Question

How do you solve (2x+8)(3x−6)=0 using any method?

Answer 1

Answer : x1 = - 4, x2 = 2

Step by step explanation :

Solving using factor method.

$(2x + 8) \times (3x - 6) = 0$

Split into possible cases. When the product of factors equals 0, a least one factor is zero.

$= > 2x + 8 = 0 \: \: - eq1 \\ \: \: \: \: \: \: \: \: 3x - 6 = 0 \: \: \: -eq2$

Solve equation 1 for x

$= > 2x + 8 = 0$

Move constant to the right side and change its sign.

$= > 2x = - 8$

Divide both sides of the equation by 2.

$= > x = - 4$

Solve equation 2 for x

$= > 3x - 6 = 0$

Move constant to the right side and change its sign.

$= > 3x = 6$

Divide both sides of the equation by 3.

$= > x = 2$

The final solutions are

$= > x1 = - 4 \\ \: \: \: \: \: \: \: \: x2 = 2$

Answer 2

Hey there!

Answer:

x = -4   or x = 2

Step-by-step explanation:

Given,

(2x + 8)(3x - 6) = 0

Using zero product rule,

2x + 8 = 0   or    3x - 6 = 0

2x = -8    or    3x = 6

x = -8/2   or x = 6/3

x = -4   or x = 2

Hope It Helps You!