Find the sum of other two zeros of the polynomial $x^{4} - 5 x^{3} + 2 x^{2} + 10 x - 8$ if it is given that two of its zeros are $- \sqrt{2}$ and $\sqrt{2}$ .

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Solution

$- \sqrt{2}, \sqrt{2}$ are two zeros it means $(x + \sqrt{2}) (x - \sqrt{2})$ are its factor
or say $(x^{2} - 2)$ is the factor hence we can write
$x^{2} - 2) x^{4} - 5 x^{3} + 2 x^{2} + 10 x - 8 - ------------------------- - (x^{2} - 5 x + 4 \pm x^{4} \mp 2 x^{2} - ----------------- - - 5 x^{3} + 4 x^{2} + 10 x \mp 5 x^{3} \mp 10 x - --------------------- - 4 x^{2} - 8 \pm 4 x^{2} \mp 8 - ----------- -$
$∴ x^{4} - 5 x^{3} + 2 x^{2} + 10 x - 8 = (x^{2} - 2) (x^{2} - 5 x + 4)$
$= (x^{2} - 2) (x - 4) (x - 1)$
Hence other two zeros are $1, 4$
$∴$ Sum $= 5$

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