Given (x^n+x-8)/(x-2) = R10
Need to find n to satisfy this relation.
Let
f(x)=(x^n+x-8)
By the remainder theorem,
Remainder of (x^n+x-8)/(x-2) = f(2)
For example, for n=2,
n=2, f(x)=x^2+x-8, Remainder f(x)/(x-2)=f(2)=2^2+2-8=-2
Similarly,
n=3, f(x)=x^3+x-8, Remainder f(x)/(x-2)=f(2)=2^3+2-8=2
n=4, f(x)=x^4+x-8, Remainder f(x)/(x-2)=f(2)=2^4+2-8=10
n=5, f(x)=x^5+x-8, Remainder f(x)/(x-2)=f(2)=2^5+2-8=26
....
The function f(x)=x^n+x-8, is monotonic increasing with n for x=2, so we know that any value of n greater than 4 will not give a remainder of f(2)/(x-2) equal to 10.
Therefore the only answer for which the remainder of the quotient f(2)=2^n+2-8 = R10 is when n=4.