Modular Root

Since $x \equiv a \pmod{x-a}$ for any integer $a$ , $f(x) \equiv f(a) \pmod{x-a}$ .

$x^3 - 3x +7 \equiv 5^3 - 3\cdot 5 +7 \equiv 117 \equiv 0 \pmod{x-5}$

$2x^2 - x +2 \equiv 2(-6)^2 -(-6) + 2 \equiv 80 \equiv 0 \pmod{x+6}$

From the first equivalence, $117$ has factors of $\left\{{1, 3, 9, 13, 39, 117}\right\}$ , and so $x \in \left\{{6, 8, 14, 18, 44, 122}\right\} = A$

Then for the latter, $80$ has $\left\{{1, 2, 4, 5, 8, 10, 16, 20, 40, 80}\right\}$ , and so $x \in \left\{{2, 4, 10, 14, 34, 74}\right\} = B$ .

$A\cap B = \left\{{14}\right\}$ .

Thus, $x = \boxed{14}$ .

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Checking answers:

$14^3 - 3\cdot 14 +7 \equiv 5^3 - 3\cdot 5 +7 \equiv 117 \equiv 0 \pmod{9}$

$2(14^2) - 14 +2 \equiv 392 -12 \equiv 380 \equiv 0 \pmod{20}$

Since, $n^3 - 3n + 7 \equiv 0$ (mod $n -5$ ) $\implies n - 5 | n^3 - 3n + 7 \\$ Similarly, $2n^2 - n + 2 \equiv 0$ (mod $n+6$ ) $\implies n + 6 |2n^2 - n + 2 \\$ Factoring $n^3 - 3n + 7$ in terms of $n - 5, \\$ $n^3 - 3n + 7 = n^3 - 5n^2 + 5n^2 - 25n + 25n - 3n + 7 \\$ $= n^2(n - 5) + 5n(n - 5) + 22n + 7 \implies n - 5|22n+7 \\$ Similarly, Factoring $2n^2 - n + 2$ in terms of $n + 6\\$ $2n^2 - n + 2 = 2n^2 + 12n - 12n -n - 78 + 78 + 2 \\$ $= 2n(n + 6) - 13(n + 6) + 80 \implies n + 6| 80 \\$ Possible divisors of 80 are - $2, 4, 5, 8, 10, 16, 20, 40, 80,\\$ From the given condition $n \geq 5, n = 14$ , satisfies both the conditions.