APPROACH #1: Factoring

Given: $\frac{2x^2+8x−24}{2x^2+20x−48}$

Factor the greatest common factor from numerator and denominator: $\frac{2(x^2+4x−12)}{2(x^2+10x−24)}$

Factor each quadratic: $\frac{2(x +6)(x - 2)}{2(x + 12)(x - 2)}$

Simplify: $\frac{x +6}{x + 12}$

Answer: E

APPROACH #2: Testing a value
Two expressions are equivalent if they evaluate to the same number for all values of x.
For example, we know that 6x + 4x is equivalent to 10x because 6x + 4x = 10x for all values of x.
So, if x = 1.1, we see that 6x + 4x = 6(1.1) + 4(1.1) = 6.6 + 4.4 = 11
Likewise, if x = 1.1, we see that 10x = 10(1.1) = 11

Let's apply this property to the given expression. We'll use a nice number to work with such as $x = 1$
We get: $\frac{2x^2+8x−24}{2x^2+20x−48} = \frac{2(1)^2+8(1)−24}{2(1)^2+20(1)−48} = \frac{-14}{-26} = \frac{7}{13}$
This tells us that the correct answer must also evaluate to be $\frac{7}{13}$ when $x=1$.

Let's plug x=1 and we chance of choice to see what we get....
A. (1−2)/(1−4) = 1/3. No good. We want 7/13. Eliminate.

B. (1−2)/(1+4) = -1/5. No good. We want 7/13. Eliminate.

C. (1+2)/(1+4) = 3/5. No good. We want 7/13. Eliminate.

D. (1+2)/(1−12) = -3/11. No good. We want 7/13. Eliminate.

E. (1+6)/(1+12) = 7/13 PERFECT

Answer: E