A) \[x+2\]
B) \[x+3\]
C) \[{{(x+2)}^{2}}\]
D) None of these
Correct Answer: A
Solution :
Let \[f(x)={{x}^{3}}+8={{x}^{3}}+{{2}^{3}}=(x+2)({{x}^{2}}-2x+4)\] |
\[=(x+2){{(x-2)}^{2}}=(x+2)(x-2)(x-2)\] |
\[g\,\,(x)={{x}^{2}}+5x+6={{x}^{2}}+3x+2x+6\] |
\[=x\,\,(x+3)+2\,\,(x+3)=(x+3)(x+2)\] |
and\[h\,\,(x)={{x}^{3}}+2{{x}^{2}}4x+8={{x}^{2}}(x+2)+4\,\,(x+2)\] |
\[=(x+2)({{x}^{2}}+4)\] |
\[\therefore \]HCF of \[\{f\,\,(x),g\,\,(x),h\,\,(x)\}=(x+2)\] |
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