eldorado21

Also if x = 2; how can remainder by -20?
We cannot plug x = 2

Remainder Theorem states that: Let p(x) be any polynomial of degree greater than or equal to 1 and let α be any real number, then, If p(x) is divided by the polynomial (x - α), then the remainder is p(α).

i.e. If we are dividing the above polynomial with (x + 1), then the remainder would be given by plugging \(x = -1\)

the problem should be solved like this
1------3x^(2n+2) (x+1)-7x^(2n+2)+5x^(2n+1)-8 /x+1
2-----3x^(2n+2) -7x^(2n+2)+5x^(2n+1)-8 /x+1
3-------3x^(2n+2) -7x^(2n+1)(x+1)-2x^(2n+1)-8 /x+1
4-----3x^(2n+2) -7x^(2n+1)-2x^(2n+1)-8 /x+1
5-----3x^(2n+2) -7x^(2n+1)-2(x^(2n+1)-4) /x+1
6-----3x^(2n+2) -7x^(2n+1)-2(x^(2n)-4)(x+1))-8x+2x^(2n) /x+1
7--------------3x^(2n+2) -7x^(2n+1)-2(x^(2n)-4)(x+1))-2x(4+x^(2n-1) /x+1
8------3x^(2n+2) -7x^(2n+1)-2(x^(2n)-4))-2x(4+x^(2n-1) /x+1
so the reminder is going to be 2x(4+x^(2n-1) /x+1
and this can continue for ever
as will as we can't know the real value of n
how is the answer -20 this is so 😂😂😂😂

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