Question

If $(x^{100}+2x^{99}+K)$ is exactly divisible by $(x+1)$, find the value of 'K'

• A. $1$
• B. $2$
• C. $-2$
• D. $-3$

Answer 1

Answer: (A)

$1$

$x^{100}+2x^{99}+k$ is exactly divisible by $(x+1)$

$\therefore x=-1$ is the root of $x^{100}+2x^{99}+k$

$\Rightarrow (-1{)}^{100}+2(-1{)}^{99}+k=0$

$\Rightarrow 1-2+k=0$

$\Rightarrow k=1$