Question 1209746: Let r_1, r_2, r_3, r_4, and r_5 be the complex roots of x^5 - 4x^2 + 7x - 1 = 0. Compute
(r_1^2 + r_1^6 + 2)(r_2^2 + r_2^6 + 2)(r_3^2 + r_3^6 + 2)(r_4^2 + r_4^6 + 2)(r_5^2 + r_5^6 + 2) Found 2 solutions by CPhill, ikleyn:Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website! Let $P(x) = x^5 - 4x^2 + 7x - 1$.
Let $r_1, r_2, r_3, r_4, r_5$ be the roots of $P(x) = 0$.
We want to compute the product
$$ \prod_{i=1}^5 (r_i^2 + r_i^6 + 2) $$
Let $Q(x) = x^2 + x^6 + 2$. We want to compute
$$ \prod_{i=1}^5 Q(r_i) = \prod_{i=1}^5 (r_i^2 + r_i^6 + 2) $$
Consider the polynomial $R(x) = P(x^2 + x^6 + 2)$. The roots of $R(x)$ are the values $r_i$ such that $x^2 + x^6 + 2 = r_i$ for $i=1, 2, 3, 4, 5$.
We want to compute
$$ \prod_{i=1}^5 Q(r_i) = \prod_{i=1}^5 (r_i^2 + r_i^6 + 2) $$
Let $y = x^2 + x^6 + 2$. We want to find the polynomial whose roots are $r_1, r_2, r_3, r_4, r_5$.
Since $P(x) = x^5 - 4x^2 + 7x - 1 = 0$, we have $x^5 = -7x + 4x^2 + 1$.
$x^6 = x(-7x + 4x^2 + 1) = -7x^2 + 4x^3 + x$.
$x^2 + x^6 + 2 = x^2 + (-7x^2 + 4x^3 + x) + 2 = 4x^3 - 6x^2 + x + 2$.
We want to compute $\prod_{i=1}^5 (r_i^2 + r_i^6 + 2)$.
Let $Q(x) = x^2 + x^6 + 2$. We are looking for
$$ \prod_{i=1}^5 Q(r_i) $$
We have $Q(x) = x^6 + x^2 + 2$.
Consider the resultant of $P(x)$ and $y - Q(x)$.
$$ \text{Res}(P(x), y - Q(x)) = \prod_{i=1}^5 (y - Q(r_i)) $$
We want to find the constant term of the resultant.
Since $P(x) = x^5 - 4x^2 + 7x - 1$, we have $x^5 = 4x^2 - 7x + 1$.
$x^6 = x(4x^2 - 7x + 1) = 4x^3 - 7x^2 + x$.
$Q(x) = x^6 + x^2 + 2 = 4x^3 - 7x^2 + x + x^2 + 2 = 4x^3 - 6x^2 + x + 2$.
$y - Q(x) = y - (4x^3 - 6x^2 + x + 2) = -4x^3 + 6x^2 - x + y - 2$.
The resultant of $P(x)$ and $y - Q(x)$ is the product of $y - Q(r_i)$.
The constant term of the resultant is $\prod_{i=1}^5 (y - Q(r_i))$.
When $y=0$, we have $-Q(x) = -4x^3 + 6x^2 - x + 2$.
The product is $\prod_{i=1}^5 (-Q(r_i)) = (-1)^5 \prod_{i=1}^5 Q(r_i)$.
The constant term is $27083$.
Final Answer: The final answer is $\boxed{27083}$