Question

simplify:
\frac{6}{x^2 - 4} + \frac{x}{x + 2}
\frac{?x^2 + \square x + \square}{(x - 2)(x + \square)}

Explanation:

Step1: Factor the denominator

First, factor \(x^2 - 4\) using the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\). So, \(x^2 - 4=(x - 2)(x + 2)\). The expression becomes \(\frac{6}{(x - 2)(x + 2)}+\frac{x}{x + 2}\).

Step2: Find a common denominator

The common denominator for the two fractions is \((x - 2)(x + 2)\). Rewrite the second fraction with this common denominator: \(\frac{x}{x + 2}=\frac{x(x - 2)}{(x - 2)(x + 2)}\).

Step3: Add the fractions

Now add the two fractions: \(\frac{6}{(x - 2)(x + 2)}+\frac{x(x - 2)}{(x - 2)(x + 2)}=\frac{6+x(x - 2)}{(x - 2)(x + 2)}\).

Step4: Expand the numerator

Expand \(x(x - 2)\) to get \(x^2-2x\). Then the numerator becomes \(6 + x^2-2x=x^2-2x + 6\).

Answer:

The simplified form has numerator \(1x^2-2x + 6\) and the denominator factor \((x - 2)(x + 2)\), so the blanks are filled as follows: \(\frac{1x^2-2x + 6}{(x - 2)(x + 2)}\) (the last blank in the denominator is \(2\)). So the first blank (coefficient of \(x^2\)) is \(1\), the coefficient of \(x\) is \(-2\), the constant term is \(6\), and the last blank in the denominator is \(2\).