To simplify the expression \( \frac{x^2 - 4}{x^2 - x - 2} \), we will first factor both the numerator and the denominator.
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Factor the numerator: \[ x^2 - 4 = (x - 2)(x + 2) \]
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Factor the denominator: \[ x^2 - x - 2 = (x - 2)(x + 1) \]
Now we can rewrite the expression as:
\[ \frac{(x - 2)(x + 2)}{(x - 2)(x + 1)} \]
We can cancel the common factor \( (x - 2) \) (assuming \( x \neq 2 \)):
\[ \frac{x + 2}{x + 1} \]
Now we will examine each option to see which one simplifies to this result.
A. \( \frac{x^2 + x - 2}{x^2 - 1} \)
Factoring the numerator \( x^2 + x - 2 = (x - 1)(x + 2) \) and the denominator \( x^2 - 1 = (x - 1)(x + 1) \):
\[ \frac{(x - 1)(x + 2)}{(x - 1)(x + 1)} \quad \text{(assuming \( x \neq 1 \))} \]
Cancelling \( (x - 1) \):
\[ \frac{x + 2}{x + 1} \]
So A simplifies to \( \frac{x + 2}{x + 1} \).
B. \( \frac{x + 2}{x^2 + 3x + 2} \)
Factoring the denominator \( x^2 + 3x + 2 = (x + 1)(x + 2) \):
\[ \frac{x + 2}{(x + 1)(x + 2)} \quad \text{(assuming \( x \neq -2 \))} \]
Cancelling \( (x + 2) \):
\[ \frac{1}{x + 1} \]
B does not simplify to \( \frac{x + 2}{x + 1} \).
C. \( \frac{x^2 - x - 2}{x^2 - 4} \)
We previously factored both:
Numerator: \( x^2 - x - 2 = (x - 2)(x + 1) \) Denominator: \( x^2 - 4 = (x - 2)(x + 2) \)
The expression becomes:
\[ \frac{(x - 2)(x + 1)}{(x - 2)(x + 2)} \quad \text{(assuming \( x \neq 2 \))} \]
Cancelling \( (x - 2) \):
\[ \frac{x + 1}{x + 2} \]
C does not simplify to \( \frac{x + 2}{x + 1} \).
D. \( \frac{x^2 - 4}{x^2 - 2x} \)
Factoring numerator: \( x^2 - 4 = (x - 2)(x + 2) \) Factoring denominator: \( x^2 - 2x = x(x - 2) \)
The expression becomes:
\[ \frac{(x - 2)(x + 2)}{x(x - 2)} \quad \text{(assuming \( x \neq 2 \))} \]
Cancelling \( (x - 2) \):
\[ \frac{x + 2}{x} \]
D does not simplify to \( \frac{x + 2}{x + 1} \).
Thus, the only expression that simplifies to an equivalent form of \(\frac{x^2 - 4}{x^2 - x - 2}\) is A.
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