Given that x-sqrt{5} is a factor of the cubic | vedclass

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(D) Let $f(x) = x^{3}-3 \sqrt{5} x^{2}+13 x-3 \sqrt{5}$.Since $(x-\sqrt{5})$ is a factor,we divide $f(x)$ by $(x-\sqrt{5})$ to find the quotient.Perfo

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$\sqrt{5}, (\sqrt{5}+\sqrt{2})$ and $(\sqrt{5}-\sqrt{2})$

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(D) Let $f(x) = x^{3}-3 \sqrt{5} x^{2}+13 x-3 \sqrt{5}$.Since $(x-\sqrt{5})$ is a factor,we divide $f(x)$ by $(x-\sqrt{5})$ to find the quotient.Performing polynomial long division:$x^{3}-3 \sqrt{5} x^{2}+13 x-3 \sqrt{5} = (x-\sqrt{5})(x^{2}-2 \sqrt{5} x+3)$.Now,we find the roots of the quadratic factor $x^{2}-2 \sqrt{5} x+3 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$:$x = \frac{2 \sqrt{5} \pm \sqrt{(-2 \sqrt{5})^{2}-4(1)(3)}}{2(1)}$$x = \frac{2 \sqrt{5} \pm \sqrt{20-12}}{2} = \frac{2 \sqrt{5} \pm \sqrt{8}}{2} = \frac{2 \sqrt{5} \pm 2 \sqrt{2}}{2} = \sqrt{5} \pm \sqrt{2}$.Thus,the zeroes are $\sqrt{5}, (\sqrt{5}+\sqrt{2})$,and $(\sqrt{5}-\sqrt{2})$.

Given that $x-\sqrt{5}$ is a factor of the cubic polynomial $x^{3}-3 \sqrt{5} x^{2}+13 x-3 \sqrt{5}$. Find all the zeroes of the polynomial.

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