Find the product of all real $x$ for which \[ 2^{3x+1} - 17 \cdot 2^{2x} + 2^{x+3} = 0. \]
1. Let \( y = 2^x \). Then, the given equation \( 2^{3x+1} - 17 \cdot 2^{2x} + 2^{x+3} = 0 \) can be rewritten in terms of \( y \).

2. Rewrite each term in the equation using \( y \):
   \[
   2^{3x+1} = 2 \cdot 2^{3x} = 2 \cdot (2^x)^3 = 2y^3,
   \]
   \[
   17 \cdot 2^{2x} = 17 \cdot (2^x)^2 = 17y^2,
   \]
   \[
   2^{x+3} = 2^3 \cdot 2^x = 8 \cdot 2^x = 8y.
   \]

3. Substitute these expressions back into the original equation:
   \[
   2y^3 - 17y^2 + 8y = 0.
   \]

4. Factor out \( y \) from the equation:
   \[
   y(2y^3 - 17y + 8) = 0.
   \]

5. This gives us the solutions:
   \[
   y = 0, \quad 2y^2 - 17y + 8 = 0.
   \]

6. Solve the quadratic equation \( 2y^2 - 17y + 8 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   y = \frac{17 \pm \sqrt{17^2 - 4 \cdot 2 \cdot 8}}{2 \cdot 2} = \frac{17 \pm \sqrt{289 - 64}}{4} = \frac{17 \pm \sqrt{225}}{4} = \frac{17 \pm 15}{4}.
   \]

7. This gives us two solutions for \( y \):
   \[
   y = \frac{17 + 15}{4} = 8, \quad y = \frac{17 - 15}{4} = \frac{1}{2}.
   \]

8. Now, convert these \( y \) values back to \( x \) using \( y = 2^x \):
   \[
   2^x = 8 \implies x = 3,
   \]
   \[
   2^x = \frac{1}{2} \implies x = -1.
   \]

9. The product of all real \( x \) that satisfy the given equation is:
   \[
   -1 \cdot 3 = -3.
   \]

The final answer is \(\boxed{-3}\).