Question

review for final (2)
question 14 of 26 (1 point) | question attempt: 1 of unlimited
solve for u .
(u² + 4u - 12)(3u + 24) = 0
(if there is more than one solution, separate them with commas.)
u =

Explanation:

Step1: Aplicar la propiedad del producto nulo

Si \( (a)(b) = 0 \), entonces \( a = 0 \) o \( b = 0 \). Aquí, \( a = u^2 + 4u - 12 \) y \( b = 3u + 24 \). Entonces:
\( u^2 + 4u - 12 = 0 \) o \( 3u + 24 = 0 \)

Step2: Resolver \( 3u + 24 = 0 \)

Restar 24 de ambos lados: \( 3u = -24 \)
Dividir por 3: \( u = \frac{-24}{3} = -8 \)

Step3: Resolver \( u^2 + 4u - 12 = 0 \) (factorizar)

Encontrar dos números que sumen 4 y multipliquen -12. Son 6 y -2.
Entonces, \( u^2 + 4u - 12 = (u + 6)(u - 2) = 0 \)
Aplicar la propiedad del producto nulo: \( u + 6 = 0 \) o \( u - 2 = 0 \)
Para \( u + 6 = 0 \): \( u = -6 \)
Para \( u - 2 = 0 \): \( u = 2 \)

Answer:

\( -8, -6, 2 \)