Write the equation of lowest degree with real coefficients if two of its roots are -1 and 1+i.

Question

The answer is x^3-x^2+2=0. I don't understand why.

Answer 1

complex roots always come in conjugate pairs, so if one is 1+i, you must have another as 1-i

so we know the equation is
(x+1)(x - (1+i))(x + 1+i) = 0
(x+1)(x-1-i)(x+1+i) = 0
(x+1)(x^2 + x + xi -x -x -i -xi - i - x^2) = 0
(x+1)(x^2 - 2x + 2) = 0
x^3 - x^2 + 2x + x^2 - 2x + 2 = 0
x^3 - x^2 + 2=0

Answer 2

I did this on paper first, but then messed up when I typed it.

the first few lines in the equations should have been:
(x+1)(x - (1+i))(x - (1-i)) = 0
(x+1)(x-1-i)(x-1+i) = 0
(x+1)(x^2 - x + xi -x +1 -i -xi + i - i^2) = 0
the rest is fine