Find all zeros of the polynomial $2x^4\ -\ 9x^3\ +\ 5x^2\ +\ 3x\ -\ 1$, if two of its zeros are $2\ +\ \sqrt{3}$ and $2\ -\ \sqrt{3}$.

Given:

Given polynomial is $2x^4\ -\ 9x^3\ +\ 5x^2\ +\ 3x\ -\ 1$ and two of its zeroes are $2\ +\ \sqrt{3}$  and  $2\ -\ \sqrt{3}$.

To do:

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

Solution:

If $2\ +\ \sqrt{3}$ and $2\ -\ \sqrt{3}$ are zeros of the given polynomial then $(x-(2+\sqrt3))(x-(2-\sqrt3))$ is a factor of it.

This implies,

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

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

$=x^2-4x+(4-3)$

$=x^2-4x+1$

Therefore,

Dividend$=2x^4-9x^3+5x^2+3x-1$

Divisor$=x^2-4x+1$

$x^2-4x+1$)$2x^4-9x^3+5x^2+3x-1$($2x^2-x-1$

Quotient$=2x^2-x-1$

To find the other zeros put $2x^2-x-1$.

$2x^2-x-1=0$

$2x^2+2x-x-1=0$

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

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

$x+1=0$ or $2x-1=0$

$x=-1$ or $2x=1$

$x=-1$ or $x=\frac{1}{2}$

All the zeros of the given polynomial are  $2\ +\ \sqrt{3}$, $2\ -\ \sqrt{3}$, $-1$ and $\frac{1}{2}$.

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Updated on: 10-Oct-2022

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