If sqrt{2} and -sqrt{2} are the zeros of p(x) | vedclass

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(B) Since $\sqrt{2}$ and $-\sqrt{2}$ are zeros of $p(x)$,$(x - \sqrt{2})(x + \sqrt{2}) = x^{2} - 2$ is a factor of $p(x)$.Dividing $p(x) = 2x^{4} + 7x

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$\frac{3}{2}, -5$

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(B) Since $\sqrt{2}$ and $-\sqrt{2}$ are zeros of $p(x)$,$(x - \sqrt{2})(x + \sqrt{2}) = x^{2} - 2$ is a factor of $p(x)$.Dividing $p(x) = 2x^{4} + 7x^{3} - 19x^{2} - 14x + 30$ by $(x^{2} - 2)$,we get the quotient $2x^{2} + 7x - 15$.To find the other zeros,we solve $2x^{2} + 7x - 15 = 0$.$2x^{2} + 10x - 3x - 15 = 0$$2x(x + 5) - 3(x + 5) = 0$$(2x - 3)(x + 5) = 0$Thus,$x = \frac{3}{2}$ and $x = -5$ are the other zeros.

If $\sqrt{2}$ and $-\sqrt{2}$ are the zeros of $p(x) = 2x^{4} + 7x^{3} - 19x^{2} - 14x + 30$,then find the other zeros of $p(x)$.

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