What is the remainder of #(x^107+3x^98-4x^4-3)/(x-1)#?

Question

1283 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-25T15:22:55

Remainder is #-3#

According to remainder theorem, when a polynomial #f(x)# is divided by #x-alpha#,

the remainder is #f(alpha#

Hence dividing #x^107+3x^98-4x^4-3# by #x-1#,

we get a remainder

#1^107+3*1^98-4*1^4-3#

= #1+3-4-3#

= #-3#