please answer this question correctly...please
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divide the polynomial by (x-2+3^0.5)or divide it by (x-2-3^0.5)
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Given f(x) = x^4 - 6x^3 - 26x^2 + 138x - 35.
Given roots 2 + root 3 and 2 - root 3.
.
Therefore x^2 - 4x + 1 is a factor of f(x).
x^2 - 2x - 35
--------------------------------------------------------
x^2 - 4x + 1) x^4 - 6x^3 - 26x^2 + 138x - 35
x^4 - 4x^3 + x^2
--------------------------------------------------------
- 2x^3 - 27x^2 + 138x
-2x^3 + 8x^2 - 2x
-----------------------------------------------------------
- 35x^2 + 140x - 35
- 35x^2 + 140x - 35
-------------------------------------------------------------
0.
Now,
x^2 - 2x - 35 = 0
x^2 - 7x + 5x - 35 = 0
x(x - 7) + 5(x- 7) = 0
(x + 5)(x - 7) = 0
x = -5,7.
Therefore the other zeroes are -5 and 7.
Hope this helps!
Given roots 2 + root 3 and 2 - root 3.
.
Therefore x^2 - 4x + 1 is a factor of f(x).
x^2 - 2x - 35
--------------------------------------------------------
x^2 - 4x + 1) x^4 - 6x^3 - 26x^2 + 138x - 35
x^4 - 4x^3 + x^2
--------------------------------------------------------
- 2x^3 - 27x^2 + 138x
-2x^3 + 8x^2 - 2x
-----------------------------------------------------------
- 35x^2 + 140x - 35
- 35x^2 + 140x - 35
-------------------------------------------------------------
0.
Now,
x^2 - 2x - 35 = 0
x^2 - 7x + 5x - 35 = 0
x(x - 7) + 5(x- 7) = 0
(x + 5)(x - 7) = 0
x = -5,7.
Therefore the other zeroes are -5 and 7.
Hope this helps!
siddhartharao77:
:-)
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