Question

Plzz ans this with whole steps...i will mark as brainliest....but only with whole procedure...​

Answer 1

Hey

Since xn−yn=(x−y)(xn−1+xn−2y+xn−3y2+⋯+xyn−2+yn−1), we see that x−y divides xn−yn for each positive integer n. In particular, x−1 divides xn−1 for each positive integer n.

So x2–1 divides each of the polynomials (x2)40−1,(x2)24−1,(x2)12−1,(x2)4–1. Hence the remainder when

x80+x48+x24+x8+1=(x80−1)+(x48−1)+(x24−1)+(x8–1)+5

is divided by x2–1 equals 5.

So we can write

x80+x48+x24+x8+1=q(x)⋅(x2–1)+5

for some polynomial q(x) with integer coefficients.

Multiplying by x gives

x81+x49+x25+x9+x=xq(x)⋅(x2–1)+5x.

Therefore the remainder is 5x.

Hope it helps you.

Mark as the brainliest.