Bairstow's method

Remarks

Differentiate $p = Bq + cx + d$ with respect to $v$: $0 = q + B \, \partial_v q + (\partial_v c) x + \partial_v d$. So $q$ leaves the remainder $-(\partial_v c) x - \partial_v d$ on division by $B$, which gives $\partial_v c = -g$ and $\partial_v d = -h$. Differentiating with respect to $u$ gives that $x q(x)$ leaves the remainder $-(\partial_u c) x - \partial_u d$; since $x(gx + h) \equiv (h - ug) x - vg$ modulo $B$, this gives the other two derivatives. The $2 \times 2$ matrix in the step is the inverse of the Jacobian matrix of $(c, d)$, times its determinant $vg^2 + h(h - ug)$. So, when this Jacobian is invertible at a quadratic factor, the method converges quadratically from nearby starting values. A quadratic factor gives two roots of $p$ by the quadratic formula, which may be a pair of complex conjugate roots, all in real arithmetic.

Going deeper

For a student

All the work is two synthetic divisions per step, so it is cheap. After a factor is found, one divides it out and repeats on the quotient of degree $n - 2$. Like Newton's method, it needs good starting values and converges only linearly at multiple factors.