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Given that is $x-\sqrt{5}$ a factor of the polynomial $x^{3} -3\sqrt{5} x^{2} -5x+15\sqrt{5}$ , find all the zeroes of the polynomials.
Given: $x-\sqrt{5}$ is a factor of the polynomial $x^{3} -3\sqrt{5} x^{2} -5x+15\sqrt{5}$.
To do: To find all the zeroes of the given polynomial.
Solution:
Let $ P( x) =x^{3} -3\sqrt{5} x^{2} -5x+15\sqrt{5}$.
$\because$ $x-\sqrt{5}$ is a factor of the given polynomial.
Now we would substitute $x=-\sqrt{5}$
$P\left( -\sqrt{5}\right) =\left( -\sqrt{5}\right)^{3} -3\sqrt{5}\left( -\sqrt{5}\right)^{2} -5\left( -\sqrt{5}\right) +15\sqrt{5}$
$=-5\sqrt{5} -15\sqrt{5} +5\sqrt{5} +15\sqrt{5}$
$=0$
$\therefore \ \left( x+\sqrt{5}\right) \ $ is a factor of $P( x) =$
$\therefore \ \left( x+\sqrt{5}\right)\left( x-\sqrt{5}\right) =x^{2} -5$ is a factor of $P( x)$
On dividing the given polynomial by $x^{2} -5$.
$\therefore \ x-3\sqrt{5} \ $ is a factor of the given polynomial.
$\therefore \ x-3\sqrt{5} =0$
$\Rightarrow x=3\sqrt{5}$
$\therefore \sqrt{5} ,\ -\sqrt{5} \ $ and $3\sqrt{5}$ are the zeroes of the given polynomial.
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