### 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, I can use the rational root theorem, which says that any rational root p/q must have p as a factor of the constant term and q as a factor of the leading coefficient.
So, in this case, the possible values of p are ±1, ±3, ±7, and ±21, and the possible values of q are ±1 and ±2.
This gives me a list of 16 possible rational roots to check, which are ±1, ±1/2, ±3, ±3/2, ±7, ±7/2, ±21, and ±21/2.
To check each root, I can use synthetic division or plug it into the polynomial and see if it makes it zero.
I will start with the simplest ones, ±1.
If I plug in x = 1, I get 2 - 9 - 32 - 21, which is -60, not zero.
So x = 1 is not a root.
If I plug in x = -1, I get -2 - 9 + 32 - 21, which is 0.
So x = -1 is a root.
This means I can factor out (x + 1) from the polynomial, and the quotient will be a quadratic polynomial.
To find the quotient, I can use synthetic division.
I write -1 in the box, and the coefficients of the cubic polynomial on the side: -1 | 2 -9 -32 -21.
I bring down the first coefficient, 2, and multiply it by -1, getting -2.
I add this to the next coefficient, -9, and get -11.
I write this under the line.
I repeat this process for the remaining coefficients.
I multiply -11 by -1, getting 11.
I add this to -32, getting -21.
I write this under the line.
I multiply -21 by -1, getting 21.
I add this to -21, getting 0.
I write this under the line.
The last number, 0, is the remainder, which should be zero if x = -1 is a root.
The other numbers, 2, -11, and -21, are the coefficients of the quotient polynomial.
So, I have (x + 1)(2x^2 - 11x - 21) = 0.
To find the other roots, I need to solve the quadratic equation 2x^2 - 11x - 21 = 0.
To solve the quadratic equation, I can use the quadratic formula, which says that the solutions are given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\] where a, b, and c are the coefficients of the quadratic polynomial.
In this case, a = 2, b = -11, and c = -21.
Plugging these into the formula, I get \[x = \frac{11 \pm \sqrt{121 + 168}}{4} = \frac{11 \pm \sqrt{289}}{4} = \frac{11 \pm 17}{4}.\]
Simplifying, I get two possible values for x: \[x = \frac{28}{4} = 7 \quad \text{or} \quad x = \frac{-6}{4} = -\frac{3}{2}.\]
These are both rational roots of the original polynomial.
So, I have found three rational roots: x = -1, x = 7, and x = -3/2.
Since the degree of the polynomial is 3, there are no more roots.