1. **State the problem:** Divide the polynomial $5x^4 - 2x^2 - 13x + 8$ by $x^2 + 1$ to find the quotient and remainder. 2. **Recall the division formula:** For polynomials $f(x)$ and $g(x)$, there exist polynomials $q(x)$ (quotient) and $r(x)$ (remainder) such that: $$f(x) = g(x) \cdot q(x) + r(x)$$ where the degree of $r(x)$ is less than the degree of $g(x)$. 3. **Set up the division:** Divide $5x^4 - 2x^2 - 13x + 8$ by $x^2 + 1$. 4. **First division step:** Divide the leading term $5x^4$ by $x^2$ to get $5x^2$. Multiply $5x^2$ by $x^2 + 1$ to get $5x^4 + 5x^2$. Subtract this from the original polynomial: $$(5x^4 - 2x^2 - 13x + 8) - (5x^4 + 5x^2) = -7x^2 - 13x + 8$$ 5. **Second division step:** Divide the leading term $-7x^2$ by $x^2$ to get $-7$. Multiply $-7$ by $x^2 + 1$ to get $-7x^2 - 7$. Subtract this from the current remainder: $$(-7x^2 - 13x + 8) - (-7x^2 - 7) = -13x + 15$$ 6. **Check remainder degree:** The remainder $-13x + 15$ has degree 1, which is less than degree 2 of divisor $x^2 + 1$, so division stops here. 7. **Final answer:** $$\text{Quotient} = 5x^2 - 7$$ $$\text{Remainder} = -13x + 15$$