Substitution is going to be the best way here, but pick up smart number..
x as 0 does not give us a denominator as 0 and also eases out calculations..
all terms with x will become 0..

\(\frac{2x^2(x + 3) -2x -6}{x^2 + 2x - 3}=\frac{-6}{-3}=2\)

It is very clear looking at the choices, at first glance itself, that only B will leave 2 as answer
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\(\frac{2x^2(x + 3) -2x -6}{x^2 + 2x - 3}\)

\(\frac{(2x^2-2)(x + 3)}{(x+3)*(x-1)}\)

\(\frac{(2x^2-2)}{(x-1)}\)

\(\frac{2(x^2-1)}{(x-1)}\)

\(\frac{2(x-1)(x+1)}{(x-1)}\)

\(2(x+1)\)

Answer: Option B
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\(\frac{2x^2(x + 3) -2x -6}{x^2 + 2x - 3}\)

= \(\frac{2x^2(x + 3) - 2 (x + 3)}{x^2 + 3x - x - 3}\)

= \(\frac{2(x^2 - 1)(x + 3)}{(x + 3)(x - 1)}\)

= \(2(x + 1)\)

= 2x + 2

Alternate approach: Put x = 0 in the given equation, we get the value as 2. Substitute x = 0 in the options, only option B remains.

Answer: B

STRATEGY: Upon reading any GMAT Problem Solving question, we should always ask, Can I use the answer choices to my advantage?
In this case, we can easily plug in values of x to test for equivalency.
Now let's give ourselves up to 20 seconds to identify a faster approach.
In this case, we can also apply some algebraic factoring techniques to simplify the given expression.
Since I feel both techniques are equally fast, I'm going to test for equivalency since I'm less likely to make mistakes with that technique.

Key concept: If two expressions are equivalent, they must evaluate to the same value for every possible value of x.
For example, since the expression 2x + 3x is equivalent to the expression 5x, the two expressions will evaluate to the same number for every value of x.
So, if x = 7, the expression 2x + 3x = 2(7) + 3(7) = 14 + 21 = 35, and the expression 5x = 5(7) = 35

Let's evaluate the given expression for \(x = 0\).

We get: \(\frac{2x^2(x + 3) -2x -6}{x^2 + 2x - 3}=\frac{2(0)^2(0 + 3) -2(0) -6}{(0)^2 + 2(0) - 3}=\frac{-6}{-3} = 2\)

We'll now evaluate each answer choice for \(x = 0\) and eliminate those that don't evaluate to \(2\)
(A) \(2(0)^2 - 2 = -2\). Doesn't evaluate to \(2\). ELIMINATE.
(B) \(2(0) + 2 = 2\). KEEP
(C) \(0 + 1 =1\). Doesn't evaluate to \(2\). ELIMINATE.
(D) \(2(0) + 6 = 6\). Doesn't evaluate to \(2\). ELIMINATE.
(E) \(0 - 1 = -1\). Doesn't evaluate to \(2\). ELIMINATE.

Answer: B
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Brent, I think you accidentally subbed in 1 instead of 0 and got the wrong answer of C.

Thanks for the heads up!
I have edited my response accordingly.

Kudos for you!!!!!
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