Answer :

Certainly! Let's simplify the expression:

[tex]\[ \frac{x^2 + x - 2}{x^2 + 3x - 4} \][/tex]

1. Factor the numerator and denominator if possible:

The numerator is [tex]\(x^2 + x - 2\)[/tex]. We need to find two numbers that multiply to [tex]\(-2\)[/tex] and add to [tex]\(1\)[/tex]. These numbers are [tex]\(2\)[/tex] and [tex]\(-1\)[/tex]:

[tex]\[ x^2 + x - 2 = (x + 2)(x - 1) \][/tex]

The denominator is [tex]\(x^2 + 3x - 4\)[/tex]. We need to find two numbers that multiply to [tex]\(-4\)[/tex] and add to [tex]\(3\)[/tex]. These numbers are [tex]\(4\)[/tex] and [tex]\(-1\)[/tex]:

[tex]\[ x^2 + 3x - 4 = (x + 4)(x - 1) \][/tex]

2. Rewrite the given expression using these factors:

[tex]\[ \frac{x^2 + x - 2}{x^2 + 3x - 4} = \frac{(x + 2)(x - 1)}{(x + 4)(x - 1)} \][/tex]

3. Simplify the fraction by canceling the common factor [tex]\((x - 1)\)[/tex]:

Since [tex]\(x \neq 1\)[/tex] (meaning [tex]\(x - 1 \neq 0\)[/tex]), we can safely cancel out the [tex]\((x - 1)\)[/tex] term from the numerator and the denominator:

[tex]\[ \frac{(x + 2)(x - 1)}{(x + 4)(x - 1)} = \frac{x + 2}{x + 4} \][/tex]

Therefore, the simplified form of the given expression is:

[tex]\[ \frac{x + 2}{x + 4} \][/tex]

Remember, this simplification is valid provided that [tex]\(x \neq 1\)[/tex] and [tex]\(x \neq -4\)[/tex], because at these values the original fraction would be undefined (the denominator would equal zero).