### Instruction: There are integers $b,c$ for which both roots of the polynomial $x^2-x-1$ are also roots of the polynomial $x^5-bx-c$. Determine the product $bc$.### Response: To find the roots of $x^2-x-1$, I can use the quadratic formula: $x = \frac{1 \pm \sqrt{5}}{2}$.
These are also roots of $x^5-bx-c$, so they must satisfy the equation $x^5-bx-c=0$.
Plugging in $x = \frac{1 + \sqrt{5}}{2}$, I get $\left(\frac{1 + \sqrt{5}}{2}\right)^5 - b\left(\frac{1 + \sqrt{5}}{2}\right) - c = 0$.
Expanding the powers of $\frac{1 + \sqrt{5}}{2}$ using the binomial theorem, I get $\frac{11 + 5\sqrt{5}}{8} - b\left(\frac{1 + \sqrt{5}}{2}\right) - c = 0$.
Multiplying both sides by 8, I get $11 + 5\sqrt{5} - 4b - 4b\sqrt{5} - 8c = 0$.
Comparing the rational and irrational parts, I get $11 - 4b - 8c = 0$ and $5 - 4b = 0$.
Solving for $b$, I get $b = \frac{5}{4}$.
Substituting this into the first equation, I get $11 - 5 - 8c = 0$.
Solving for $c$, I get $c = \frac{3}{4}$.
Therefore, the product $bc$ is $\frac{5}{4} \cdot \frac{3}{4} = \frac{15}{16}$.