Question

factor completely (3x^3 + 12x^2 + 18x).

• (x^2 + 4x + 6)
• (3(x^3 + 4x^2 + 6x))
• (3x(x^2 + 4x + 6))
• (x(3x^2 + 12x + 18))

question 6 (multiple choice worth 1 points)
(02.01r)

what is the completely factored form of (t^4 - 16)?

• ((t^2 + 4)(t^2 - 4))
• ((t^2 + 4)(t + 2)(t - 2))
• ((t^2 + 4)(t - 2)(t - 2))
• ((t + 2)(t + 2)(t + 2)(t - 2))

Explanation:

Factor out the greatest common factor

Using the Greatest Common Factor knowledge point

$$
LATEXBLOCK0
$$

Check for further factorization

The remaining trinomial is \(x^2 + 4x + 6\).
We look for two integers that multiply to \(6\) and add to \(4\).
The factors of \(6\) are:

• \(1 \times 6\) (sum = \(7\))
• \(2 \times 3\) (sum = \(5\))

No such integers exist, so the trinomial cannot be factored further over the integers.
The completely factored form is:

$$ 3x(x^2 + 4x + 6) $$

Apply difference of squares to the second expression

Using the Difference of Squares knowledge point

$$
LATEXBLOCK1
$$

Factor the remaining binomial completely

Using the Difference of Squares knowledge point

$$
LATEXBLOCK2
$$

Combining these gives:

$$ (t^2 + 4)(t + 2)(t - 2) $$

Answer:

Question 1

• (A) \(x^2 + 4x + 6\)
• (B) \(3(x^3 + 4x^2 + 6x)\)
• (C) \(3x(x^2 + 4x + 6)\) (Correct answer)
• (D) \(x(3x^2 + 12x + 18)\)

Question 2

• (A) \((t^2 + 4)(t^2 - 4)\)
• (B) \((t^2 + 4)(t + 2)(t - 2)\) (Correct answer)
• (C) \((t^2 + 4)(t - 2)(t - 2)\)
• (D) \((t + 2)(t + 2)(t + 2)(t - 2)\)