Heuristic Solution: Determine the product of $q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)$.

Given $p(x)=x^5+x^2+1$ have roots $r_1,\,r_2,\,r_3,\,r_4,\,r_5$. Let $q(x)=x^2-2$. Determine the product of $q(r_1)\cdot q(r_2)\cdot q(r_3)\cdot q(r_4)\cdot q(r_5)$.

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If $y=x^2-2$, then $x=(y+2)^{\frac{1}{2}}$, and

$\begin{align*}x^5+x^2+1&=((y+2)^{\frac{1}{2}})^5+((y+2)^{\frac{1}{2}})^2+1\\&=(y+2)^{\frac{5}{2}}+y+2+1\\&=(y+2)^{\frac{5}{2}}+y+3\end{align*}$

But if $(y+2)^{\frac{5}{2}}+y+3=0$ then we have $(y+2)^5=(y+3)^2$, so that $y^5+\cdots+(2^5-3^2)=0$.

Therefore, the product of the roots of the polynomial $q(x)$ is the negative of the constant term, i.e. $-(2^5-3^2)=-23$.