Find all zeros of the polynomial $2x^4\ +\ 7x^3\ -\ 19x^2\ -\ 14x\ +\ 30$, if two of its zeros are $\sqrt{2}$ and $-\sqrt{2}$.

Given:

Given polynomial is $2x^4\ +\ 7x^3\ -\ 19x^2\ -\ 14x\ +\ 30$ 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 $(x+\sqrt{2})(x-\sqrt{2})$ is a factor of it.

This implies,

$(x-\sqrt2)(x+\sqrt2)=x^2-(\sqrt{2})^2=x^2-2$

Therefore,

Dividend$=2x^4+7x^3-19x^2-14x+30$

Divisor$=x^2-2$

$x^2-2$)$2x^4+7x^3-19x^2-14x+30$($2x^2+7x-15$

Quotient$=2x^2+7x-15$

To find the other zeros, put $2x^2+7x-15$.

$2x^2+10x-3x-15=0$

$2x(x+5)-3(x+5)=0$

$(x+5)(2x-3)=0$

$x+5=0$ or $2x-3=0$

$x=-5$ or $2x=3$

$x=-5$ or $x=\frac{3}{2}$

All the zeros of the given polynomial are $-\sqrt{2}$, $\sqrt{2}$, $-5$ and $\frac{3}{2}$.

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Updated on: 10-Oct-2022

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