$$\textcolor[rgb]{0,1,0}{1}y^3+3y^2-121y-\textcolor[rgb]{0,1,0}{363}$$
Now the factors of this must also be factors of 363/1 that is factors of 363
The factors of 363 are 3 and 121 = 3 and 11 and 11
So the only integers that can be factors here are
$$\pm3,\;\pm11,\;\pm121,\;\pm33,\;\pm363 \quad $ I think that about covers it$$$
Now
$${{\mathtt{3}}}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{121}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{363}} = -{\mathtt{672}}$$ so x=3 is not a root, so (x-3) is not a factor
$${\left(-{\mathtt{3}}\right)}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\left(-{\mathtt{3}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{121}}{\mathtt{\,\times\,}}\left(-{\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{363}} = {\mathtt{0}}$$ so x=-3 is a root so (x+3) is a factor
Polynomial division will give you that
$$(y^3+3y^2-121y-363)\div (x+3)= x^2-121$$
and
$$x^2-121=(x-11)(x+11)$$
so
$$\\y^3+3y^2-121y-363=(x+3)(x-11)(x+11)\\\\
$The roots will be -3, -11,and\;11$$$
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