# what is the remainder when x+x^9+x^25+x^49+x^81 is divided by x^3-x?

You should know that degree of reminder is less then the degree of `x^3-x` , hence, the reminder will be a quadratic such that:

`r(x) = ax^2 + bx + c`

You need to find a,b,c to determine the reminder, hence, you need to write the reminder theorem such that:

`x+x^9+x^25+x^49+x^81 = (x^3-x)q(x) + r(x)`

`x+x^9+x^25+x^49+x^81 = (x^3-x)q(x) + ax^2 + bx + c`

Notice that if you substitute any of the roots of x^3-x in relation above, the part `(x^3-x)q(x)`  cancels, hence, you need to find the roots of x^3-x such that:

`(x^3-x) = 0`

Factoring out x yields:

`x(x^2 - 1)= 0 => x_1 = 0`

`x^2 - 1 = 0 => (x-1)(x+1) = 0 => x_2 = 1 ; x_3 = -1`

Hence, you need to substitute each root of x^3-x, one by one, in equation `x+x^9+x^25+x^49+x^81 = (x^3-x)q(x) + ax^2 + bx + c`  such that:

`x = -1`

`-1 + (-1)^9 + (-1)^25 + (-1)^49 + (-1)^81 = (-1+1)q(-1) + a*(-1)^2 + b*(-1) + c`

`-1 - 1 - 1 - 1 - 1 = 0*q(-1) + a - b + c`

`-5 = a - b + c`

`x = 0 `

`0+ (0)^9 + (0)^25 + (0)^49 + ()^801 = (0-0)q(0) + a*(0)^2 + b*(0) + c =gt 0 = c`

`x = 1`

`1 + (1)^9 + (1)^25 + (1)^49 + (1)^81 = (1-1)q(1) + a*(1)^2 + b*(1) + c`

`5 = a + b + c`

You need to consider the three equations such that:

`{(a - b + c = -5) , (c = 0), (a + b + c = 5):}` `=>`  `{(a-b = -5),(a+b = 5):}` `=> 2a = 0 => a = 0`

`0 + b = 5 => b = 5`

Hence, evaluating the reminder under the given conditions yields `r(x) = 5x` .

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