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Question 1172125: Write the equation of 5th degree polynomial in standard form given that 3 is the only real zero and 2+i is an imaginary zero.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Unfortunately there isn't enough information to form a 5th degree polynomial.
This is because we have 1 real root, and 2 complex roots (2+i and 2-i). In total we have 1+2 = 3 roots.
We would need to have five roots to form a 5th degree polynomial.
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We could form a cubic (3rd degree polynomial) like so
x = 3 or x = 2+i or x = 2-i
x-3 = 0 or x-2 = i or x-2 = -i
x-3 = 0 or (x-2)^2 = i^2
x-3 = 0 or (x-2)^2 = -1
x-3 = 0 or (x-2)^2+1 = 0
(x-3)((x-2)^2+1) = 0
(x-3)(x^2-4x+4+1) = 0
(x-3)(x^2-4x+5) = 0
x(x^2-4x+5)-3(x^2-4x+5) = 0
x^3-4x^2+5x-3x^2+12x-15 = 0
x^3-7x^2+17x-15 = 0
The cubic function
f(x) = x^3-7x^2+17x-15
has the three roots
x = 3, x = 2+i, x = 2-i
So once again we're 2 roots short of forming a 5th degree polynomial.
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