### 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 cubic polynomial, I can try to factor it using the rational root theorem, which says that any rational root must be a factor of the constant term divided by a factor of the leading coefficient.
So, the possible rational roots are $\pm 1, \pm 3, \pm 9$.
I can test each one using synthetic division or plugging into the polynomial.
If I plug in 1, I get $1-5+3+9=8$, which is not zero.
So, 1 is not a root.
If I plug in -1, I get $-1+5-3+9=10$, which is also not zero.
So, -1 is not a root either.
If I plug in 3, I get $27-45+9+9=0$, which is zero.
So, 3 is a root.
This means that the polynomial is divisible by $x-3$, and I can use synthetic division to find the quotient.
The synthetic division table looks like this:

\begin{array}{r|r}
3 & 1 & -5 & 3 & 9\\
\hline
& & 3 & -6 & -9\\
\hline
& 1 & -2 & -3 & 0\\
\end{array}
The quotient is $x^2-2x-3$, which is a quadratic polynomial that I can factor further.
I can use the AC method or the quadratic formula, but I notice that $x^2-2x-3=(x-3)(x+1)$ by inspection.