consider the equation (4x^4 - 8x^3 + 4x^2 - 128x - 960 = 0). the real solutions are (5) and (-3). what are the nonreal solutions?

(-2i, 2i)
(-4i, 4i)
(-8i, 8i)
(-16i, 16i)

Question

consider the equation (4x^4 - 8x^3 + 4x^2 - 128x - 960 = 0). the real solutions are (5) and (-3). what are the nonreal solutions?

(-2i, 2i)
(-4i, 4i)
(-8i, 8i)
(-16i, 16i)

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Polynomial Division", "Roots of Polynomials", "Quadratic Formula with Complex Roots" ], "new_concepts": [], "current_concepts": [ "Polynomial Division", "Roots of Polynomials", "Quadratic Formula with Complex Roots" ] } ## Simplify the polynomial equation $$ \begin{aligned} &4x^4 - 8x^3 + 4x^2 - 128x - 960 = 0 \ &x^4 - 2x^3 + x^2 - 32x - 240 = 0 \end{aligned} $$ ## Divide by the known real roots $$ \begin{aligned} &(x - 5)(x + 3) = x^2 - 2x - 15 \ &\frac{x^4 - 2x^3 + x^2 - 32x - 240}{x^2 - 2x - 15} = x^2 + 16 \end{aligned} $$ ## Solve for the nonreal roots $$ \begin{aligned} &x^2 + 16 = 0 \ &x^2 = -16 \ &x = \pm 4i \end{aligned} $$

(A) $-2i, 2i$

(B) $-4i, 4i$

(C) $-8i, 8i$

(D) $-16i, 16i$

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

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QUESTION IMAGE

Question

Explanation:

Simplify the polynomial equation

Divide by the known real roots

Solve for the nonreal roots

Answer: