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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Updated on: 10-Oct-2022

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