Problem source

\begin{questionparts}
\item
Show that $x-3$ is a factor of 
\begin{equation}
x^3-5x^2+2x^2y+xy^2-8xy-3y^2+6x+6y \;.
\tag{$*$}
\end{equation}
Express ($ * $) in the form 
$(x-3)(x+ay+b)(x+cy+d)$ where $a$, $b$, $c$ and $d$ 
are integers to be determined. 
\item Factorise 
$6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10$ into three linear factors.
\end{questionparts}

Solution source

\begin{questionparts}
\item Let $f(x,y) = x^3-5x^2+2x^2y+xy^2-8xy-3y^2+6x+6y$, then
\begin{align*}
f(3,y) &= 27 - 5 \cdot 9 +18y + 3y^2-24y-3y^2+18 + 6y \\
&= 0
\end{align*}, therefore $x-3$ is a factor of $f(x,y)$.

\begin{align*}
f(x,y) &= x^3-5x^2+6x+y(2x^2-8x+6) + y^2(x-3) \\
&= (x-3)(x^2-2x)+y(x-3)(2x-2)+y^2(x-3) \\
&= (x-3)(x^2-2x+2y(x-1)+y^2) \\
&= (x-3)(x+y)(x+y-2)
\end{align*}

\item Let $g(x,y) = 6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10$, notice that $g(x,-2) = 0$, so $y+2$ is a factor,

\begin{align*}
g(x,y) &= 6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10 \\
&= x^2(2+y) + x(12-4y-5y^2) + 6y^3-y^2-21y+10 \\
&= x^2(y+2) + x(y+2)(6-5y) + (y+2)(6y^2-13y+5) \\
&= (y+2)(x^2+(6-5y)x+(6y^2-13y+5)) \\
&= (y+2)(x-2y +1)(x-3y+5)
\end{align*}

\end{questionparts}