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10th Class Mathematics Related to Competitive Exam Question Bank Algebra

  • question_answer
    If \[2{{x}^{3}}+4{{x}^{2}}+2ax+b\] is exactly divisible by \[{{x}^{2}}-1\],then the value of a and b respectively will be

    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                 \[a=-1,\,\,b=-4\]


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