A) \[-1,3\]
B) \[2,5\]
C) \[-1,-6\]
D) \[-3,2\]
Correct Answer: C
Solution :
\[{{x}^{4}}+{{x}^{3}}+8{{x}^{2}}+ax+b\]is exactly divisible by \[{{x}^{2}}+1\] \[\Rightarrow \] Remainder must be zero. \[(a-1)\,x+(b-7)=0\] \[\Rightarrow \] \[a-1=0\] and \[b-7=0\] \[\Rightarrow \] \[a=1\] and \[b=7\] Now, \[a{{x}^{2}}+bx+6\]becomes\[{{x}^{2}}+7x+6\]. \[{{x}^{2}}+7x+6={{x}^{2}}+6x+x+6=0\] \[\Rightarrow \] \[x(x+6)+1(x+6)=0\] \[\Rightarrow \] \[(x+1)(x+6)=0\] \[\Rightarrow \] \[x=-1,\,-6\]You need to login to perform this action.
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