Find the other zeroes of the polynomial 2x²-11x³+13x-7, it being given that two of its zeroes are (3+√2) and (3-√2).
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If α and β are zeroes of the polynomial then x2 -(α+ β)x + αβ .
α = (3+√2) and β = (3 - √2).
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x3 - 7x2 + 13x
x3- 6x2 + 7x (substract)
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- x2 + 6x - 7
- x2 + 6x - 7 (substract)
-----------------------------
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)
---------------------------------
- 6x2 + 7x
- 6x2 + 3x (substract)
---------------------------------
4x - 2
4x - 2 (substract)
---------------------------------
0
∴ g(x) = 2x2 - 3x + 2.
hope it helps............chill
all d best for board exam
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
-----------------------------
x3 - 7x2 + 13x
x3- 6x2 + 7x (substract)
--------------------------------------
- x2 + 6x - 7
- x2 + 6x - 7 (substract)
-----------------------------
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)
---------------------------------
- 6x2 + 7x
- 6x2 + 3x (substract)
---------------------------------
4x - 2
4x - 2 (substract)
---------------------------------
0
∴ g(x) = 2x2 - 3x + 2.
hope it helps............chill
all d best for board exam
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