A) 4
B) \[-\,4\]
C) 2
D) 0
Correct Answer: B
Solution :
[b] The expression \[3{{x}^{3}}-\,k{{x}^{2}}+4x+16\] is divisible by \[x-\,\frac{k}{2}.\] Then, \[x=\frac{k}{2}\]satisfy the equation \[\Rightarrow 3\,\,{{\left( \frac{k}{2} \right)}^{3}}-k\,\,{{\left( \frac{k}{2} \right)}^{2}}+4\,\,\left( \frac{k}{2} \right)+16=0\] \[\Rightarrow \,\,\,\frac{3{{k}^{3}}-2{{k}^{3}}+16k+128}{8}=0\] \[\Rightarrow \,\,\,{{k}^{3}}+16k+128=0\] \[\Rightarrow (k+4)\,\,({{k}^{2}}-4k+32)=0\] \[\Rightarrow \,\,\,k+4=0\] \[\Rightarrow \,\,\,k=-\,4\]
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