Question:
If $\left(x^{100}+2 x^{99}+k\right)$ is divisible by $(x+1)$, then the value of $k$ is
(a) 1
(b) 2
(c) −2
(d) −3
Solution:
(a) 1
Let:
$p(x)=x^{100}+2 x^{99}+k$
Now,
$x+1=0 \Rightarrow x=-1$
$(x+1)$ is divisible by $p(x)$
$\therefore p(-1)=0$
$\Rightarrow(-1)^{100}+2 \times(-1)^{99}+k=0$
$\Rightarrow 1-2+k=0$
$\Rightarrow-1+k=0$
$\Rightarrow k=1$
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