Question

Question #c7d51

Answer 1

`r(x) = (2^51-1)x+2-2^51`

Explanation:

`x^51=p_(49)(x)(x^2-3x+2)+r(x)`

Here `p_(49)(x)` represents a polynomial of degree `49` and `r(x)` is the division remainder. The remainder degree is always lower than the divisor. So `p_49(x)` is the quotient, `(x^2-3x+2)` is the divisor, and `r(x)` the remainder. Note that

degree[`p_49(x)`] = `49`
degree[`(x^2-3x+2)`] = `2`
degree[`r(x)`] = `1`

then

degree[`p_(49)(x)(x^2-3x+2)`] = `51`

but `(x^2-3x+2)=(x-1)(x-2)` and

the general `1` degree polynomial representing `r(x)` is `ax+b` so

`x^51=p_(49)(x)(x-1)(x-2)+ax+b` The constants `a,b` are obtained because we know two conditions to be obeyed.

`x=1->1^51=a+b` and
`x=2->2^51=2a+b`

Solving for `a, b` we get

`r(x) = (2^51-1)x+2-2^51`