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:

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

$$
LATEXBLOCK0
$$

$$ f(x) = (x + 2)(x^2 + 6x + 6) $$

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.

$$ \text{The complete set of zeros is } \{-2, -3 + \sqrt{3}, -3 - \sqrt{3}\} $$

Answer:

• A. \(\{1, -1, -12\}\)
• B. \(\{2, -6 + \sqrt{3}, -6 - \sqrt{3}\}\)
• C. \(\{-2, -3 + \sqrt{3}, -3 - \sqrt{3}\}\) (Correct answer)
• D. \(\{-2, -6 + \sqrt{12}, -6 - \sqrt{12}\}\)