Use the table to determine a solution of the equation. Use synthetic division to verify that this number is a solution of the equation. Then solve the polynomial equation. \[ 24 x^{3}-34 x^{2}+11 x-1=0 \] \begin{tabular}{|c|c|c|} \hline $\mathbf{X}$ & \multicolumn{2}{|c|}{$\mathbf{Y}_{1}=24 \mathbf{x}^{3}-34 \mathbf{x}^{2}+11 \mathrm{x}-1$} \\ \hline-3 & -988 & \\ -2 & -351 & \\ -1 & -70 & \\ 0 & -1 & \\ 1 & 0 & \\ 2 & 77 & \\ 3 & 374 & \\ \hline \end{tabular} Which solution of the equation can be determined from the table? The solution is

Step 1 ：From the table, we can see that when \(x = 1\), the value of the function \(Y1 = 24x^3 - 34x^2 + 11x - 1\) equals 0. Therefore, \(x = 1\) is a solution of the equation.

Step 2 ：To verify this solution using synthetic division, we set up the synthetic division as follows: \[\begin{array}{c|cc cc} 1 & 24 & -34 & 11 & -1 \\ & & 24 & -10 & 1 \\ \hline & 24 & -10 & 1 & 0 \end{array}\]

Step 3 ：The bottom row represents the coefficients of the quotient polynomial, and the last number in the bottom row is the remainder. Since the remainder is 0, \(x = 1\) is indeed a solution of the equation.

Step 4 ：Now, to solve the polynomial equation, we use the quotient from the synthetic division \((24x^2 - 10x + 1 = 0)\). This is a quadratic equation, which we can solve using the quadratic formula: \[x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4*24*1}}{2*24}\]

Step 5 ：Solving the above equation gives: \[x = \frac{10 \pm \sqrt{100 - 96}}{48}\] \[x = \frac{10 \pm \sqrt{4}}{48}\]

Step 6 ：So the solutions are \(x = \frac{1}{4}\) and \(x = \frac{1}{2}\).

Step 7 ：Therefore, the solutions of the original polynomial equation are \(\boxed{x = 1, x = \frac{1}{4}, x = \frac{1}{2}}\).