question_answer

Find the roots of \[a{{x}^{2}}+bx+6,\]if the polynomial \[{{x}^{4}}+{{x}^{3}}+8{{x}^{2}}+ax+b\]is exactly divisible by\[{{x}^{2}}+1\].

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\[{{x}^{2}}+7x+6\]. \[{{x}^{2}}+7x+6={{x}^{2}}+6x+x+6=0\] \[\Rightarrow \]            \[x(x+6)+1(x+6)=0\] \[\Rightarrow \] \[(x+1)(x+6)=0\] \[\Rightarrow \] \[x=-1,\,-6\]