Question

algebra: concepts and connections - plc investigating rational and irrational numbers which rational expression has a value of 0 when ( x = -2 )? (\frac{2x - 4}{x^2 - 4}) (\frac{-x^2 - 4}{3x + 6}) (\frac{5x - 5}{-3x^2 + 6}) (\frac{2x^2 - 8}{4x - 8})

Explanation:

Step1: Evaluate first expression at $x=-2$

Substitute $x=-2$ into $\frac{2x-4}{x^2-4}$:
$\frac{2(-2)-4}{(-2)^2-4} = \frac{-4-4}{4-4} = \frac{-8}{0}$ (undefined)

Step2: Evaluate second expression at $x=-2$

Substitute $x=-2$ into $\frac{-x^2-4}{3x+6}$:
$\frac{-(-2)^2-4}{3(-2)+6} = \frac{-4-4}{-6+6} = \frac{-8}{0}$ (undefined)

Step3: Evaluate third expression at $x=-2$

Substitute $x=-2$ into $\frac{5x-5}{-3x^2+6}$:
$\frac{5(-2)-5}{-3(-2)^2+6} = \frac{-10-5}{-12+6} = \frac{-15}{-6} = \frac{5}{2}$ (not 0)

Step4: Evaluate fourth expression at $x=-2$

Substitute $x=-2$ into $\frac{2x^2-8}{4x-8}$:
$\frac{2(-2)^2-8}{4(-2)-8} = \frac{8-8}{-8-8} = \frac{0}{-16} = 0$

Answer:

$\frac{2x^2-8}{4x-8}$