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.
Factorise
\(6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10\) into three linear factors.
Solution:
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*}
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*}