(c) Let \( f(x)=6 x^{3}+19 x^{2}+2 x-3 \) be a polynomial. Find all rational roots of the equation \( f(x)=0 \).

To find the rational roots of the polynomial \( f(x) = 6x^3 + 19x^2 + 2x - 3 \), we can use the Rational Root Theorem. This theorem states that any rational solution of \( f(x) = 0 \) must be of the form \( \frac{p}{q} \), where \( p \) is a factor of the constant term (-3), and \( q \) is a factor of the leading coefficient (6). The factors of -3 (the constant term) are \( \pm 1, \pm 3 \), and the factors of 6 (the leading coefficient) are \( \pm 1, \pm 2, \pm 3, \pm 6 \). Thus, the possible rational roots are: \[ \pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{3}, \pm \frac{3}{3} = \pm 1, \pm \frac{1}{6} \] Now we will test these potential roots in \( f(x) \) to find any valid rational roots. 1. Testing \( x = 1 \): \[ f(1) = 6(1)^3 + 19(1)^2 + 2(1) - 3 = 6 + 19 + 2 - 3 = 24 \quad (\text{not a root}) \] 2. Testing \( x = -1 \): \[ f(-1) = 6(-1)^3 + 19(-1)^2 + 2(-1) - 3 = -6 + 19 - 2 - 3 = 8 \quad (\text{not a root}) \] 3. Testing \( x = 3 \): \[ f(3) = 6(3)^3 + 19(3)^2 + 2(3) - 3 = 6(27) + 19(9) + 6 - 3 = 162 + 171 + 6 - 3 = 336 \quad (\text{not a root}) \] 4. Testing \( x = -3 \): \[ f(-3) = 6(-3)^3 + 19(-3)^2 + 2(-3) - 3 = 6(-27) + 19(9) - 6 - 3 = -162 + 171 - 6 - 3 = 0 \quad (\text{root found!}) \] 5. Testing \( x = \frac{1}{2} \): \[ f\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^3 + 19\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) - 3 = \frac{6}{8} + \frac{19}{4} + 1 - 3 = \frac{3}{4} + \frac{19}{4} + 1 - 3 = \frac{22}{4} + 1 - 3 = \frac{22}{4} + \frac{4}{4} - \frac{12}{4} = \frac{14}{4} \quad (\text{not a root}) \] 6. Testing \( x = -\frac{1}{2} \): \[ f\left(-\frac{1}{2}\right) = 6\left(-\frac{1}{2}\right)^3 + 19\left(-\frac{1}{2}\right)^2 + 2\left(-\frac{1}{2}\right) - 3 = -\frac{6}{8} + \frac{19}{4} - 1 - 3 = -\frac{3}{4} + \frac{19}{4} - \frac{4}{4} - \frac{12}{4} = \frac{-3 + 19 - 4 - 12}{4} = \frac{0}{4} = 0 \quad (\text{another root found!}) \] Now, we have found two rational roots: \( x = -3 \) and \( x = -\frac{1}{2} \). Thus, the rational roots of the equation \( f(x) = 0 \) are: \[ x = -3, \quad x = -\frac{1}{2} \]