Using the factor theorem, factorize ${\mathrm{x}}^3-6{\mathrm{x}}^2+3\mathrm{x}+10$.

Given expression: ${\mathrm{x}}^3-6{\mathrm{x}}^2+3\mathrm{x}+10$
At $\mathbf{x}\mathbf{=}\mathbf{1},$
$(1{)}^3-6(1{)}^2+3\left(1\right)+10=1-6+3+10=\mathbf{8}$

Thus, (x - 1) is not the factor as the value of the expression is not equal to zero at $\mathrm{x}=1$.

At $\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{1},$
$(-1{)}^3-6(-1{)}^2+3(-1)+10=-1-6-3+10=\mathbf{0}$

Thus, $\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}$ is one factor.

At $\mathbf{x}\mathbf{=}\mathbf{2},$

$(2{)}^3-6(2{)}^2+3\left(2\right)+10=8-24+6+10=\mathbf{0}$

Thus, $\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}$ is one factor.

At $\mathbf{x}\mathbf{=}\mathbf{5},$

$(5{)}^3-6(5{)}^2+3\left(5\right)+10=125-150+15+10=\mathbf{0}$

Thus, $\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{)}$ is one factor.

So, the zeros are: $\mathrm{x}=-1, 2, 5$

Hence, ${\mathrm{x}}^3-6{\mathrm{x}}^2+3\mathrm{x}+10=\left(x+1\right)\left(x-2\right)\left(x-5\right)$