If one root of $x^3-6x^2+3x+10=0$ is 2, then the remaining two roots are

The given equation is $x^3-6x^2+3x+10=0$

Given that one root is 2

Consider the coefficients of given equation

And do synthetic division, we get

$\begin{array}{c}2\overline{)\begin{array}{cccc}1& -6& 3& 10\\ 0& 2& -8& -10\end{array}}\\ \begin{array}{ccc}1& -4& -5\end{array}\overline{)0}\end{array}$

$\Rightarrow \left(x-2\right)\left(x^2-4x-5\right)=0$

$\Rightarrow \left(x-2\right)\left(x+1\right)\left(x-5\right)=0$

$\Rightarrow x=2,-1,5$

$\therefore$  the remaining roots are 5, $-$ 1