O-Level Polynomials and partial fractions: worked solution

Question

Resolve (3x² + 5x − 2)/[(x+1)(x²+4)] into partial fractions.

First decide on the correct form. The denominator has a linear factor (x + 1) and the quadratic factor (x² + 4). That quadratic is irreducible, because x² + 4 = 0 gives x² = −4, which has no real root, so it cannot be split further. A linear factor needs a constant on top; an irreducible quadratic factor needs a linear expression on top. So write (3x² + 5x − 2)/[(x+1)(x²+4)] ≡ A/(x+1) + (Bx + C)/(x²+4). Multiply both sides by (x+1)(x²+4) to clear all the denominators: 3x² + 5x − 2 ≡ A(x²+4) + (Bx + C)(x+1). This is an identity, so it holds for every value of x. Step 1: choose x = −1, because that makes (x + 1) zero and wipes out the (Bx + C) term. 3(−1)² + 5(−1) − 2 = A((−1)² + 4) 3 − 5 − 2 = A(1 + 4) −4 = 5A A = −4/5 Step 2: expand the right-hand side and compare coefficients. A(x²+4) + (Bx + C)(x+1) = Ax² + 4A + Bx² + Bx + Cx + C = (A + B)x² + (B + C)x + (4A + C) Compare the x² terms: 3 = A + B, so B = 3 − A = 3 − (−4/5) = 3 + 4/5 = 19/5. Compare the constant terms: −2 = 4A + C = 4(−4/5) + C = −16/5 + C, so C = −2 + 16/5 = −10/5 + 16/5 = 6/5. Check with the x terms: B + C = 19/5 + 6/5 = 25/5 = 5, which matches the 5x on the left, so the values are right. Put the values back in: = (−4/5)/(x+1) + ((19/5)x + 6/5)/(x²+4) Tidy each fraction by taking the 1/5 outside: = −4/[5(x+1)] + (19x + 6)/[5(x²+4)]