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Question

# Find all the zeros of (2x4 − 3x3 − 5x2 + 9x − 3), it being given that two of its zeros are $\sqrt{3}$ and $-\sqrt{3}$.

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Solution

## $\begin{array}{l}\text{The given polynomial is}f\left(x\right)=2{x}^{4}-3{x}^{3}-5{x}^{2}+9x-3.\\ \text{Since}\sqrt{3}\text{and}-\sqrt{3}\text{are the zeroes of}f\left(x\right),\text{it follows that each one of}\left(x-\sqrt{3}\right)\text{and}\left(x+\sqrt{3}\right)\text{is a factor of}f\left(x\right).\\ \\ \text{Consequently,}\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=\left({x}^{2}-3\right)\text{is a factor of}f\left(x\right).\\ \text{On dividing}f\left(x\right)\text{by}\left({x}^{2}-3\right)\text{, we get:}\end{array}$ $\begin{array}{l}\\ \therefore f\mathit{\left(}x\mathit{\right)}\mathit{=}\mathit{0}\\ \mathit{=}\mathit{>}2{x}^{\mathit{4}}\mathit{-}3{x}^{\mathit{3}}\mathit{-}5{x}^{\mathit{2}}\mathit{+}9x\mathit{-}3\mathit{=}\mathit{0}\\ \begin{array}{l}\mathit{=}\mathit{>}\mathit{\left(}{x}^{\mathit{2}}\mathit{-}3\mathit{\right)}\mathit{\left(}2{x}^{\mathit{2}}\mathit{-}3x\mathit{+}1\mathit{\right)}\mathit{=}\mathit{0}\\ \mathit{\text{=>}}\mathit{\left(}{x}^{\mathit{2}}\mathit{-}3\mathit{\right)}\mathit{\left(}2{x}^{\mathit{2}}\mathit{-}2x\mathit{-}x\mathit{+}1\mathit{\right)}\mathit{=}\mathit{0}\\ \begin{array}{l}=>\left(x-\sqrt{\mathrm{3}}\right)\left(x+\sqrt{\mathrm{3}}\right)\left(\mathrm{2}x-\mathrm{1}\right)\left(x-\mathrm{1}\right)=0\\ =>x=\sqrt{3}\mathrm{or}x=-\sqrt{3}3\mathrm{or}x=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{or}\mathit{}x=\mathrm{1}\end{array}\end{array}\\ \end{array}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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