SOLUTION: Write a polynomial function with rational coefficients in standard form with the given zeros. 2-i, square root of 5

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Question 940564: Write a polynomial function with rational coefficients in standard form with the given zeros. 2-i, square root of 5
Found 2 solutions by MathLover1, KMST:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
given zeros:
2-i, if given 2-i there must be 2%2Bi too because complex roots always come in pairs
sqrt%285%29
to find a polynomial function use zero product rule:
%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B2%5D%29=0 plug in given zeros

%28x-%282-i%29%29%28x-%282%2Bi%29%29%28x-sqrt%285%29%29=0

%28x-2%2Bi%29%28x-2-i%29%28x-sqrt%285%29%29=0

%28x%5E2-2x-xi-2x%2B4%2B2i%2Bxi-2i-i%5E2%29%28x-sqrt%285%29%29=0



%28x%5E2-4x%2B4-%28-1%29%29%28x-sqrt%285%29%29=0

%28x%5E2-4x%2B4%2B1%29%28x-sqrt%285%29%29=0

%28x%5E2-4x%2B5%29%28x-sqrt%285%29%29=0

x%5E3-4x%5E2-x%5E2%2Asqrt%285%29%2B4xsqrt%285%29%2B5x-5sqrt%285%29=0

x%5E3-%284%2Bsqrt%285%29%29x%5E2%2B%284sqrt%285%29%2B5%29x-5sqrt%285%29=0







Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If the polynomial must have real coefficients ,
having an imaginary root requires
also having the conjugate imaginary number as a root.
If the polynomial has an irrational root,
to have rational coefficients,
it must also have the conjugate irrational number as a root.
So your polynomial must have as roots
2-i , 2%2Bi , sqrt%285%29 , and -sqrt%285%29 .
The simplest one would be

f%28x%29=%28x-2%2Bi%29%2A%28x-2-i%29%2A%28x%5E2-5%29
f%28x%29=%28%28x-2%29%5E2-i%5E2%29%2A%28x%5E2-5%29
f%28x%29=%28x%5E2-4x%2B4%2B1%29%2A%28x%5E2-5%29
f%28x%29=%28x%5E2-4x%2B5%29%2A%28x%5E2-5%29
f%28x%29=x%5E4-4x%5E3%2B20x-25