If the real roots of the equation [tex]\( 3x^4 - 6x^3 + 3x^2 - 54x - 216 = 0 \)[/tex] are -2 and 4, what are the nonreal roots?

A. [tex]\(-3i, 3i\)[/tex]
B. [tex]\(-3i\sqrt{3}, 3i\sqrt{3}\)[/tex]
C. [tex]\(-9i, 9i\)[/tex]
D. [tex]\(-6i\sqrt{6}, 6i\sqrt{6}\)[/tex]



Answer :

To determine the nonreal roots of the polynomial [tex]\(3x^4 - 6x^3 + 3x^2 - 54x - 216 = 0\)[/tex] given that its real roots are -2 and 4, we can proceed with the following steps:

1. Recognize that the polynomial can be factored using its real roots. Thus, the polynomial can be expressed as follows:
[tex]\[ (x + 2)(x - 4)(\text{quadratic polynomial with nonreal roots}) \][/tex]

2. Assume the quadratic factor has nonreal roots. Let the quadratic polynomial have the form:
[tex]\[ ax^2 + bx + c \][/tex]

3. Multiply the factors:
[tex]\[ 3(x + 2)(x - 4)(ax^2 + bx + c) \][/tex]

4. To find the quadratic polynomial, we compare the product coefficients with those of the original polynomial [tex]\(3x^4 - 6x^3 + 3x^2 - 54x - 216\)[/tex].

5. The quadratic polynomial will yield complex conjugate pairs since the coefficients of the polynomial are real numbers.

6. The nonreal roots of this polynomial turn out to be:
[tex]\[ -3i \quad \text{and} \quad 3i \][/tex]

Thus, the nonreal roots of the given polynomial [tex]\(3x^4 - 6x^3 + 3x^2 - 54x - 216 = 0\)[/tex] are:
[tex]\[ (-2.7755575615628914e-17 + 2.9999999999999987i) \quad \text{and} \quad (-2.7755575615628914e-17 - 2.9999999999999987i) \][/tex]
which are approximately:
[tex]\[ 3i \quad \text{and} \quad -3i \][/tex]

Therefore, the correct nonreal roots are:
[tex]\[ -3i \quad \text{and} \quad 3i \][/tex]

Hence, the correct answer is:
[tex]\[ -3i, 3i \][/tex]