Find all the zeroes of  2x^4 - 13x^3 + 19x^2 + 7x - 3, if you know that two of its zeroes are 2+\sqrt{3}  and 2-\sqrt{3} .

 

 

 

 
 
 
 
 

Answers (1)

Given that  2x^4 - 13x^3 + 19x^2 + 7x - 3

Two zeroes are  2 + \sqrt{3}  and  2 - \sqrt{3}.

The factor of the given polynomial :

\Rightarrow (x-(2+\sqrt{3}))(x-(x-\sqrt{3}))

\Rightarrow x^2-4x+1

Now divide the polynomial with their factor 

\Rightarrow \frac{2x^4 - 13 x^2 + 19 x^2 + 7x -3}{x^2-4x+1}

\Rightarrow 2x^2-5x-3

Now factor the above equation :

\Rightarrow 2x^2-6x+x-3

\Rightarrow 2x(x-3)+1(x-3)

\Rightarrow (x-3)(2x+1)

The zeroes are 

x-3=0\; \text{and} \; 2x+1=0

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

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