Answer:
Let \[P(x)={{x}^{4}}-{{x}^{3}}-3{{x}^{2}}+3x\] Given, \[0,\sqrt{3},\,\,-\sqrt{3}\] are three zeroes, so \[x=0\], \[x=\sqrt{3}\] and \[x=-\sqrt{3}\] \[\Rightarrow \] \[(x-\sqrt{3})=0\] and \[x+\sqrt{3}=0\] Here, \[x(x+\sqrt{3})(x-\sqrt{3})\] will also be the factor of \[P(x)\] Or, \[x({{x}^{2}}-3)\] will be the factor of \[P(x)\]. then quotient \[=(x-1)\] So fourth zero \[\Rightarrow x-1=0\] \[x=1\] Hence four zeroes will be \[1,0,\sqrt{3},-\sqrt{3}\].
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