Question

multiply: $(2a - 3)(4a^2 + 6a + 9)$
$\bigcirc\\ 8a^3 - 24a^2 - 27$
$\bigcirc\\ 8a^3 + 24a^2 + 27$
$\bigcirc\\ 8a^3 + 27$
$\bigcirc\\ 8a^3 - 27$
1 pts
question 2
factor: $x^3 - 8$
$\bigcirc\\ (x - 2)(x^2 - 2x - 4)$
$\bigcirc\\ (x - 2)(x^2 - 2x + 4)$
$\bigcirc\\ (x - 2)^3$
$\bigcirc\\ (x^2 + 2x - 4)$

Explanation:

Step1: Distribute first term

$2a \cdot (4a^2 + 6a + 9) = 8a^3 + 12a^2 + 18a$

Step2: Distribute second term

$-3 \cdot (4a^2 + 6a + 9) = -12a^2 - 18a - 27$

Step3: Combine all terms

$8a^3 + 12a^2 + 18a - 12a^2 - 18a - 27$

Step4: Simplify by canceling terms

$8a^3 + (12a^2-12a^2) + (18a-18a) -27 = 8a^3 -27$

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Step1: Recognize difference of cubes

$x^3 - 8 = x^3 - 2^3$

Step2: Apply difference of cubes formula

The formula for $a^3 - b^3 = (a-b)(a^2+ab+b^2)$, so substitute $a=x, b=2$:
$(x-2)(x^2 + (x)(2) + 2^2) = (x-2)(x^2 + 2x + 4)$

Answer:

Question 1: $\boldsymbol{8a^3 - 27}$
Question 2: $\boldsymbol{(x-2)(x^2 + 2x + 4)}$