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

Answers (1)

Answer. $\sqrt{5},\sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}$

Solution.

The given cubic polynomial is $x^3 -3 \sqrt{5}x^2 + 13x - 3\sqrt{5}$

Hence $x-\sqrt{5}$ is a factor

Use divided algorithm

$x^3 - 3\sqrt{3} x^2 + 13x - 3\sqrt{5}=(x-\sqrt{5}) (x^{2}-2\sqrt{5}+3)$

$=(x-\sqrt{5}) (x^{2}-2\sqrt{5}+3)$

$x^{2}-2\sqrt{5}+3$

a = 1, b = $-2\sqrt{5}$ , c = 3

$x=\frac{-b\pm \sqrt{b^{2}}-4ac}{2a}$

$x=\frac{2\sqrt{5}\pm \sqrt{20-12}}{2}$

$x=\frac{2\sqrt{5}\pm 2\sqrt{2}}{2}=\sqrt{5}\pm \sqrt{2}$

$x=\sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}$

Hence, $\sqrt{5},\sqrt{5}+\sqrt{2}, \sqrt{5}-\sqrt{2}$ are the zeroes of the polynomial.

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