# 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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Given:

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

To do:

We have to find all the zeros of the given polynomial.

Solution:

$x-\sqrt{5}$ is a factor of the given polynomial.

On applying the division algorithm,

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

Divisor$=x-\sqrt{5}$

$x-\sqrt5$)$x^3-3\sqrt{5}x^2+13x-3\sqrt{5}$($x^2-2\sqrt{5}x+3$

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

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

$-2\sqrt{5}x^2+10x$

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$3x-3\sqrt{5}$

$3x-3\sqrt{5}$

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$0$

Therefore,

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

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

To get the other zeros, put $x^2-2\sqrt{5}x+3=0$.

$x^2-2\sqrt{5}x+3=0$

$x=\frac{-( -2\sqrt{5}) \pm \sqrt{( -2\sqrt{5})^{2}-4( 1)( 3)}}{2( 1)}$

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

$x=\frac{2\sqrt{5} \pm \sqrt{8}}{2}$

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

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

All the zeros of the given polynomial are $\sqrt{5}$, $\sqrt{5}+\sqrt{2}$ and $\sqrt{5}-\sqrt{2}$.

Updated on 10-Oct-2022 13:27:09