Find all zeros of the polynomial $x^3\ +\ 3x^2\ -\ 2x\ -\ 6$, if two of its zeros are $-\sqrt{2}$ and $\sqrt{2}$.

Given:

Given polynomial is $x^3\ +\ 3x^2\ -\ 2x\ -\ 6$ and two of its zeroes are $-\sqrt{2}$  and  $\sqrt{2}$.

To do:

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

Solution:

If $-\sqrt{2}$  and  $\sqrt{2}$ are zeros of the given polynomial then $(x-\sqrt2)(x+\sqrt2)$ is a factor of it.

This implies,

$(x-\sqrt2)(x+\sqrt2)=x^2-(\sqrt2)^2=x^2-2$

Therefore,

Dividend$=x^3+3x^2-2x-6$

Divisor$=x^2-2$

$x^2-2$)$x^3+3x^2-2x-6$($x+3$

Quotient$=x+3$

To find the other zeros put $x+3$.

$x+3=0$

$x=-3$

All the zeros of the given polynomial are $-\sqrt2$, $\sqrt2$ and $-3$. 

Tutorialspoint

Updated on: 10-Oct-2022

57 Views

Kickstart Your Career

Get certified by completing the course

Get Started