A) \[({{x}^{2}}+1)\,(x-1)\]
B) \[({{x}^{2}}+1)\]
C) \[({{x}^{2}}+1)\,(x+1)\]
D) \[(x+1)\]
Correct Answer: C
Solution :
\[{{x}^{4}}-1=({{x}^{2}}-1)\,({{x}^{2}}+1)=(x-1)\,(x+1)\] \[({{x}^{2}}+1)\] Now \[{{x}^{4}}-2{{x}^{3}}-2{{x}^{2}}-2x-3\] Putting \[x=-1\]in this equation gives 0, so \[(x+1)\] is a factor, divide \[{{x}^{4}}-2{{x}^{3}}-2{{x}^{2}}-2x-3\]by \[(x+1)\] gives \[{{x}^{3}}-3{{x}^{2}}+x-3\] Now put \[x=3,\]gives 0, so another factor is \[(x-3),\] divide \[(x-3)\] gives \[{{x}^{2}}+1\]which cannot be farther divided So\[{{x}^{4}}-2{{x}^{3}}-2{{x}^{2}}-2x-3=({{x}^{2}}+1)(x+1)(x-3)\] Now common factors in both expressions are \[({{x}^{2}}+1)\,(x+1)\] which is the HCF.
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