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9th Class Mathematics Polynomials Question Bank Polynomials

  • question_answer
    The product of \[\left( \mathbf{x-}\frac{\mathbf{1}}{\mathbf{x}} \right)\left( \mathbf{x+}\frac{\mathbf{1}}{\mathbf{x}} \right)\left( {{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{1}}{{{\mathbf{x}}^{\mathbf{2}}}} \right)\left( {{\mathbf{x}}^{\mathbf{4}}}\mathbf{+}\frac{\mathbf{1}}{{{\mathbf{x}}^{\mathbf{4}}}} \right)\]is

    A)  \[\left( {{x}^{8}}-\frac{1}{{{x}^{8}}} \right)\]           

    B)  \[\left( {{x}^{4}}-\frac{1}{{{x}^{4}}} \right)\]      

    C)  \[\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)\]           

    D)  \[\left( {{x}^{8}}+\frac{1}{{{x}^{8}}} \right)\]

    Correct Answer: A

    Solution :

    (a):We have, \[\left( x-\frac{1}{x} \right)\left( x+\frac{1}{x} \right)\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)\] \[=\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)\] \[=\left\{ {{\left( {{x}^{2}} \right)}^{2}}-{{\left( \frac{1}{{{x}^{2}}} \right)}^{2}} \right\}\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)\]        


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