Question

the equation $x^3 = 1$ has one real and two complex roots. the product of the complex roots is:
a. $i$ b. $3 + i$ c. $1$ d. $-1$ e. none of these

#3 *
given that $2i$ is a root of $x^3 + 3x^2 + 4x + 12 = 0$, find the other roots and identify the one true statement.

Explanation:

Step1: Rewrite equation to standard form

$x^3 - 1 = 0$

Step2: Factor using difference of cubes

$(x-1)(x^2+x+1)=0$

Step3: Identify complex roots' quadratic

The complex roots come from $x^2+x+1=0$

Step4: Use product of roots formula

For $ax^2+bx+c=0$, product is $\frac{c}{a}$. Here, $\frac{1}{1}=1$

Step1: Use conjugate root theorem

Since $2i$ is a root, $-2i$ is also a root.

Step2: Form quadratic factor from roots

$(x-2i)(x+2i)=x^2+4$

Step3: Divide cubic by quadratic factor

$\frac{x^3+3x^2+4x+12}{x^2+4}=x+3$

Step4: Find remaining root

Set $x+3=0$, so $x=-3$

Step5: List all roots

Roots are $2i$, $-2i$, $-3$

Answer:

C. 1

---