Y^3+3y^2-121y-363=0

$$y^3+3y^2-121y-363\rightarrow(y+3)(y+11)(y-11)$$

 

So the solutions are y = -3;  y = -11;  y = 11

.

 

$$\textcolor[rgb]{0,1,0}{1}y^3+3y^2-121y-\textcolor[rgb]{0,1,0}{363}$$

 

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

$$\\y^3+3y^2-121y-363=(x+3)(x-11)(x+11)\\\\
$The roots will be -3, -11,and\;11$$$

.