1. We are given the polynomial equation $$6x^6-25x^5+31x^4-31x^2+25x-6=0$$ and asked to solve it using synthetic division. 2. First, try to find rational roots using the Rational Root Theorem: possible roots are factors of the constant term (6) over factors of the leading coefficient (6), i.e., $$\pm1, \pm2, \pm3, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{1}{6}$$. 3. Test root $$x=1$$ by synthetic division: Coefficients: 6 | -25 | 31 | 0 (for missing x^3) | -31 | 25 | -6 Bring down 6. Multiply 6*1=6; add to -25 = -19. Multiply -19*1 = -19; add to 31 = 12. Multiply 12*1=12; add to 0 =12. Multiply 12*1=12; add to -31 = -19. Multiply -19*1=-19; add to 25 =6. Multiply 6*1=6; add to -6 =0. Remainder is 0, so $$x=1$$ is a root. 4. The depressed polynomial after division is: $$6x^5 - 19x^4 + 12x^3 + 12x^2 - 19x + 6 = 0$$ 5. Repeat synthetic division on the depressed polynomial to factor further. Try $$x=1$$ again: Coefficients: 6 | -19 | 12 | 12 | -19 | 6 Bring down 6. 6*1=6; add to -19 = -13. -13*1=-13; add to 12 = -1. -1*1 = -1; add to 12 = 11. 11*1=11; add to -19 = -8. -8*1=-8; add to 6 = -2. Remainder not zero, so $$x=1$$ is not a root here. 6. Try $$x=2$$: Bring down 6. 6*2=12; add to -19 = -7. -7*2=-14; add to 12 = -2. -2*2 = -4; add to 12 = 8. 8*2=16; add to -19 = -3. -3*2 = -6; add to 6 = 0. Remainder 0, so $$x=2$$ is a root. 7. The quotient polynomial is: $$6x^4 -7x^3 -2x^2 + 8x -3 = 0$$ 8. We now have factored the original polynomial as: $$(x-1)(x-2)(6x^4 -7x^3 - 2x^2 + 8x -3) = 0$$ 9. Continue factoring the quartic polynomial. Try possible rational roots again. Try $$x=\frac{1}{3}$$: Multiply coefficients by 1/3: Bring down 6. 6*(1/3)=2; add to -7 = -5. -5*(1/3) = -5/3; add to -2 = $$-2 - \frac{5}{3} = -\frac{11}{3}$$ -11/3*(1/3) = -11/9; add to 8 = $$8 - \frac{11}{9} = \frac{61}{9}$$ 61/9*(1/3) = 61/27; add to -3 = $$-3 + \frac{61}{27} = -\frac{20}{27}$$ remainder not zero. 10. Try $$x=3$$: Bring down 6. 6*3=18; add to -7=11. 11*3=33; add to -2=31. 31*3=93; add to 8=101. 101*3=303; add to -3=300 remainder not zero. 11. Try $$x=\frac{1}{2}$$: Bring down 6. 6*0.5=3; add to -7 = -4. -4*0.5 = -2; add to -2 = -4. -4*0.5 = -2; add to 8 = 6. 6*0.5=3; add to -3 = 0 remainder zero. 12. So $$x=\frac{1}{2}$$ is a root. The depressed cubic is: $$6x^3 - 4x^2 - 4x + 6 = 0$$ 13. Factor polynomial completely: $$(x-1)(x-2)(x - \frac{1}{2})(6x^3 - 4x^2 - 4x + 6) = 0$$ 14. Now solve cubic $6x^3 - 4x^2 - 4x + 6 = 0$. Try rational root $$x=1$$: 6(1) - 4(1) - 4(1) + 6 = 6 -4 -4 +6 = 4 \neq 0. Try $$x=-1$$: 6(-1)^3 -4(-1)^2 -4(-1) +6 = -6 -4 +4 +6 = 0 remainder zero. 15. So $$x=-1$$ is a root. Synthetic division gives: $$6x^2 + 2x - 6 = 0$$ 16. Solve quadratic: $$6x^2 + 2x - 6 = 0$$ Use quadratic formula: $$x= \frac{-2 \pm \sqrt{2^2 - 4 \times 6 \times (-6)}}{2 \times 6} = \frac{-2 \pm \sqrt{4 + 144}}{12} = \frac{-2 \pm \sqrt{148}}{12}$$ $$=\frac{-2 \pm 2\sqrt{37}}{12} = \frac{-1 \pm \sqrt{37}}{6}$$ 17. The complete solutions are: $$x=1, x=2, x=\frac{1}{2}, x=-1, x=\frac{-1 + \sqrt{37}}{6}, x=\frac{-1 - \sqrt{37}}{6}$$