Question

Find all the zeroes of the polynomial 2x4-9x3+5x2+3x-1,if two of its zeroes are 2+underroot3 and 2-root3

Answer 1

...........................................................................................................................Good evening, the answer of this question is here............................................
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p(x) = 2x4-9x³+5x²+3x-1
α = 2+√3
β = 2-√3
α+β = 2+√3+2-√3
α+β = 4
αβ = (2+√3)(2-√3)
αβ = (2)²-(√3)²
αβ = 4 - 3
αβ = 1
quad equation 
g(x) = x²-(α+β)x+αβ
g(x) = x²-4x+1
p(x)÷g(x)
2x4-9x³+5x²+3x-1/x²-4x+1
(x-2+√3) (x-2-√3) (2x²-x+1)
(x-2+√3) (x-2-√3) [2x²-(2-1)x+1]
(x-2+√3) (x-2-√3) (2x²-2x-x+1)
(x-2+√3) (x-2-√3) [2x(x-1)-1(x-1)]
(x-2+√3) (x-2-√3) (x-1)(2x-1)
So, other zeroes are 
α = x-1 = 0
α = 1
β = 2x-1= 0
β =1/2
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Thank u and I think i helped in you this question. Good Night...........................
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Answer 2

Answer:

All zeroes of p ( x ) are 1 , - 1 / 2 , 2 + √ 3 and 2 - √ 3

Step-by-step explanation:

Given :

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

Two zeroes of p ( x ) :

2 + √ 3 and 2 - √ 3

p ( x ) = ( x - 2 - √ 3 ) ( x - 2 + √ 3 ) × g ( x )

g ( x ) = p ( x ) ÷ ( x² - 4 x + 1 )

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

$\displaystyle{\begin{tabular}{c | c | c}(x^2-4x+1) & \overline{2x^4-9x^3+5x^2+3x-1}} & 2x^2-x-1\\ & 2x^4-8x^3+2x^2 \ \ \ \ \ \ \ \ \ \ \ \\ \\ & ( - ) ( + ) ( - ) \ \ \ \ \ \ \ \\ \\ & \overline{-x^3+3x^2+3x} \\ \\ & - x^3+4x^2-x\\ \\ & ( + ) ( - ) ( + ) \\ \\ & \overline{-x^2 + 4x-1} \\ \\ & -x^2 + 4x-1 \\ \\ & ( + ) ( - ) ( + ) \\ \\ & \overline{0}\endtabular$

We got ,

g ( x ) = 2 x² - x - 1

By splitting mid term

2 x² - 2 x + x - 1

2 x ( x - 1 ) + ( x - 1 )

( x - 1 ) ( 2 x + 1 )

Now zeroes of g ( x )

x = 1 or x = - 1 / 2 .