$- \sqrt{2}$ and $\sqrt{2}$ are two of the zeros of $p (x) = 2 x^{4} + 7 x^{3} - 8 x^{2} - 14 x + 30$ .
Find the remaining zeros of $p (x)$ .

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Solution

Since two zeroes are $\sqrt{2}$ and $\sqrt{2}$
$∴ (x - \sqrt{2}) (x + \sqrt{2}) = x^{2} - 2$ is a factor of the given polynomial,
Now, we divide the given polynomial by $x^{2} - 2$
$x^{2} - 2) ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ {2 x}^{4} + {7 x}^{3} - {19 x}^{2} - 14 x + 30 ({2 x}^{2} + 7 x - 15 {2 x}^{4} - {4 x}^{2}_________________{7 x}^{3} - {15 x}^{2} - 14 x + 30 {7 x}^{3} - 14 x_- {15 x}^{2} + 30 - {15 x}^{2} + 30_0_________________$

$∴ 2 x^{4} + 7 x^{3} - 19 x^{2} - 14 x + 30 = (x^{2} - 2) (2 x^{2} + 7 x - 15)$
Now, $2 x^{2} + 7 x - 15$
$= 2 x^{2} + 10 x - 3 x - 15$
$= 2 y (x - 5) - 3 (x + 5) = (x + 5) (2 x - 3)$
So. its are given by $x = - 5$ and $x = 3 / 2$ Hence the zeroes of the given polynomial are:
$\sqrt{2}, (- \sqrt{2}), (- 5)$ and $\frac{3}{2}$ Ans.

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