Question

find a rational zero of the polynomial function and use it to find all the zeros of the function.

\\(f(x) = x^3 + 8x^2 + 18x + 12\\)

a. \\(\\{1, -1, -12\\}\\)
b. \\(\\{2, -6 + \sqrt{3}, -6 - \sqrt{3}\\}\\)
c. \\(\\{-2, -3 + \sqrt{3}, -3 - \sqrt{3}\\}\\)
d. \\(\\{-2, -6 + \sqrt{12}, -6 - \sqrt{12}\\}\\)

Explanation:

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<reasoning>

Find a rational zero

Using the Rational Root Theorem knowledge point
\[
\text{Possible rational roots of } f(x) = x^3 + 8x^2 + 18x + 12 \text{ are factors of } 12: \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12
\]
\[
f(-2) = (-2)^3 + 8(-2)^2 + 18(-2) + 12 = -8 + 32 - 36 + 12 = 0
\]
\[
\text{Thus, } x = -2 \text{ is a rational zero.}
\]

Divide the polynomial

Using the Synthetic Division knowledge point
\[

$$\begin{array}{r|rrrr} -2 & 1 & 8 & 18 & 12 \\ & & -2 & -12 & -12 \\ \hline & 1 & 6 & 6 & 0 \end{array}$$

\]
\[
f(x) = (x + 2)(x^2 + 6x + 6)
\]

Solve the quadratic factor

Using the Quadratic Formula knowledge point
\[
x^2 + 6x + 6 = 0
\]
\[
x = \frac{-6 \pm \sqrt{6^2 - 4(1)(6)}}{2(1)} = \frac{-6 \pm \sqrt{12}}{2} = \frac{-6 \pm 2\sqrt{3}}{2} = -3 \pm \sqrt{3}
\]

List all zeros

Combine the rational zero and the remaining zeros.
\[
\text{The complete set of zeros is } \{-2, -3 + \sqrt{3}, -3 - \sqrt{3}\}
\]
</reasoning>

<answer>
<mcq-option>A. \(\{1, -1, -12\}\)</mcq-option>
<mcq-option>B. \(\{2, -6 + \sqrt{3}, -6 - \sqrt{3}\}\)</mcq-option>
<mcq-correct>C. \(\{-2, -3 + \sqrt{3}, -3 - \sqrt{3}\}\)</mcq-correct>
<mcq-option>D. \(\{-2, -6 + \sqrt{12}, -6 - \sqrt{12}\}\)</mcq-option>
</answer>

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Answer:

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<reasoning>

Find a rational zero

Using the Rational Root Theorem knowledge point
\[
\text{Possible rational roots of } f(x) = x^3 + 8x^2 + 18x + 12 \text{ are factors of } 12: \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12
\]
\[
f(-2) = (-2)^3 + 8(-2)^2 + 18(-2) + 12 = -8 + 32 - 36 + 12 = 0
\]
\[
\text{Thus, } x = -2 \text{ is a rational zero.}
\]

Divide the polynomial

Using the Synthetic Division knowledge point
\[

$$\begin{array}{r|rrrr} -2 & 1 & 8 & 18 & 12 \\ & & -2 & -12 & -12 \\ \hline & 1 & 6 & 6 & 0 \end{array}$$

\]
\[
f(x) = (x + 2)(x^2 + 6x + 6)
\]

Solve the quadratic factor

Using the Quadratic Formula knowledge point
\[
x^2 + 6x + 6 = 0
\]
\[
x = \frac{-6 \pm \sqrt{6^2 - 4(1)(6)}}{2(1)} = \frac{-6 \pm \sqrt{12}}{2} = \frac{-6 \pm 2\sqrt{3}}{2} = -3 \pm \sqrt{3}
\]

List all zeros

Combine the rational zero and the remaining zeros.
\[
\text{The complete set of zeros is } \{-2, -3 + \sqrt{3}, -3 - \sqrt{3}\}
\]
</reasoning>

<answer>
<mcq-option>A. \(\{1, -1, -12\}\)</mcq-option>
<mcq-option>B. \(\{2, -6 + \sqrt{3}, -6 - \sqrt{3}\}\)</mcq-option>
<mcq-correct>C. \(\{-2, -3 + \sqrt{3}, -3 - \sqrt{3}\}\)</mcq-correct>
<mcq-option>D. \(\{-2, -6 + \sqrt{12}, -6 - \sqrt{12}\}\)</mcq-option>
</answer>

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