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

Question

1476 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

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

= $1+3-4-3$

= $-3$

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