Multiple choice

If $x=1+i$ is a roots of the equation $x^3-ix+1-i=0$, then the other real root is

  1. $0$
  2. $1$
  3. $-1$

  4. None of the above.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Dividing the cubic by the known factor x - (1 + i) leaves x^2 + (1 + i)x + i. This factors as (x + 1)(x + i), so the other real root is -1.

AI explanation

If x equals 1 plus i is a root, substitute x equals 1 into the polynomial to check for real roots, yielding 1 cubed minus i times 1 plus 1 minus i, which does not equal zero. Substitute x equals negative 1 into the polynomial to get negative 1 cubed minus i times negative 1 plus 1 minus i, which equals negative 1 plus i plus 1 minus i, giving exactly zero. Therefore, the other real root is negative 1.

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