1. **State the problem:** We need to divide the polynomial $$x^4 + 4x^3 - 24x^2 + 8x + 20$$ by $$x^2 - 2x - 3$$ and express the result in the form $$q(x) + \frac{r(x)}{b(x)}$$ where $$q(x)$$ is the quotient and $$r(x)$$ is the remainder. 2. **Recall the division algorithm for polynomials:** When dividing a polynomial $$f(x)$$ by a polynomial $$g(x)$$, there exist unique polynomials $$q(x)$$ and $$r(x)$$ such that $$f(x) = q(x)g(x) + r(x)$$ and the degree of $$r(x)$$ is less than the degree of $$g(x)$$. 3. **Perform polynomial long division:** - Divide the leading term $$x^4$$ by $$x^2$$ to get $$x^2$$. - Multiply $$x^2$$ by $$x^2 - 2x - 3$$ to get $$x^4 - 2x^3 - 3x^2$$. - Subtract this from the original polynomial: $$ (x^4 + 4x^3 - 24x^2 + 8x + 20) - (x^4 - 2x^3 - 3x^2) = 6x^3 - 21x^2 + 8x + 20 $$. 4. **Repeat division with new polynomial:** - Divide $$6x^3$$ by $$x^2$$ to get $$6x$$. - Multiply $$6x$$ by $$x^2 - 2x - 3$$ to get $$6x^3 - 12x^2 - 18x$$. - Subtract: $$ (6x^3 - 21x^2 + 8x + 20) - (6x^3 - 12x^2 - 18x) = -9x^2 + 26x + 20 $$. 5. **Repeat division again:** - Divide $$-9x^2$$ by $$x^2$$ to get $$-9$$. - Multiply $$-9$$ by $$x^2 - 2x - 3$$ to get $$-9x^2 + 18x + 27$$. - Subtract: $$ (-9x^2 + 26x + 20) - (-9x^2 + 18x + 27) = 8x - 7 $$. 6. **Determine quotient and remainder:** - Quotient $$q(x) = x^2 + 6x - 9$$. - Remainder $$r(x) = 8x - 7$$. - Since degree of remainder $$1 < 2$$ (degree of divisor), division is complete. 7. **Express final answer:** $$\frac{x^4 + 4x^3 - 24x^2 + 8x + 20}{x^2 - 2x - 3} = x^2 + 6x - 9 + \frac{8x - 7}{x^2 - 2x - 3}$$. This matches the form $$q(x) + \frac{r(x)}{b(x)}$$ as requested.