5. \( \frac{x^{2}-x+2 x}{x^{2}-2 x-3} \div \frac{x^{2}-4 x+3}{1-x^{2}} \)
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First, let's simplify the expression step by step. We start with: \[ \frac{x^{2}-x+2x}{x^{2}-2x-3} \div \frac{x^{2}-4x+3}{1-x^{2}} \] This simplifies to: \[ \frac{x^{2}+x}{x^{2}-2x-3} \div \frac{x^{2}-4x+3}{1-x^{2}} \] Performing the division of fractions involves multiplying by the reciprocal: \[ = \frac{x^{2}+x}{x^{2}-2x-3} \times \frac{1-x^{2}}{x^{2}-4x+3} \] Next, we will factor the polynomials in the expression where possible. 1. **Factor \(x^{2}+x\)**: \[ x^{2} + x = x(x + 1) \] 2. **Factor \(x^{2}-2x-3\)**: \[ x^{2}-2x-3 = (x-3)(x+1) \] 3. **Factor \(1 - x^{2}\)**: \[ 1 - x^{2} = (1 - x)(1 + x) \] 4. **Factor \(x^{2}-4x+3\)**: \[ x^{2}-4x+3 = (x-3)(x-1) \] Now substituting the factored forms back in: \[ = \frac{x(x+1)}{(x - 3)(x + 1)} \times \frac{(1 - x)(1 + x)}{(x - 3)(x - 1)} \] We can cancel the \(x + 1\) common term: \[ = \frac{x \cdot (1 - x)(1 + x)}{(x - 3)^{2}(x - 1)} \] Now expand the remaining numerator: \[ = \frac{x(1 - x^2)}{(x - 3)^{2}(x - 1)} = \frac{x(1 - x)(1 + x)}{(x - 3)^{2}(x - 1)} \] At this point, the expression is simplified as much as possible. Hence, the final result is: \[ \frac{x(1 - x)(1 + x)}{(x - 3)^{2}(x - 1)} \]
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