + 3 p = -5$: $$2 \left(\frac{-6 - 3 p}{2}\right) + 3 p = -5$$ $$-6 - 3 p + 3 p = -5$$ $$-6 = -5$$ Contradiction again. 27. **Try $n=1$, $q=6$:** $3 q + 2 n + m p = 18 + 2 + m p = 20 + m p = -37$ $m q + n p = 6 m + p = -5$ $3 p + 2 m = -6$ 28. **From $3 p + 2 m = -6$, express $m = \frac{-6 - 3 p}{2}$.** Substitute into $6 m + p = -5$: $$6 \left(\frac{-6 - 3 p}{2}\right) + p = -5$$ $$3 (-6 - 3 p) + p = -5$$ $$-18 - 9 p + p = -5$$ $$-18 - 8 p = -5$$ $$-8 p = 13$$ $$p = -\frac{13}{8}.$$ 29. **Find $m$:** $$m = \frac{-6 - 3(-13/8)}{2} = \frac{-6 + 39/8}{2} = \frac{-48/8 + 39/8}{2} = \frac{-9/8}{2} = -\frac{9}{16}.$$ 30. **Check $20 + m p = -37$:** $$20 + \left(-\frac{9}{16}\right) \left(-\frac{13}{8}\right) = 20 + \frac{117}{128} = 20.914 > -37,$$ no match. 31. **Try $n=6$, $q=1$:** $3 q + 2 n + m p = 3 + 12 + m p = 15 + m p = -37$ $m q + n p = m + 6 p = -5$ $3 p + 2 m = -6$ 32. **From $3 p + 2 m = -6$, express $m = \frac{-6 - 3 p}{2}$.** Substitute into $m + 6 p = -5$: $$\frac{-6 - 3 p}{2} + 6 p = -5$$ Multiply both sides by 2: $$-6 - 3 p + 12 p = -10$$ $$-6 + 9 p = -10$$ $$9 p = -4$$ $$p = -\frac{4}{9}.$$ 33. **Find $m$:** $$m = \frac{-6 - 3(-4/9)}{2} = \frac{-6 + 4/3}{2} = \frac{-18/3 + 4/3}{2} = \frac{-14/3}{2} = -\frac{7}{3}.$$ 34. **Check $15 + m p = -37$:** $$15 + \left(-\frac{7}{3}\right) \left(-\frac{4}{9}\right) = 15 + \frac{28}{27} = 15.037 > -37,$$ no match. 35. **Since factoring is complicated, use numerical or approximate methods for quartic roots.** 36. **Summary:** One root is $x = -1$. The remaining quartic $6x^4 - 6x^3 - 37x^2 - 5x + 6 = 0$ can be solved numerically or with advanced methods.