Solve \( 3 x^{2}+17 x-6=0 \) \[ \begin{array}{r} 3 x^{2}+18 x-x-6=0 \\ 3 x(x+6)-1(x+6)=0 \\ (3 x-1)(x+6)=0 \\ 3 x-1=0 \text { or } x+6=0 \end{array} \] Based on the work shown to the left, which of these values are possible solutions of the oquation? Chock all of tho bores that apply \( x=-8 \) \( x=8 \) \( x=-\frac{1}{3} \) \( x=\frac{1}{3} \) \( x=0 \) DOIE

Let's check the factored form of the quadratic equation: \( (3x-1)(x+6)=0 \). From this, we set each factor equal to zero to find the potential solutions: 1. \( 3x - 1 = 0 \) gives \( x = \frac{1}{3} \). 2. \( x + 6 = 0 \) gives \( x = -6 \). So, the possible solutions from the equation are \( x = \frac{1}{3} \) and \( x = -6 \). Now, let’s evaluate the options provided: - \( x = -8 \) (not a solution) - \( x = 8 \) (not a solution) - \( x = -\frac{1}{3} \) (not a solution) - \( x = \frac{1}{3} \) (solution) - \( x = 0 \) (not a solution) Therefore, the only value that is a possible solution from the options given is \( x = \frac{1}{3} \). In summary: - Checked values: Only \( x = \frac{1}{3} \) is a solution. - The original factorization technique effectively revealed the solutions, underscoring the usefulness of factoring in solving quadratic equations!