Question

(\frac{3x + 5}{x^{-2}-4})

Explanation:

Step1: Simplify the denominator

Recall that \(x^{-2}=\frac{1}{x^{2}}\), so the denominator \(x^{-2}-4=\frac{1}{x^{2}} - 4\). We can rewrite \(4\) as \(\frac{4x^{2}}{x^{2}}\) to get a common denominator: \(\frac{1 - 4x^{2}}{x^{2}}\). Notice that \(1-4x^{2}\) is a difference of squares, \(1-4x^{2}=(1 - 2x)(1 + 2x)\). So the denominator becomes \(\frac{(1 - 2x)(1 + 2x)}{x^{2}}\).

Step2: Rewrite the fraction

The original expression \(\frac{3x + 5}{x^{-2}-4}\) can be rewritten as \(\frac{3x + 5}{\frac{(1 - 2x)(1 + 2x)}{x^{2}}}\). Dividing by a fraction is the same as multiplying by its reciprocal, so this becomes \((3x + 5)\times\frac{x^{2}}{(1 - 2x)(1 + 2x)}=\frac{x^{2}(3x + 5)}{(1 - 2x)(1 + 2x)}\). We can also factor out a negative sign from \(1 - 2x\) to get \(1 - 2x=-(2x - 1)\), so the expression can be written as \(\frac{x^{2}(3x + 5)}{-(2x - 1)(2x + 1)}=-\frac{x^{2}(3x + 5)}{(2x - 1)(2x + 1)}\) (or we can leave it in the form \(\frac{x^{2}(3x + 5)}{(1 - 2x)(1 + 2x)}\) depending on the required simplification).

If we assume the problem is to simplify the rational expression, the simplified form is \(\frac{x^{2}(3x + 5)}{(1 - 2x)(1 + 2x)}\) (or equivalent forms after factoring the negative sign).

Answer:

\(\frac{x^{2}(3x + 5)}{(1 - 2x)(1 + 2x)}\) (or \(-\frac{x^{2}(3x + 5)}{(2x - 1)(2x + 1)}\))