Question

find all the zeros of the polynomial 2x4-5x3-12x2+11x-2 if two of it's zeroes are (2+root3)

Answer 1

here is your right answer.

Answer 2

Answer:

-2 , $\frac{-1}{2}$ , $2+\sqrt{3}$ and $2-\sqrt{3}$.

Step-by-step explanation:

Dividend = $2x^4-5x^3-12x^2+11x-2$

Since we are given that  two of it's zeroes are $2+\sqrt{3}$ and $2-\sqrt{3}$

So, $(x - 2-\sqrt{3}) ( x - 2 + \sqrt{3} )$

Using identity $(x+y)(x-y)=x^2-y^2$

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

$x^2 + 1 - 4x$

Now

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

Since we know that :

$Dividend =(Divisor \times Quotient)+Remainder$

$2x^4-5x^3-12x^2+11x-2=(x^2 + 1 - 4x \times 2x^2)+(3x^3-14x^2+11x-2)$

$2x^4-5x^3-12x^2+11x-2=(x^2 + 1 - 4x \times 2x^2+3x)+(-2x^2+8x-2)$

$2x^4-5x^3-12x^2+11x-2=(x^2 + 1 - 4x \times 2x^2+3x-2)+0$

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

Factorize the quotient

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

$2x^2+4x-x-2 =0$

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

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

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

Hence all the zeros of the polynomial $2x^4-5x^3-12x^2+11x-2$ is -2 , $\frac{-1}{2}$ , $2+\sqrt{3}$ and $2-\sqrt{3}$.