Solve the equation. Write your answers in exact, \[ 48 p^{3}+16 p^{2}-3 p-1=0 \] The solution set is \( \{\square\} \). \( \square, \ldots, \square \)

To solve the equation \( 48 p^{3}+16 p^{2}-3 p-1=0 \), we can first look for rational roots using the Rational Root Theorem. After testing some values, we find that \( p = \frac{1}{4} \) is a root. We can then use synthetic division to divide the polynomial by \( p - \frac{1}{4} \). After performing the division, we are left with a quadratic equation: \[ 48 p^2 + 24 p + 4 = 0 \] We can simplify this by dividing every term by 4: \[ 12 p^2 + 6 p + 1 = 0 \] Next, we can apply the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 12 \), \( b = 6 \), and \( c = 1 \): \[ p = \frac{-6 \pm \sqrt{6^2 - 4(12)(1)}}{2(12)} \] \[ p = \frac{-6 \pm \sqrt{36 - 48}}{24} \] \[ p = \frac{-6 \pm \sqrt{-12}}{24} \] \[ p = \frac{-6 \pm 2i\sqrt{3}}{24} \] \[ p = \frac{-1 \pm \frac{i\sqrt{3}}{6}}{4} \] So the complete solution set is: \[ \left\{ \frac{1}{4}, \frac{-1}{4} + \frac{i\sqrt{3}}{12}, \frac{-1}{4} - \frac{i\sqrt{3}}{12} \right\} \]