Let *p*(*x*) = *x* ^{51} and *q*(*x*) = *x* ^{2 }– 3*x* + 2 = (*x *– 1) (*x *– 2)

When *p*(*x*) is divided by *q*(*x*), then by division algorithm there exists *Q*(*x*) and *R*(*x*) = ax+b such that

x^{51} = *Q*(*x*)(x-1)(x-2) + ax +b where *Q*(*x*) is result and ax+b is remainder

For x = 1 you get

1^51 = *Q*(1)(1-1)(1-2) + a+b then a+b = 1

and

2^51 = *Q*(2)(2-1)(2-2) + 2a + b then 2a+b = 2^51

Now solve

a+b = 1

2a+b = 2^51

a = 2^(51) - 1

b = 2- 2^(51)

Remainder is x(2^51 - 1) + 2-2^51