Given the polynomial:
$$10x^5+37x^4-83x^3-83x^2+37x+10$$
where $a, b, c, d, e$ are roots of the above polynomial and $a < b < c < d < e$, what is the value of $(a + b)(b + c)(c + d)(d + e)(e + a)$? I know that you can use synthetic division to find the roots, but is there a different way to solving this problem? I've tried to do some factorization with Vieta's, but I couldn't find any representation of the expression in terms that Vieta's would fully help with this problem.
Building on from the comments:
The coefficients of this polynomial are symmetric so if $x_0$ is a root, so is $1/x_0$ ($x = 0$ is not a root).
Now using Descartes' rule of signs, the coefficients of $f(x)$ have $2$ sign changes, and $f(-x) = -10x^5 + 37x^4 + 83x^3 - 83x^2 - 37x + 10$ has $3$ sign changes. Since the question tells you that the roots can be ordered, all of the roots must be real. Hence we can conclude that $f(x)$ has exactly $2$ positive roots and $3$ negative roots.
So because of the condition $a < b < c < d < e$, it turns out that $ac = 1$ (there are two negative roots which are not $-1$ and the negative roots must also multiply to $1$) and $de = 1$. From this it follows that $b = -1$, and so we need to find:
$$(a - 1)(-1 + 1/a)(1/a + d)(d + 1/d)(1/d + a)$$
Expanding this is going to be very messy, so we can use Vieta to note the sum of the roots $a - 1 + 1/a + d + 1/d = -37/10$ or $a + 1/a + d + 1/d = -27/10. \tag{1}$
Similarly, for the $x^2$ term, $-(-83/10) = 83/10$ is the sum of the product of all combinations of $3$ roots. Since $abcde = 1 \cdot 1 \cdot -1 = -1$, this is:
$$-(1/(ab) + 1/(bc) + 1/(cd) + 1/(de) + 1/(ea) + 1/(ac) + 1/(bd) + 1/(ce) + 1/(da) + 1/(be))$$
where there are ${5 \choose 3} = 10$ terms in total.
Thus:
$$-(-1/a - a + a/d + 1 + d/a + 1 - 1/d + ad + 1/(ad) - d) = 83/10$$ $$-(1/a + a + d + 1/d) + a/d + 1 + d/a + 1 + ad + 1/(ad) = -83/10$$ $$a/d + d/a + ad + 1/(ad) = -83/10 - 27/10 - 2 = -13 \tag{2}$$
Now, if we let $a + 1/a = u$, $d + 1/d = v$, then $uv = ad + a/d + d/a + 1/(ad)$ which is exactly what we have already!
Therefore, $u + (-13)/u = -2.7$, and solving this quadratic leads to $u = -5.2, v = 2.5$ or vice versa. Solving the other two quadratics gives the roots as $-5, -1, -0.2, 0.5, 2$.
Addendum: Descartes' rule of signs only tells us that there can be $2$ or $0$ positive real roots, and $3$ or $1$ negative real roots. Therefore, it is possible that the roots are complex. This can happen when we have $x = -1$ as the only real root and the roots satisfy $r_1 r_2 = -1, r_3 r_4 = -1$ first, and also are in conjugate pairs:
The condition in the question $a < b < c < d < e$ is designed to rule out this scenario.