SOLUTION: write a polynomial P with the lowest possible degree that has the given solutions 0,2,1+i,1-i

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Question 1072115: write a polynomial P with the lowest possible degree that has the given solutions
0,2,1+i,1-i
Found 2 solutions by josgarithmetic, Edwin McCravy:
Answer by josgarithmetic(39838) About Me  (Show Source):
You can put this solution on YOUR website!
x%28x-2%29%28x-%281%2Bi%29%29%28x-%281-i%29%29

-
x%28x-2%29%28x-1-i%29%28x-1%2Bi%29
x%28x-2%29%28%28x-1%29-i%29%28%28x-1%29%2Bi%29
x%28x-2%29%28%28x-1%29%5E2-i%5E2%29
x%28x-2%29%28x%5E2-2x%2B1%2B1%29
x%28x-2%29%28x%5E2-2x%2B2%29
or further steps in multiplying if wanted

Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!


x = 0;    x = 2;      x = 1+1;   x = 1-i

Get zero on the right, on the last three:

x = 0;  x-2 = 0;  x-1-i = 0; x-1+i = 0

Multiply all four left sides together 

x(x-2)(x-1-i)(x-1+i)

and set it equal to all four right sides
multiplied together: (0)(0)(0)(0) = 0

x(x-2)(x-1-i)(x-1+i) = 0

(x²-2x)[(x-1)-i][(x-1+i] = 0

(x²-2x)[(x-1)²-i²] = 0

(x²-2x)[x²-2x+1-(-1)] = 0

(x²-2x)[x²-2x+1+1] = 0

(x²-2x)[x²-2x+2] = 0

x⁴-2x³+2x²-2x³+4x²-4x = 0

x⁴-4x³+6x²-4x = 0

P(x) = x⁴-4x³+6x²+4x²-4x

Edwin