find all solutions to the equation (3x^3 + 12x = 2x^2 + 8).

* (-2, -\frac{2}{3}, 2)
* (-2, \frac{2}{3}, 2)
* (-\frac{2}{3}, -2i, 2i)
* (\frac{2}{3}, -2i, 2i)

Question

find all solutions to the equation (3x^3 + 12x = 2x^2 + 8).

* (-2, -\frac{2}{3}, 2)
* (-2, \frac{2}{3}, 2)
* (-\frac{2}{3}, -2i, 2i)
* (\frac{2}{3}, -2i, 2i)

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Roots of Polynomials", "Complex Roots of Polynomials" ], "new_concepts": [], "current_concepts": [ "Roots of Polynomials", "Complex Roots of Polynomials", "Polynomial Factoring" ] } ## Rearrange the polynomial equation $$ 3x^3 - 2x^2 + 12x - 8 = 0 $$ ## Factor by grouping $$ \begin{aligned} x^2(3x - 2) + 4(3x - 2) &= 0 \ (3x - 2)(x^2 + 4) &= 0 \end{aligned} $$ ## Solve for x $$ \begin{aligned} 3x - 2 = 0 &\implies x = \frac{2}{3} \ x^2 + 4 = 0 &\implies x^2 = -4 \implies x = \pm 2i \end{aligned} $$

(A) $-2, -\frac{2}{3}, 2$

(B) $-2, \frac{2}{3}, 2$

(C) $-\frac{2}{3}, -2i, 2i$

(D) $\frac{2}{3}, -2i, 2i$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Complex Roots of Polynomials" ] }

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Question

Explanation:

Rearrange the polynomial equation

Factor by grouping

Solve for x

Answer: