# Factoring a and quartic

While scanning this online pdf, I came across a possible mistake?

From $(12)$ to $(14)$, the PDF starts with $$p\{[p+a_2-(a_3^2/4)]^2-4a_0\}=\{(a_3/2)[p+a_2-(a_3^2/4)]-a_1\}^2\tag{12}$$

And ends up with the cubic $$p_1^3-a_2p_1^2+(a_1a_2-4a_0)p_1+4a_0F_1-a_1^2=0\tag{14}$$ where $F_1=a_2-(a_3^2/4)$ and $p_1=p+F_1$ With $a_3,a_2,a_1,a_0$ coefficients of the quartic such that it is $x^4+a_3x^3+a_2x^2+a_1x+a_0=0$. But with their example, starting with a quartic $x^4-x^3-19x^2-11x+30=0$ they found that it as a corresponding cubic of $$p_1^3+19p_1^2-109p_1-2431=0$$ but when I computed $a_1a_2-4a_0$, I get a different answer. Namely, $(-19)\cdot(-11)-4\cdot 30=89$ Which is no where near $-109$.

Am I doing something wrong, or does the PDF have a little mistake?

Am I doing something wrong, or does the PDF have a little mistake?

I think the latter. The following should be correct : $$p_1^3-a_2p_1^2+(a_1a_\color{red}{3}-4a_0)p_1+4a_0F_1-a_1^2=0\tag{14}$$ since from $(12)$ we get $$(p_1-F_1)(p_1^2-4a_0)=\left(\frac{a_3}{2}p_1-a_1\right)^2,$$ i.e. $$p_1^3-4a_0p_1-F_1p_1^2+4a_0F_1=\frac{a_3^2}{4}p_1^2-a_1a_\color{red}{3}p_1+a_1^2$$