A) \[({{x}^{2}}+1)\,\,(x-1)~~\]
B) \[({{x}^{2}}+1)\]
C) \[({{x}^{2}}+1)\,\,(x+1)\]
D) \[(x+1)\]
Correct Answer: C
Solution :
[c] \[{{x}^{4}}-\,1=({{x}^{2}}-\,1)\,\,({{x}^{2}}+1)=(x-\,1)\,\,(x+1)\] \[({{x}^{2}}+1)\,\,Now\,\,{{x}^{4}}-\,2{{x}^{3}}-\,2{{x}^{2}}-\,2x-\,3\] 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 further 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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