### Instruction: Find all roots of the polynomial $x^4+2x^3-7x^2-8x+12$. Enter your answer as a list of numbers separated by commas.### Response: To find the roots of a polynomial, I need to factor it into linear factors of the form $(x-a)$, where $a$ is a root.
I can use the rational root theorem to find possible candidates for rational roots.
The rational root theorem says that if a polynomial with integer coefficients has a rational root, then it 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.
In this case, the constant term is $12$ and the leading coefficient is $1$, so the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$.
I can check each of these values by plugging them into the polynomial and seeing if I get zero.
I can use synthetic division to make the calculations faster.
I start with $x=1$.
I write $1$ in the corner and the coefficients of the polynomial in a row: $1, 2, -7, -8, 12$.
I bring down the first coefficient, $1$, and write it below the line.
I multiply $1$ by $1$ and write the result, $1$, above the second coefficient.
I add $1$ and $2$ and write the result, $3$, below the line.
I repeat this process for the rest of the coefficients, multiplying by $1$ and adding the result to the next coefficient.
I get $1, 3, -4, -12, 0$ below the line.
The last number, $0$, is the remainder when I divide the polynomial by $(x-1)$.
Since it is zero, that means $(x-1)$ is a factor and $x=1$ is a root.
I can also use the other numbers below the line, $1, 3, -4, -12$, as the coefficients of a quotient polynomial, $x^3+3x^2-4x-12$.
I can repeat the same process with this polynomial and another possible rational root, say $x=-2$.
I write $-2$ in the corner and the coefficients of the polynomial in a row: $1, 3, -4, -12$.
I bring down the first coefficient, $1$, and write it below the line.
I multiply $1$ by $-2$ and write the result, $-2$, above the second coefficient.
I add $-2$ and $3$ and write the result, $1$, below the line.
I repeat this process for the rest of the coefficients, multiplying by $-2$ and adding the result to the next coefficient.
I get $1, 1, -6, 0$ below the line.
The last number, $0$, is the remainder when I divide the polynomial by $(x+2)$.
Since it is zero, that means $(x+2)$ is a factor and $x=-2$ is a root.
I can also use the other numbers below the line, $1, 1, -6$, as the coefficients of a quotient polynomial, $x^2+x-6$.
I can factor this polynomial further by looking for two numbers that multiply to $-6$ and add to $1$.
I see that $3$ and $-2$ work, so I can write $x^2+x-6 = (x+3)(x-2)$.
This means that $x=-3$ and $x=2$ are also roots of the original polynomial.
Therefore, the four roots are $x=1, x=-2, x=-3, x=2$.