Could you tell me how can i calculate the :
gcd(2n^3-14n+2, n+3) ?

Question

Could you tell me how can i calculate the :
gcd(2n^3-14n+2, n+3) ?

zehanz · Answer

First, try to factor 2n^3-14n+2.
If you are lucky, n+3 is one of the factors.
In that case, n+3 itself is the gcd.
If n+3 is NOT a factor (and this is the case, I'm afraid), then the gcd is 1.

You can see whether n+3 is a factor of 2n^3-14n+2 by doing a long or synthetic division.
If you need help with that, just ask.

anonymous · Answer

not a factor

anonymous · Answer

$$(n+3)(an^2+bn+c)=2n^3-14n+2$$
so 
$$an^3+(b+3a)n^2+(c+3b)n+3c = 2n^3-14n+2$$
$$a=2, b+3a=0, c+3b=-14, 3c=2$$
unfortunately n+3 is not a factor

anonymous · Answer

so the gcd is 1

zehanz · Answer

I agree with @AntarAzri!
It is a good way to check if n+3 is a factor: not a long division or a synthetic division, but with the aid of a few simple equations!