Question 1072115
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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

x&#8308;-2x³+2x²-2x³+4x²-4x = 0

x&#8308;-4x³+6x²-4x = 0

P(x) = x&#8308;-4x³+6x²+4x²-4x

Edwin</b></pre></font>