Find the value

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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