How to slove this type of quention, like I have the solution, but I would appreciate if someone can explain the logic, or if you have a simple approach.

I would have used the remainder theorem. Since the divisor has a degree of 2, the maximum degree of the remainder must be 1. So, ax+b, lets say. Even if the remainder is just a constant, 'a' will simply be zero in the remainder.

The divisor can be factorized into (x-1)(x-2) giving us two values (1 and 2) which can be used to find the two unknown variables in the remainder (a and b).

f(1)=7 and f(2)=127. a+b=7 and 2a+b=127. Now it is just a matter of solving the simultaneous equations.

If the divisor was with a degree of three, we could have assumed the remainder to have a power of 2 (ax^2+bx+c), and so on.

First factor the divisor:

x² − 3x + 2 = (x − 1)(x − 2)

When a polynomial is divided by a quadratic, the remainder will be of the form

R(x) = ax + b

Let

P(x) = x⁶ + x⁵ + x⁴ + x³ + x² + x + 1

Since (x − 1) and (x − 2) are factors of the divisor, we use x = 1 and x = 2.

P(1) = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7

So a + b = 7

P(2) = 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127

So 2a + b = 127

Now subtract the equations:

(2a + b) − (a + b) = 127 − 7

a = 120

Substitute back:

120 + b = 7

b = −113

So the remainder is

120x − 113

i think long division will be the most probable

maybe you can try long division method.