Given that $\sqrt{2}$ is a zero of the cubic polynomial $6x^3\ +\ \sqrt{2}x^2\ -\ 10x\ -\ 4\sqrt{2}$, find its other two zeroes.

Given:

Given polynomial is $6x^3\ +\ \sqrt{2}x^2\ -\ 10x\ -\ 4\sqrt{2}$ and one of its zeros is $\sqrt2$.

To do:

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

Solution:

If $a$ is a zero of $f(x)$ then $(x-a)$ is a factor of $f(x)$.

Therefore,

$x-\sqrt{2}$ is a factor of the given polynomial.

On applying the division algorithm,

Dividend$=6x^3+\sqrt{2}x^2-10x-4\sqrt{2}$

Divisor$x-\sqrt{2}$

$x-\sqrt2$)$6x^3+\sqrt{2}x^2-10x-4\sqrt{2}$($6x^2+7\sqrt{2}x+4$

                    $6x^3-6\sqrt{2}x^2$

                  -----------------------------

                      $7\sqrt{2}x^2-10x-4\sqrt{2}$

                      $7\sqrt{2}x^2-14x$

                     -----------------------

                                         $4x-4\sqrt{2}$                                    

                                         $4x-4\sqrt{2}$

                                     ----------------------

                                                 $0$  

Therefore,

Quotient$=6x^2+7\sqrt{2}x+4$

$6x^3+\sqrt{2}x^2-10x-4\sqrt{2}=(x-\sqrt{2})(6x^2+7\sqrt{2}x+4)$

To get the other zeros, put $6x^2+7\sqrt{2}x+4=0$.

$6x^2+7\sqrt{2}x+4=0$

$6x^2+3\sqrt{2}x+4\sqrt{2}x+4=0$

$3x(2x+\sqrt2)+2\sqrt2(2x+\sqrt2)=0$

$(3x+2\sqrt2)(2x+\sqrt2)=0$

$3x+2\sqrt2=0$ or $2x+\sqrt2=0$

$3x=-2\sqrt2$ or $2x=-\sqrt2$

$x=-\frac{2\sqrt2}{3}$ or $x=-\frac{\sqrt2}{2}$

All the zeros of the given polynomial are $\sqrt{2}$, $-\frac{2\sqrt2}{3}$ and $-\frac{\sqrt2}{2}$.

Tutorialspoint

Updated on: 10-Oct-2022

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