SOLUTION: I need to know what the polynomial of 2,1-i,2-√3 is in standard form it has a leading coefficient of 1.

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Question 1196595: I need to know what the polynomial of 2,1-i,2-√3 is in standard form it has a leading coefficient of 1.
Found 3 solutions by MathLover1, MathTherapy, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

given roots:
x%5B1%5D=2
x%5B2%5D=1-i -> complex roots always come in pairs, you also have x%5B3%5D=1%2Bi
x%5B4%5D=2-sqrt%283%29

so, we have 4th degree polynomial

f%28x%29=%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B3%5D%29%28x-x%5B4%5D%29

f%28x%29=%28x-2%29%28x-%281-i%29%29%28x-%281%2Bi%29%29%28x-%282-sqrt%283%29%29%29

f%28x%29=%28x-2%29%28x-1%2Bi%29%28x-1-i%29%28x-2%2Bsqrt%283%29%29..........multiply all and you will get







Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
I need to know what the polynomial of 2,1-i,2-√3 is in standard form it has a leading coefficient of 1.
This is NOT a 4th degree polynomial, but a 5th degree one.



    Multiply that out!

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.

The problem's formulation in the post is incorrect.

A typical  (and a standard)  correct formulation would be as follows:

        Find a polynomial of the lowest degree with integer coefficients
        and the leading coefficient equal to 1,  if it has the roots  2,  1-i,  2-√3.


With this corrected formulation,  the  " solution "  by  @MathLover1  is  IRRELEVANT,
and the solution by  @MathTherapy  is correct.


So,  first  FAULT  is due to the person who posted incorrect formulation.
Next fault is due to  @MathLover1.


"I sat on the bank of the river and waited for the right solution to come."