find the cubic polynomoal whose three zeroes are 1/1,1,-2
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Hope it helps...................................
For cubic polynomials we have the formula =
x^3 -(alpha +beeta)x^2 + (alpha×beeta +beeta ×gama+gama ×alpha)x - (alpha ×beeta ×gama)
Now put the values of the zeroes respectively.
x^3 - 1/1x^2 +1x -(-2)
Now solve it .
x^3 -1x^2 +x +2
x^3 - x^2+x+2 is your answer....
Please mark my answer as brainlist....
For cubic polynomials we have the formula =
x^3 -(alpha +beeta)x^2 + (alpha×beeta +beeta ×gama+gama ×alpha)x - (alpha ×beeta ×gama)
Now put the values of the zeroes respectively.
x^3 - 1/1x^2 +1x -(-2)
Now solve it .
x^3 -1x^2 +x +2
x^3 - x^2+x+2 is your answer....
Please mark my answer as brainlist....
Anonymous:
Hey Ridhi , have a look at your answer once.There is a mistake .
No there is not
the formula of cubic polynomial is x³ - ( α + ß + Γ ) x² + ( αß + ßΓ + αΓ )x - αßΓ
i have written that only
- + - = +
so in the last ans changed
You haven't written gama in the coefficient of x62
sorry , x^2
x^3 - x^2+x+2........
ok
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