Multiple choice

The sum of non-real roots of the polynomial equation $x^3 + 3x^2 + 3x + 3 =0$ :

  1. Equal $0$
  2. Lies between $0$ and $1$
  3. Lies between $-1$ and $0$
  4. Has absolute value bigger than $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation is x^3 + 3x^2 + 3x + 3 = 0, which is (x+1)^3 + 2 = 0. So (x+1)^3 = -2. The roots are x = -1 + cube_root(-2), -1 + cube_root(-2) * omega, -1 + cube_root(-2) * omega^2. The real root is -1 - 2^(1/3). The sum of all roots is -3. The sum of non-real roots is -3 - (-1 - 2^(1/3)) = -2 + 2^(1/3). Since 2^(1/3) is approx 1.26, the sum is approx -0.74, which lies between -1 and 0.

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