1. **State the problem:** Express the rational function $$\frac{5x^2 + 2}{2x^2 + x}$$ as a sum of partial fractions. 2. **Factor the denominator:** $$2x^2 + x = x(2x + 1)$$ 3. **Set up the partial fractions:** Since the denominator factors into linear terms, we write: $$\frac{5x^2 + 2}{x(2x + 1)} = \frac{A}{x} + \frac{B}{2x + 1}$$ 4. **Multiply both sides by the denominator to clear fractions:** $$5x^2 + 2 = A(2x + 1) + Bx$$ 5. **Expand the right side:** $$5x^2 + 2 = 2Ax + A + Bx = (2A + B)x + A$$ 6. **Equate coefficients of like powers of $x$:** - Coefficient of $x^2$: Left side has 5, right side has 0, so $5 = 0$ (contradiction) This indicates the degree of numerator is not less than denominator, so we must perform polynomial division first. 7. **Perform polynomial division:** Divide $5x^2 + 2$ by $2x^2 + x$: $$\frac{5x^2 + 2}{2x^2 + x} = Q(x) + \frac{R(x)}{2x^2 + x}$$ Divide leading terms: $$\frac{5x^2}{2x^2} = \frac{5}{2}$$ Multiply divisor by $\frac{5}{2}$: $$\frac{5}{2}(2x^2 + x) = 5x^2 + \frac{5}{2}x$$ Subtract: $$\left(5x^2 + 2\right) - \left(5x^2 + \frac{5}{2}x\right) = -\frac{5}{2}x + 2$$ So: $$\frac{5x^2 + 2}{2x^2 + x} = \frac{5}{2} + \frac{-\frac{5}{2}x + 2}{2x^2 + x}$$ 8. **Express the remainder fraction as partial fractions:** $$\frac{-\frac{5}{2}x + 2}{x(2x + 1)} = \frac{A}{x} + \frac{B}{2x + 1}$$ Multiply both sides by denominator: $$-\frac{5}{2}x + 2 = A(2x + 1) + Bx = (2A + B)x + A$$ Equate coefficients: - Coefficient of $x$: $-\frac{5}{2} = 2A + B$ - Constant term: $2 = A$ 9. **Solve for $A$ and $B$:** From constant term: $$A = 2$$ Substitute into $x$ coefficient: $$-\frac{5}{2} = 2(2) + B = 4 + B$$ $$B = -\frac{5}{2} - 4 = -\frac{5}{2} - \frac{8}{2} = -\frac{13}{2}$$ 10. **Write the final partial fraction decomposition:** $$\frac{5x^2 + 2}{2x^2 + x} = \frac{5}{2} + \frac{2}{x} - \frac{13}{2(2x + 1)}$$ **Answer:** $$\boxed{\frac{5x^2 + 2}{2x^2 + x} = \frac{5}{2} + \frac{2}{x} - \frac{13}{2(2x + 1)}}$$