Partial Fraction 28Dbe1

1. **State the problem:** We want to decompose the rational function $$\frac{5x^2 + 10x + 12}{(x + 2)(x^2 - x + 6)}$$ into partial fractions of the form $$\frac{f(x)}{x + 2} + \frac{g(x)}{x^2 - x + 6}$$ where \(f(x)\) is a polynomial of degree less than 1 (a constant) and \(g(x)\) is a polynomial of degree less than 2 (a linear polynomial). 2. **Set up the form of the partial fractions:** Since \(x + 2\) is linear, \(f(x) = A\) (a constant). Since \(x^2 - x + 6\) is quadratic and irreducible, \(g(x) = Bx + C\) (linear). So, $$\frac{5x^2 + 10x + 12}{(x + 2)(x^2 - x + 6)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 - x + 6}$$ 3. **Multiply both sides by the denominator \((x + 2)(x^2 - x + 6)\):** $$5x^2 + 10x + 12 = A(x^2 - x + 6) + (Bx + C)(x + 2)$$ 4. **Expand the right side:** $$A(x^2 - x + 6) = Ax^2 - Ax + 6A$$ $$(Bx + C)(x + 2) = Bx^2 + 2Bx + Cx + 2C = Bx^2 + (2B + C)x + 2C$$ So, $$5x^2 + 10x + 12 = Ax^2 - Ax + 6A + Bx^2 + (2B + C)x + 2C$$ 5. **Combine like terms:** $$5x^2 + 10x + 12 = (A + B)x^2 + (-A + 2B + C)x + (6A + 2C)$$ 6. **Equate coefficients of powers of \(x\):** - Coefficient of \(x^2\): $$5 = A + B$$ - Coefficient of \(x\): $$10 = -A + 2B + C$$ - Constant term: $$12 = 6A + 2C$$ 7. **Solve the system of equations:** From the first equation: $$B = 5 - A$$ Substitute \(B\) into the second: $$10 = -A + 2(5 - A) + C = -A + 10 - 2A + C = 10 - 3A + C$$ So, $$C = 10 - 3A + C - 10 = 3A$$ Wait, rearranged correctly: $$10 = 10 - 3A + C \implies 0 = -3A + C \implies C = 3A$$ Substitute \(C = 3A\) into the third equation: $$12 = 6A + 2(3A) = 6A + 6A = 12A$$ So, $$12 = 12A \implies A = 1$$ Then, $$B = 5 - 1 = 4$$ $$C = 3(1) = 3$$ 8. **Write the final partial fraction decomposition:** $$f(x) = A = 1$$ $$g(x) = Bx + C = 4x + 3$$ **Answer:** $$\frac{5x^2 + 10x + 12}{(x + 2)(x^2 - x + 6)} = \frac{1}{x + 2} + \frac{4x + 3}{x^2 - x + 6}$$