Cubic Factorization Over Complex Numbers!

If the cubic is to factor in Complex [ $x; y$ ], then one of the factors must be linear. Without loss of generality, we may assume this factor is of the form $x-py-q$ where $p$ and $q$ are complex numbers.

Substituting $x = py +q$ in the original cubic, we get a polynomial in $y$ that must be identically $0$ . Thus each of its coefficients must be $0$ . This gives us the four equations:

$1 + p + p^3 = 0$ $4+ 2p+ 3p^2+q + 3p^2q = 0$ $3 + Aq + 3q^2+q^3 = 0$ $B+Ap+ 2q +6pq + 3pq^2 = 0$

Solving these equations simultaneously, yields $A = 4$ and $B = 5$ .

As a check we note that the resulting polynomial can be written as $(x + y +2)(y+1)^2 + (x+1)^3$ . The reducibility of this polynomial will not change if we let $x = X-1 \text{ and } y = Y-1$ . This produces the polynomial $X^3 +XY^2+Y^3$ . Letting $z = X/Y$ shows that this polynomial factors over Complex [ $x; y$ ], since $z^3 + z +1$ factors over Complex [ $z$ ].