Question

find all solutions to the equation \\(x^3 + 16x - 3x^2 = 48\\).

-4, -3, 4
-4, 3, 4
3, -4i, 4i
-3, -4i, 4i

Explanation:

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<reasoning>

Rearrange the polynomial equation

\[
x^3 - 3x^2 + 16x - 48 = 0
\]

Factor by grouping

\[

$$\begin{aligned} x^2(x - 3) + 16(x - 3) &= 0 \\ (x^2 + 16)(x - 3) &= 0 \end{aligned}$$

\]

Solve for all roots

\[

$$\begin{aligned} x - 3 = 0 &\implies x = 3 \\ x^2 + 16 = 0 &\implies x^2 = -16 \implies x = \pm 4i \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \(-4, -3, 4\)</mcq-option>
<mcq-option>(B) \(-4, 3, 4\)</mcq-option>
<mcq-correct>(C) \(3, -4i, 4i\)</mcq-correct>
<mcq-option>(D) \(-3, -4i, 4i\)</mcq-option>
</answer>

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Answer:

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<reasoning>

Rearrange the polynomial equation

\[
x^3 - 3x^2 + 16x - 48 = 0
\]

Factor by grouping

\[

$$\begin{aligned} x^2(x - 3) + 16(x - 3) &= 0 \\ (x^2 + 16)(x - 3) &= 0 \end{aligned}$$

\]

Solve for all roots

\[

$$\begin{aligned} x - 3 = 0 &\implies x = 3 \\ x^2 + 16 = 0 &\implies x^2 = -16 \implies x = \pm 4i \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \(-4, -3, 4\)</mcq-option>
<mcq-option>(B) \(-4, 3, 4\)</mcq-option>
<mcq-correct>(C) \(3, -4i, 4i\)</mcq-correct>
<mcq-option>(D) \(-3, -4i, 4i\)</mcq-option>
</answer>

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