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

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Answer 1 · 2018-01-25T15:22:55

Remainder is $- 3$ $-3$

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

the remainder is $f (α$ $f(alpha$

Hence dividing $x^{107} + 3 x^{98} - 4 x^{4} - 3$ $x^107+3x^98-4x^4-3$ by $x - 1$ $x-1$ ,

we get a remainder

$1^{107} + 3 \cdot 1^{98} - 4 \cdot 1^{4} - 3$ $1^107+3*1^98-4*1^4-3$

= $1 + 3 - 4 - 3$ $1+3-4-3$

= $- 3$ $-3$

What is the remainder of (x^107+3x^98-4x^4-3)/(x-1)x107+3x98−4x4−3x−1(x^107+3x^98-4x^4-3)/(x-1)?