How to prove that $\gcd(2n^3-14n+2,n+3) = \gcd(n+3,10)$?

Question

We have $\alpha=2n^3-14n+2$ and $\beta=n+3$,

How can we prove that the $\gcd(\alpha,\beta)=\gcd(\beta,10)$,

and what are the possible values for this $\gcd$.

Answer 1

$2n^3-14n+2 = 2n^2(n+3)-6n(n+3)+4(n+3)-10$ so it easily follows that $\gcd(\alpha, \beta) = \gcd(\beta, -10) = \gcd(\beta, 10)$.

The value of this gcd is $d$ such that $d|10$.

Answer 2

Hint:

$$2n^3-14n+2=(n+3)(2n^2-6n+4)-10$$

Answer 3

Hint $\ $ By Euclid, $\,\gcd(n\!-\!a,\,f(n)) = \gcd(n\!-\!a,\,f(n)\ {\rm mod}\ n\!-\!a) = \gcd(n\!-\!a,\color{#c00}{f(a)})\ $ for any polynomial $\,f(n)\,$ with integer coefficients, since $\,{\rm mod} \ n\!-\!a\!:\ n\equiv a\,\Rightarrow\, f(n)\equiv \color{#c00}{f(a)}\ $ by the Polynomial Congruence Rule (or by the Polynomial Remainder Theorem).

Yours is special case $\ f(n) = 2n^3-14n+2,\,\ a=-3,\ $ so $\,\color{#c00}{f(a)} = f(-3) = \color{#c00}{-10}\ $ so the above yields that $\, \gcd(n\!+\!3,f(n)) = \gcd(n\!+\!3,-10) = \gcd(n\!+\!3,10).$