Question

Find all zeroes of the polynomial (2x⁴-9x³ + 5x ² + 3x - 1) if
two of its zeroes are (2+√3) and (2-√3)​

Answer 1

Answer:

1 and - 1/2

Step-by-step explanation:

$2 {x}^{4} - 9 {x}^{3} + 5 {x}^{2} + 3x - 1 = 0 \\ (x - 1)((x + \frac{1}{2} )(x - ( 2 + \sqrt{3} )(x - (2 - \sqrt{3} ) = 0 \\ x = 1or \: x = \frac{ - 1}{2} and \: two \: are \: given \: in \: question \:$

Answer 2

Given that :

x=(2+√3) and ,X=(2-√3)

now ,

{(x-2)-√3} × {(x-2)+√3}

hence, 

{ x -(2 -√3)}{x -(2+√3)} is a factor of given polynomial . 

{ x² -(2+√3)x -(2-√3)x +(2-√3)(2+√3)} is a factor of given polynomial .

{ x²-(4)x + 1} is a factor of given polynomial .

hence, x²-4x +1 is divisible by given polynomial

Now,

x² -4x +1 ) 2x⁴ -9x³+ 5x² +3x -1( 2x²-x -1

2x⁴ -8x³ +2x² 

_______________

-x³ +3x² +3x 

-x³ +4x² -x

______________

-x² + 4x -1 

-x² +4x -1 

____________

0000

Then,

( 2x² -x -1) is a factor of the two polynomial

 

now, 

2x² -x -1 =0

2x² -2x +x -1 =0

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

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

x = -1/2 and 1 

So,

-1/2 and 1 are two roots of given polynomial .

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