Obtain all zeroes of the polynomial $f(x)\ =\ x^4\ –\ 3x^3\ –\ x^2\ +\ 9x\ –\ 6$, if the two of its zeroes are  $-\sqrt{3}$ and $\sqrt{3}$.

Given:

$f(x)\ =\ x^4\ –\ 3x^3\ –\ x^2\ +\ 9x\ –\ 6$ and the two of its zeroes are $-\sqrt{3}$ and $\sqrt{3}$.

To do:

We have to find all the zeros of $f(x)$.

Solution:

If $-\sqrt{3}$ and $\sqrt{3}$ are zeros of $f(x)$ then $(x+\sqrt3)(x-\sqrt3)$ is a factor of $f(x)$.

This implies,

$(x+\sqrt3)(x-\sqrt3)=x^2-(\sqrt3)^2=x^2-3$

Therefore,

Dividend$f(x)\ =\ x^4\ –\ 3x^3\ –\ x^2\ +\ 9x\ –\ 6$

Divisor$=x^2-3$

$x^2-3$)$x^4-3x^3-x^2+9x-6$($x^2-3x+2$

 Quotient$=x^2-3x+2$

$f(x)=(x^2-3)(x^2-3x+2)$

To find the other zeros put $x^2-3x+2=0$.

$x^2-3x+2=0$

$x^2-2x-x+2=0$

$x(x-2)-1(x-2)=0$

$(x-2)(x-1)=0$

$x-2=0$ and $x-1=0$

$x=2$ and $x=1$

All the zeros of $f(x)$ are $1$, $2$, $-\sqrt{3}$ and $\sqrt{3}$.

Tutorialspoint

Updated on: 10-Oct-2022

34 Views

Kickstart Your Career

Get certified by completing the course

Get Started