### Instruction: Find all the rational roots of
\[2x^3 - 9x^2 - 32x - 21 = 0.\]Enter all the rational roots, separated by commas.### Response: To find the rational roots of a polynomial equation, I can use the rational root theorem, which says that any rational root of the form p/q, where p and q are relatively prime integers, must satisfy two conditions: p must be a factor of the constant term, and q must be a factor of the leading coefficient.
In this case, the constant term is -21 and the leading coefficient is 2, so the possible values of p are $\pm 1, \pm 3, \pm 7, \pm 21$ and the possible values of q are $\pm 1, \pm 2$.
This gives me a list of 16 possible rational roots to test: $\pm 1, \pm 3, \pm 7, \pm 21, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{7}{2}, \pm \frac{21}{2}$.
I can use synthetic division or a calculator to check which of these values make the polynomial zero.
I'll start with the simplest ones, $\pm 1$.
Plugging in x = 1, I get $2 - 9 - 32 - 21 = -60$, which is not zero, so x = 1 is not a root.
Plugging in x = -1, I get $-2 - 9 + 32 - 21 = 0$, which is zero, so x = -1 is a root.
This means that x + 1 is a factor of the polynomial, and I can use synthetic division to find the other factor.
Dividing the polynomial by x + 1, I get $2x^2 - 11x - 21$.
This is a quadratic polynomial, and I can use the quadratic formula to find its roots.
The roots are $\frac{11 \pm \sqrt{121 + 168}}{4} = \frac{11 \pm \sqrt{289}}{4} = \frac{11 \pm 17}{4}$.
Simplifying, I get $x = -\frac{3}{2}$ and $x = 7$.
These are both rational roots, and they are already in the list of possible rational roots.
Therefore, I have found all the rational roots of the polynomial equation.
They are -1, -3/2, and 7.