### Instruction: Find all the rational roots of
\[2x^3 - 9x^2 - 32x - 21 = 0.\]Enter all the rational roots, separated by commas.### Response: I know that the rational root theorem says that any rational root of the polynomial must be of the form $\pm p/q$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.
So, the possible values of $p$ are $\pm 1, \pm 3, \pm 7, \pm 21$, and the possible values of $q$ are $\pm 1, \pm 2$.
That means there are 16 possible rational roots to try: $\pm 1, \pm 1/2, \pm 3, \pm 3/2, \pm 7, \pm 7/2, \pm 21, \pm 21/2$.
To check each one, I can plug it into the polynomial and see if it makes it zero.
Alternatively, I can use synthetic division to reduce the degree of the polynomial by one each time I find a root.
I'll start with the easiest ones: $\pm 1$.
If I plug in $x = 1$, I get $2 - 9 - 32 - 21 = -60$, which is not zero.
So $x = 1$ is not a root.
If I plug in $x = -1$, I get $-2 - 9 + 32 - 21 = 0$, which is zero.
So $x = -1$ is a root.
To use synthetic division, I write the coefficients of the polynomial in a row, and then write the root I found below and to the left of them.
Then I bring down the first coefficient, which is 2, and multiply it by the root, which is -1, and write the result below and to the right of the 2.
I get -2, and I add it to the second coefficient, which is -9, and write the result below it.
I get -11, and I multiply it by the root, which is -1, and write the result below and to the right of the -11.
I get 11, and I add it to the third coefficient, which is -32, and write the result below it.
I get -21, and I multiply it by the root, which is -1, and write the result below and to the right of the -21.
I get 21, and I add it to the fourth coefficient, which is -21, and write the result below it.
I get 0, which confirms that $x = -1$ is a root.
The other numbers in the bottom row are the coefficients of the quotient polynomial, which is $2x^2 - 11x - 21$.
This is a quadratic polynomial, so I can try to factor it by looking for two numbers that multiply to -42 (the product of the leading and constant coefficients) and add to -11 (the middle coefficient).
I see that -14 and 3 work, so I can write $2x^2 - 11x - 21 = (2x + 3)(x - 7)$.
This means that the other two roots of the original polynomial are $x = -3/2$ and $x = 7$.
I don't need to check the other possible rational roots, since I already found all three of them.