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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.
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$=x-\sqrt{5}$
$x-\sqrt5$)$x^3-3\sqrt{5}x^2+13x-3\sqrt{5}$($x^2-2\sqrt{5}x+3$
$x^3-\sqrt{5}x^2$
-----------------------------------------
$-2\sqrt{5}x^2+13x-3\sqrt{5}$
$-2\sqrt{5}x^2+10x$
---------------------------------
$3x-3\sqrt{5}$
$3x-3\sqrt{5}$
----------------------
$0$
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 $\sqrt{5}-\sqrt{2}$.
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