Given that sqrt{2} is a zero of the cubic pol | vedclass

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(B) Let $f(x) = 6x^{3}+\sqrt{2}x^{2}-10x-4\sqrt{2}$.Since $\sqrt{2}$ is a zero of $f(x)$,$(x-\sqrt{2})$ is a factor of $f(x)$.Dividing $f(x)$ by $(x-\

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$-\frac{1}{\sqrt{2}}, -\frac{4}{3\sqrt{2}}$

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(B) Let $f(x) = 6x^{3}+\sqrt{2}x^{2}-10x-4\sqrt{2}$.Since $\sqrt{2}$ is a zero of $f(x)$,$(x-\sqrt{2})$ is a factor of $f(x)$.Dividing $f(x)$ by $(x-\sqrt{2})$:$6x^{3}+\sqrt{2}x^{2}-10x-4\sqrt{2} = (x-\sqrt{2})(6x^{2}+7\sqrt{2}x+4)$.Now,factorize the quadratic polynomial $6x^{2}+7\sqrt{2}x+4$:$6x^{2}+4\sqrt{2}x+3\sqrt{2}x+4 = 2\sqrt{2}x(\sqrt{3}x+2) + ...$ (Wait,let's simplify correctly).$6x^{2}+7\sqrt{2}x+4 = 6x^{2}+3\sqrt{2}x+4\sqrt{2}x+4 = 3\sqrt{2}x(\sqrt{2}x+1) + 4(\sqrt{2}x+1) = (3\sqrt{2}x+4)(\sqrt{2}x+1)$.Setting these factors to zero:$\sqrt{2}x+1 = 0 \implies x = -\frac{1}{\sqrt{2}}$.$3\sqrt{2}x+4 = 0 \implies x = -\frac{4}{3\sqrt{2}}$.Thus,the other two zeroes are $-\frac{1}{\sqrt{2}}$ and $-\frac{4}{3\sqrt{2}}$.

Given that $\sqrt{2}$ is a zero of the cubic polynomial $6x^{3}+\sqrt{2}x^{2}-10x-4\sqrt{2}$,find its other two zeroes.

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