### Instruction: Find all roots of the polynomial $x^3+x^2-4x-4$. Enter your answer as a list of numbers separated by commas.### Response: I know that if a polynomial has a root $r$, then $(x-r)$ is a factor of the polynomial.
So I want to find a value of $r$ that makes the polynomial zero.
One way to do that is to try some small integers, since they are easy to plug in and check.
I'll start with $r=1$.
If I plug in $x=1$, I get $1^3+1^2-4(1)-4 = 1+1-4-4 = -6$, which is not zero.
So $r=1$ is not a root.
I'll try $r=-1$ next.
If I plug in $x=-1$, I get $(-1)^3+(-1)^2-4(-1)-4 = -1+1+4-4 = 0$.
So $r=-1$ is a root, and $(x+1)$ is a factor.
Now I can use long division or synthetic division to divide the original polynomial by $(x+1)$ and get a quotient polynomial of lower degree.
I'll use synthetic division, which is faster and easier for me.
I write the coefficients of the polynomial in a row, and write $-1$ in the corner, since that's the root I'm dividing by.
Then I bring down the first coefficient, which is $1$, and multiply it by $-1$, and write the result below the second coefficient, which is also $1$.
I get $-1$, and I add it to $1$, and get $0$.
Then I multiply $0$ by $-1$, and write the result below the third coefficient, which is $-4$.
I get $0$, and I add it to $-4$, and get $-4$.
Then I multiply $-4$ by $-1$, and write the result below the fourth coefficient, which is $-4$.
I get $4$, and I add it to $-4$, and get $0$.
This means that the remainder is zero, and the quotient is the polynomial with the coefficients in the bottom row, except the last one.
So the quotient polynomial is $x^2-4$, and I can write the original polynomial as $(x+1)(x^2-4)$.
Now I want to find the roots of $x^2-4$, which is a quadratic polynomial.
I can use the quadratic formula, or I can factor it further.
I notice that $x^2-4$ is a difference of squares, so I can factor it as $(x+2)(x-2)$.
Therefore, the original polynomial is $(x+1)(x+2)(x-2)$, and the roots are the values of $x$ that make any of these factors zero.
So the roots are $x=-1$, $x=-2$, and $x=2$.
I can check that these are indeed the roots by plugging them back into the original polynomial and verifying that they make it zero.