Partly Factored Polynomials

Quartic equations of the shape $(ax+b)(cx+d)(px+q)(rx+s)+k=0$ ($k\ne0$) can sometimes be rewritten so the four linear factors pair up into two quadratics that share the same leading and linear terms — say $(\alpha x^2+\beta x+\lambda)(\alpha x^2+\beta x+\mu)+k=0$. Substituting $y=\alpha x^2+\beta x$ then collapses the quartic to a quadratic in $y$: $(y+\lambda)(y+\mu)+k=0$.

How to find the right pairing. Among the four linear factors, look for the way of splitting them into two pairs whose products share the same $x^2$-coefficient and $x$-coefficient (only the constant term differs) — e.g. pairing $(x-2)(x-3)=x^2-5x+6$ with $(x-7)(x+2)=x^2-5x-14$, both sharing $x^2-5x$, rather than pairing the factors in the order the equation happens to print them.

Worked illustration (Example 3.23 pattern). $(x-2)(x-7)(x-3)(x+2)+19=0$ can be regrouped as $(x-2)(x-3)(x-7)(x+2)+19=0$ [re-pairing the given factors so two of them share a linear part], i.e. $(x^2-5x+6)(x^2-5x-14)+19=0$. With $y=x^2-5x$: $(y+6)(y-14)+19=0 \Rightarrow y^2-8y-65=0 \Rightarrow y=13$ or $y=-5$. Each $y$-value gives a quadratic in $x$ ($x^2-5x-13=0$ and $x^2-5x+5=0$) to solve by the usual formula, yielding all four roots of the quartic. …