Partial Fraction 7B833E

1. **State the problem:** We want to find the partial fraction decomposition of the rational function $$\frac{x^2 + 12}{x^3 + x^2}$$ in the form $$\frac{f(x)}{x} + \frac{g(x)}{x^2} + \frac{h(x)}{x + 1}$$ where $f(x)$, $g(x)$, and $h(x)$ are polynomials to be determined. 2. **Factor the denominator:** The denominator is $$x^3 + x^2 = x^2(x + 1)$$ This shows the denominator has factors $x$, $x^2$, and $x + 1$. 3. **Set up the partial fractions:** Since $x^2$ is a repeated factor, the decomposition includes terms with $x$ and $x^2$ in the denominator: $$\frac{x^2 + 12}{x^2(x + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1}$$ where $A$, $B$, and $C$ are constants (since the factors are linear). 4. **Multiply both sides by the denominator $x^2(x + 1)$ to clear fractions:** $$x^2 + 12 = A \cdot x (x + 1) + B (x + 1) + C x^2$$ 5. **Expand the right side:** $$A x (x + 1) = A x^2 + A x$$ $$B (x + 1) = B x + B$$ So, $$x^2 + 12 = A x^2 + A x + B x + B + C x^2$$ 6. **Group like terms:** $$x^2 + 12 = (A + C) x^2 + (A + B) x + B$$ 7. **Equate coefficients of corresponding powers of $x$ on both sides:** - Coefficient of $x^2$: $1 = A + C$ - Coefficient of $x$: $0 = A + B$ - Constant term: $12 = B$ 8. **Solve the system of equations:** From $12 = B$, we get $$B = 12$$ From $0 = A + B$, substitute $B=12$: $$0 = A + 12 \implies A = -12$$ From $1 = A + C$, substitute $A = -12$: $$1 = -12 + C \implies C = 13$$ 9. **Write the final partial fraction decomposition:** $$\frac{x^2 + 12}{x^3 + x^2} = \frac{-12}{x} + \frac{12}{x^2} + \frac{13}{x + 1}$$ So, $$f(x) = -12$$ $$g(x) = 12$$ $$h(x) = 13$$