jmc

Squaring $y=x^2-8$, we obtain $y^2=x^4-16x^2+64$. Setting the right-hand sides equal to each other, we find $$\begin{array}{rl}-5x+44& =x^4-16x^2+64\Rightarrow \\ 0& =x^4-16x^2+5x+20\Rightarrow \\ & =x^2(x^2-16)+5(x+4)\Rightarrow \\ & =x^2(x-4)(x+4)+5(x+4)\Rightarrow \\ & =(x+4)(x^3-4x^2+5).\end{array}$$ Therefore, one of the solutions has an $x$-value of $-4$. Then there is the polynomial $x^3-4x^2+5$. The only possible rational roots are now $\pm 1$ and $\pm 5$. Using synthetic or long division, it can be determined that $(x+1)$ is a factor: $$(x+1)(x^2-5x+5)=x^3-4x^2+5$$ Therefore, one of the solutions has an $x$-value of $-1$. Because $x^2-5x+5$ does not factor easily, we use the quadratic formula to get $$\begin{array}{rl}x& =\frac{5\pm \sqrt{25-4\cdot 1\cdot 5}}{2}\Rightarrow \\ & =\frac{5\pm \sqrt{5}}{2}.\end{array}$$ The four values for $x$ are then $-4,-1,\frac{5\pm \sqrt{5}}{2}$. Squaring each: $$(-4{)}^2=16$$ $$(-1{)}^2=1$$ $${\left(\frac{5+\sqrt{5}}{2}\right)}^2=\frac{25+10\sqrt{5}+5}{4}=\frac{15+5\sqrt{5}}{2}$$ $${\left(\frac{5-\sqrt{5}}{2}\right)}^2=\frac{25-10\sqrt{5}+5}{4}=\frac{15-5\sqrt{5}}{2}$$ And subtracting $8$: $$16-8=8$$ $$1-8=-7$$ $$\frac{15+5\sqrt{5}}{2}-\frac{16}{2}=\frac{-1+5\sqrt{5}}{2}$$ $$\frac{15-5\sqrt{5}}{2}-\frac{16}{2}=\frac{-1-5\sqrt{5}}{2}$$ Therefore, the four solutions are $$(-4,8),(-1,-7),$$ $$\left(\frac{5+\sqrt{5}}{2},\frac{-1+5\sqrt{5}}{2}\right),\left(\frac{5-\sqrt{5}}{2},\frac{-1-5\sqrt{5}}{2}\right).$$

Multiplying the $y$-coordinates: $$8\cdot -7\cdot \frac{-1+5\sqrt{5}}{2}\cdot \frac{-1-5\sqrt{5}}{2}=\frac{-56(1-25\cdot 5)}{4}=\boxed{1736}.$$