sale rapido

Critical points: $8x^7+4ax^3=0\implies 4x^3(2x^4+a)=0\implies x\in\left\{0,\pm\sqrt[4]{-\frac{a}{2}}\right\}$. $y(0)=1$. $y\left(\pm\sqrt[4]{-\frac{a}{2}}\right)=\frac{a^2}{4}-a\cdot\frac{a}{2}+1=1-\frac{a^2}{4}$. Four roots if $\frac{a^2}{4}>1\wedge a<0\implies a<-2$.

Why not use the Viete relations, because: $a+a+r+a+2r+a+3r=0$, hence $a=-2r$ and write for example the second Viete relation, to find $r$ and $a$.