1. **State the problem:** Simplify the expression $$\frac{2x^4 + 2x^3 - 7x^2 + 5x + 10}{x^2 - 1}$$. 2. **Recall the formula and rules:** The denominator can be factored using the difference of squares formula: $$a^2 - b^2 = (a - b)(a + b)$$. 3. **Factor the denominator:** $$x^2 - 1 = (x - 1)(x + 1)$$ 4. **Factor the numerator if possible:** Group terms: $$2x^4 + 2x^3 - 7x^2 + 5x + 10 = (2x^4 + 2x^3) + (-7x^2 + 5x + 10)$$ Factor out common terms: $$2x^3(x + 1) - (7x^2 - 5x - 10)$$ Try to factor the quadratic $$7x^2 - 5x - 10$$: Find two numbers that multiply to $$7 \times (-10) = -70$$ and add to $$-5$$. These are $$-10$$ and $$7$$. Rewrite: $$7x^2 - 10x + 7x - 10 = (7x^2 - 10x) + (7x - 10)$$ Factor each group: $$5x(7x - 10) + 1(7x - 10) = (5x + 1)(7x - 10)$$ Since the sign is negative in front, we have: $$-(7x^2 - 5x - 10) = -(5x + 1)(7x - 10)$$ So numerator becomes: $$2x^3(x + 1) - (5x + 1)(7x - 10)$$ 5. **Rewrite the expression:** $$\frac{2x^3(x + 1) - (5x + 1)(7x - 10)}{(x - 1)(x + 1)}$$ 6. **Check for common factors:** The numerator does not factor nicely to cancel with denominator factors. 7. **Perform polynomial division:** Divide numerator by denominator: Divide $$2x^4 + 2x^3 - 7x^2 + 5x + 10$$ by $$x^2 - 1$$. - First term: $$2x^4 / x^2 = 2x^2$$ - Multiply divisor by $$2x^2$$: $$2x^4 - 2x^2$$ - Subtract: $$ (2x^4 + 2x^3 - 7x^2) - (2x^4 - 2x^2) = 2x^3 - 5x^2$$ - Bring down $$+5x$$ - Next term: $$2x^3 / x^2 = 2x$$ - Multiply divisor by $$2x$$: $$2x^3 - 2x$$ - Subtract: $$ (2x^3 - 5x^2 + 5x) - (2x^3 - 2x) = -5x^2 + 7x$$ - Bring down $$+10$$ - Next term: $$-5x^2 / x^2 = -5$$ - Multiply divisor by $$-5$$: $$-5x^2 + 5$$ - Subtract: $$ (-5x^2 + 7x + 10) - (-5x^2 + 5) = 7x + 5$$ 8. **Write the result:** $$2x^2 + 2x - 5 + \frac{7x + 5}{x^2 - 1}$$ **Final answer:** $$\boxed{2x^2 + 2x - 5 + \frac{7x + 5}{x^2 - 1}}$$