### Instruction: Find all the integer roots of
\[2x^3 + 13x^2 - 23x + 8 = 0.\]Enter all the integer roots, separated by commas.### Response: I know that if x is an integer root of a polynomial with integer coefficients, then x must divide the constant term.
In this case, the constant term is 8, so the possible integer roots are $\pm 1, \pm 2, \pm 4, \pm 8$.
I can use synthetic division to test each of these candidates and see if they are actually roots.
Synthetic division is a shortcut for dividing a polynomial by a linear factor of the form x - a, where a is some constant.
If the remainder is zero, then x - a is a factor of the polynomial and a is a root.
To use synthetic division, I write the coefficients of the polynomial in a row, followed by a vertical bar and a zero.
Then I write the value of a below the bar and bring down the first coefficient.
Then I multiply a by the first coefficient and write the product below the second coefficient.
Then I add the second coefficient and the product and write the sum below the bar.
I repeat this process until I reach the end of the row.
The last number below the bar is the remainder.
If it is zero, then a is a root.
Let me try this with a = 1.
I write 2  13  -23  8 | 0 and bring down the 2.
I write 1 below the bar and multiply 1 by 2 and write 2 below the 13.
I add 13 and 2 and write 15 below the bar.
I multiply 1 by 15 and write 15 below the -23.
I add -23 and 15 and write -8 below the bar.
I multiply 1 by -8 and write -8 below the 8.
I add 8 and -8 and write 0 below the bar.
The remainder is zero, so 1 is a root of the polynomial.
This also means that x - 1 is a factor of the polynomial.
To find the other factor, I look at the numbers below the bar, except the last one.
They form the coefficients of the quotient polynomial, which is 2x^2 + 15x - 8.
I can use the quadratic formula to find the roots of this polynomial.
The roots are x = \frac{-15 \pm \sqrt{289}}{4}.
The square root of 289 is 17, so the roots are x = \frac{-15 + 17}{4} and x = \frac{-15 - 17}{4}.
Simplifying, I get x = \frac{1}{2} and x = -8.
However, x = \frac{1}{2} is not an integer, so I discard it.
Therefore, the only integer roots of the original polynomial are 1 and -8.