9. $[\mathbf{7}]$ Find the product of all real $x$ for which
$$
2^{3 x+1}-17 \cdot 2^{2 x}+2^{x+3}=0 \text {. }
$$
Answer: -3 We can re-write the equation as $2^{x}\left(2 \cdot\left(2^{x}\right)^{2}-17 \cdot\left(2^{x}\right)+8\right)=0$, or $2 \cdot\left(2^{x}\right)^{2}-17 \cdot\left(2^{x}\right)+8=0$. Make the substitution $y=2^{x}$. Then we have $2 y^{2}-17 y+8=0$, which has solutions (by the quadratic formula) $y=\frac{17 \pm \sqrt{289-64}}{4}=\frac{17 \pm 15}{4}=8, \frac{1}{2}$, so $2^{x}=8, \frac{1}{2}$ and $x=3,-1$. The product of these numbers is -3 .