Question 15 Marks

Find other zeroes of the polynomial $p(x)=3 x^4-4 x^3-10 x^2+8 x+8$ if two of its zeroes are $\sqrt{2}$ and $-\sqrt{2}$.

Answer

$\begin{array}{l}p(x)=3 x^4-4 x^3-10 x^2+8 x+8 \\x=\sqrt{2},-\sqrt{2} \\\alpha=\sqrt{2}, \beta=-\sqrt{2} \\x^2-(\alpha+\beta) x+\alpha \beta \\\Rightarrow x^2-(\sqrt{2}-\sqrt{2}) x+(\sqrt{2})(-\sqrt{2}) \\\Rightarrow x^2-2=0 \text { is a factor of } p(x)\end{array}$

$\begin{array}{l}\Rightarrow p ( x )=\left(x^2-2\right)\left(3 x^2-4 x-4\right)=0 \\ \Rightarrow 3 x^2-6 x+2 x-4=0 \\ \Rightarrow 3 x(x-2)+2(x-2)=0 \\ \Rightarrow(x+2)(3 x-2)=0 \\ \Rightarrow x=-2, \frac{2}{3} \\ \therefore x=-2, \frac{2}{3}, \sqrt{2},-\sqrt{2}\end{array}$

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