O-Level Polynomials and partial fractions: worked solution

Question

Solve 3x³ − 11x² + 3x + 2 = 0, giving any non-integer roots in surd form.

Step 1: Hunt for a first root using the factor theorem. Let f(x) = 3x³ − 11x² + 3x + 2. The factor theorem says that if f(p/q) = 0 then (qx − p) is a factor of f(x). Any rational root must have p a factor of the constant term 2 and q a factor of the leading coefficient 3, so the candidates to test are ±1, ±2, ±1/3, ±2/3 Test them in turn: f(1) = 3 − 11 + 3 + 2 = −3, not zero f(2) = 3(8) − 11(4) + 6 + 2 = 24 − 44 + 8 = −12, not zero f(−1) = −3 − 11 − 3 + 2 = −15, not zero f(2/3) = 3(8/27) − 11(4/9) + 3(2/3) + 2 = 8/9 − 44/9 + 2 + 2 = −36/9 + 4 = −4 + 4 = 0 ✓ Step 2: Turn that root into a factor. Since f(2/3) = 0 with 2/3 = p/q where p = 2 and q = 3, the factor is (3x − 2). Step 3: Divide f(x) by (3x − 2) to find the other factor. By long division: 3x³ ÷ 3x = x²; x²(3x − 2) = 3x³ − 2x²; subtracting leaves −9x² + 3x + 2 −9x² ÷ 3x = −3x; −3x(3x − 2) = −9x² + 6x; subtracting leaves −3x + 2 −3x ÷ 3x = −1; −1(3x − 2) = −3x + 2; subtracting leaves 0 So 3x³ − 11x² + 3x + 2 = (3x − 2)(x² − 3x − 1) (Check by expanding: (3x − 2)(x² − 3x − 1) = 3x³ − 9x² − 3x − 2x² + 6x + 2 = 3x³ − 11x² + 3x + 2 ✓) Step 4: Solve the quadratic factor. x² − 3x − 1 = 0 has no integer factor pair (nothing multiplies to −1 and adds to −3), so use the quadratic formula with a = 1, b = −3, c = −1: x = (−b ± √(b² − 4ac))/(2a) x = (3 ± √((−3)² − 4(1)(−1)))/2 x = (3 ± √(9 + 4))/2 x = (3 ± √13)/2 13 has no square factor, so √13 will not simplify - this is the required surd form. Step 5: Collect all three roots. From 3x − 2 = 0: x = 2/3 From the quadratic: x = (3 + √13)/2 or x = (3 − √13)/2 x = 2/3, x = (3 + √13)/2 or x = (3 − √13)/2.