Rewrite the expression $\frac{2x^2+2x}{-x^2+1}$ into an equivalent expression having a denominator of $x-1$

• A. $\(\frac{2x}{x-1}\)$
• B. $-\(\frac{2x}{x-1}\)$
• C. $\(\frac{2x^2}{x-1}\)$
• D. $-\(\frac{2x^2}{x-1}\)$

Answer: (B)

Verified step by step guidance

1. Start with the original expression: \(\frac{2x^2 + 2x}{-x^2 + 1}\).
2. Factor the numerator and denominator separately. For the numerator, factor out the common factor \$2x$ to get \(2x(x + 1)\). For the denominator, recognize that $-x^2 + 1$ can be rewritten as $-(x^2 - 1)$, and then factor $x^2 - 1$ as a difference of squares: $(x - 1)(x + 1)$. So the denominator becomes $-(x - 1)(x + 1)$.
3. Rewrite the expression using these factorizations: \(\frac{2x(x + 1)}{-(x - 1)(x + 1)}\).
4. Cancel the common factor \((x + 1)\) from numerator and denominator, leaving \(\frac{2x}{-(x - 1)}\).
5. Rewrite the expression to have the denominator \(x - 1\) by factoring out the negative sign from the denominator, resulting in \(-\frac{2x}{x - 1}\).