### Instruction: Find all roots of the polynomial $x^3-5x^2+3x+9$. Enter your answer as a list of numbers separated by commas. If a root occurs more than once, enter it as many times as its multiplicity.### Response: To find the roots of a polynomial, I need to factor it into linear and/or irreducible quadratic factors.
I can try to use the rational root theorem, which 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 $9$ and the leading coefficient is $1$, so the possible rational roots are $\pm 1, \pm 3, \pm 9$.
I can use synthetic division or the remainder theorem to check if any of these are actually roots.
For example, to check if $3$ is a root, I can plug in $x=3$ and see if the polynomial evaluates to zero.
I get $3^3-5(3^2)+3(3)+9 = 27-45+9+9 = 0$, so $3$ is a root.
This means that $(x-3)$ is a factor of the polynomial, and I can divide the polynomial by $(x-3)$ to get a quadratic quotient.
I can use long division or synthetic division to do this.
I'll use synthetic division.
I write $3$ in a box and the coefficients of the polynomial in a row: $1, -5, 3, 9$.
I bring down the first coefficient, $1$, and write it below the line.
I multiply it by $3$ and write the result, $3$, above the second coefficient, $-5$.
I add $-5$ and $3$ and write the result, $-2$, below the line.
I multiply it by $3$ and write the result, $-6$, above the third coefficient, $3$.
I add $3$ and $-6$ and write the result, $-3$, below the line.
I multiply it by $3$ and write the result, $-9$, above the last coefficient, $9$.
I add $9$ and $-9$ and write the result, $0$, below the line.
This confirms that $3$ is a root and $(x-3)$ is a factor.
The numbers below the line are the coefficients of the quotient polynomial, which is $x^2-2x-3$.
I can factor this quadratic polynomial by looking for two numbers that multiply to $-3$ and add to $-2$.
These are $-3$ and $1$.
So the quadratic polynomial factors into $(x-3)(x+1)$.
This means that the original polynomial factors into $(x-3)(x-3)(x+1)$.
The roots are the values of $x$ that make each factor zero, so they are $3, 3, -1$.
The root $3$ has multiplicity two, because it occurs twice in the factorization.
The final answer is $3, 3, -1$.