Find all zeros of the polynomial $2x^3\ +\ x^2\ -\ 6x\ -\ 3$, if two of its zeros are $-\sqrt{3}$ and $\sqrt{3}$.
Given:
Given polynomial is $2x^3\ +\ x^2\ -\ 6x\ -\ 3$ and two of its zeroes are $-\sqrt3$ and $\sqrt3$.
To do:
We have to find all the zeros of the given polynomial.
Solution:
If $-\sqrt3$ and $\sqrt3$ are zeros of the given polynomial then $(x-\sqrt3)(x+\sqrt3)$ is a factor of it.
This implies,
$(x-\sqrt3)(x+\sqrt3)=x^2-(\sqrt{3})^2=x^2-3$
Therefore,
Dividend$=2x^3+x^2-6x-3$
Divisor$=x^2-3$
$x^2-3$)$2x^3+x^2-6x-3$($2x+1$
$2x^3 -6x$
----------------------
$x^2-3$
$x^2-3$
---------------
$0$
Quotient$=2x+1$
To find the other zero put $2x+1$.
$2x+1=0$
$2x=-1$
$x=-\frac{1}{2}$
All the zeros of the given polynomial are $-\sqrt3$, $\sqrt3$ and $-\frac{1}{2}$.
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