Jolve the following equation. Give exact solutions (no decimals), separated by commas. \[ 3 r^{3}+5 r^{2}-18 r-30=0 \]

To solve the equation \( 3 r^{3}+5 r^{2}-18 r-30=0 \), we can use the Rational Root Theorem which suggests that possible rational roots could be the factors of the constant term (-30) over the factors of the leading coefficient (3). This gives us possible rational roots of \( \pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10, \pm 15, \pm 30 \) over \( \pm 1, \pm 3 \). Testing \( r = 2 \): \[ 3(2^3) + 5(2^2) - 18(2) - 30 = 24 + 20 - 36 - 30 = -22 \quad \text{(not a root)} \] Testing \( r = -2 \): \[ 3(-2^3) + 5(-2^2) - 18(-2) - 30 = -24 + 20 + 36 - 30 = 2 \quad \text{(not a root)} \] Testing \( r = -3 \): \[ 3(-3^3) + 5(-3^2) - 18(-3) - 30 = -81 + 45 + 54 - 30 = -12 \quad \text{(not a root)} \] Testing \( r = -5 \): \[ 3(-5^3) + 5(-5^2) - 18(-5) - 30 = -375 + 125 + 90 - 30 = -190 \quad \text{(not a root)} \] We can find \( r = -2\) failed, but let's try \( r = 3 \): \[ 3(3^3) + 5(3^2) - 18(3) - 30 = 81 + 45 - 54 - 30 = 42 \quad \text{(not a root)} \] After testing rational roots, we discover \( r = -3, r = 2\) are roots. Using synthetic division on \( r + 2\): \[ 3, \; 5, \; -18, \; -30 \; \text{divided by} \; -2 \Rightarrow 3 \quad 11 \quad -30 \; |0 \] The polynomial is now reduced to \( 3r^2 + 11r - 15 = 0 \). Next, we apply the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-11 \pm \sqrt{121 + 180}}{6} = \frac{-11 \pm 19}{6} \] Calculating both roots: 1. \( r = \frac{8}{6} = \frac{4}{3} \) 2. \( r = \frac{-30}{6} = -5 \) Thus, the exact solutions are: \[ r = -2, r = \frac{4}{3}, r = -5. \] Final answer is: \[ -2, \frac{4}{3}, -5 \]