obtain all zeros of the polynomial 2x4-11x3+7x2+13x-7 , it begins given that the two of its zeroes are 3+√2 and 3-√2.
Answers
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The given polynomial is f(x) = 2x^4 - 11x^3 + 7x^2 + 13x - 7
since , (3+√3) and (3-√2) is the zeroes of f(x)
it is follows that x - 3+√3 and x - 3- √2 are the factor of f ( x).
Dividing p(x) by x^2 - 6x +7
we , get remainder is 0
=> f (x) =0
=> (x^2 - 6x + 7)(2x^2 + x - 1)
=> (x + 3 + √2)(x + 3 - √2)(2x-1)(2x+1)=0
=>x = -3 -√2. , x = -3+√2 or x =1/2 or x = -1
HENCE , the zeroes of the given polynomial are ( -3-√2) , (-3+√2) , 1/2 and -1
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Sol:
If α and β are zeroes of the polynomial then x2 -(α+ β)x + αβ .
α = (3+√2) and β = (3 - √2).
α+ β = 6 and αβ = 7.
∴ x2 - 6x + 7.
Given polynomial 2x4-11x3+ 7x2 + 13x - 7.
x2 - 6x + 7 ) 2x4-11x3+ 7x2 + 13x - 7 ( 2x2 + x -1
2x4-12x3+14x2 (substract)
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x3 - 7x2 + 13x
x3- 6x2 + 7x (substract)
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- x2 + 6x - 7
- x2 + 6x - 7 (substract)
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0
∴ The Quotient is 2x2 + x -1
= 2x2 + 2x - x -1
= 2x(x + 1)-1( x + 1)
= ( 2x -1)( x +1).
∴ x = 1/2 , -1 are the other zeros of the polynomial.
2)
Using remainder theorem f(x) = g(x) (2x-1) + (x+3)
4x3-8x2 + 8x + 1 = g(x) (2x-1) + (x+3)
4x3-8x2 + 7x - 2 = g(x) (2x-1).
g(x) = (4x3-8x2 + 7x - 2) / (2x -1).
2x -1 ) 4x3-8x2 + 7x - 2 ( 2x2 - 3x + 2
4x3-2x2 (substract)
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- 6x2 + 7x
- 6x2 + 3x (substract)
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4x - 2
4x - 2 (substract)
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0
∴ g(x) = 2x2 - 3x + 2.