A) 1, 2
B) - 1, 4
C) 1, - 2
D) -1, - 4
Correct Answer: D
Solution :
Since \[f(x)=2{{x}^{3}}+4{{x}^{2}}+2ax+b\]is exactly divisible by\[{{x}^{2}}-1=(x-1)(x+1)\] \[\therefore \] \[f(1)=0\]and\[f(-1)=0\] These give \[2+4+2a+6=0\] or \[2a+b+6=0\] ... (i) and \[-2+4-2a+6=0\] or \[2a-6-2=0\] ? (ii) Solving equations (i) and (u), we get \[a=-1,\,\,b=-4\]You need to login to perform this action.
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