Is there alternative factoring of a quintic equation?

It seems that the claim of the paper is not true.

The author finally got the following equation $$b^4-2b^3\bigg(\frac{q-p+\sqrt{q^2-q}}{q-\sqrt{q^2-q}}\bigg)+q-p+\sqrt{q^2-q}=0$$

Here, let us consider one example. We have $$x^5-31x+30=(x^3+3x^2+7x+15)(x^2-3x+2)$$ This means that one of the possible values of $b$ is $3$ for $(p,q)=(-31,30)$.

However, the above equation does not have a solution $b=3$ for $(p,q)=(-31,30)$.

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Since we have $$(x^3+bx^2+cx+d)(x^2+ex+f)$$ $$=x^5+(b+e)x^4+(eb+c+f)x^3+(bf+ec+d)x^2+(cf+ed)x+df=0$$ if we compare this with $x^5+px+q$, then we get the following system $$\begin{cases}b+e=0 \\eb+c+f=0 \\bf+ec+d=0 \\cf+ed=p \\df=q\end{cases}$$ from which we want to represent $b,c,d,e,f$ by $p,q$.

Now, we have $$\begin{align}&\begin{cases}b+e=0 \\eb+c+f=0 \\bf+ec+d=0 \\cf+ed=p \\df=q\end{cases}\\\\&\stackrel{\text{eliminating $e$}}{\implies} \begin{cases}e=-b \\(-b)b+c+f=0 \\bf+(-b)c+d=0 \\cf+(-b)d=p \\df=q\end{cases} \\\\&\stackrel{\text{eliminating $f$}}{\implies}\begin{cases}e=-b \\df=q \\-b^2d+cd+q=0 \\bq-bcd+d^2=0 \\bd^2=cq-pd \end{cases} \\\\&\stackrel{\text{eliminating $b$}}{\implies} \begin{cases}e=-b \\df=q \\bd^2=cq-pd \\c^2dq^2-cqd^2p+d^3p^2-d^2pcq-cd^5-qd^4=0 \\c^2dq^2-cq^3+dpq^2-cd^2pq-d^4q=0 \end{cases} \\\\&\stackrel{\text{eliminating $c$}}{\implies} \begin{cases}e=-b \\df=q \\bd^2=cq-pd \\c(-d^2pq-d^5+q^3)=dpq^2-d^3p^2 \\d^{10} + p qd^7 + p^3d^6 - 2 q^3d^5 - p^2 q^2 d^4- p q^4d^2 + q^6=0 \end{cases}\end{align}$$

So, we have to solve the following equation for $d$ : $$d^{10} + p qd^7 + p^3d^6 - 2 q^3d^5 - p^2 q^2 d^4- p q^4d^2 + q^6=0$$ whose degree is $10$.

In conclusion, if we want to find $b,c,d,e,f$ such that $$x^5+px+q=(x^3+bx^2+cx+d)(x^2+ex+f)$$ then, in general, we have to solve an equation whose degree is $10$.