O-Level Polynomials and partial fractions: worked solution

Question

Explain why f(x) = 3x³ + 2x² + 16 = 0 has only one real root. [5]

Every cubic has at least one real root, so to show there is exactly one we factorise f(x) and prove the remaining quadratic factor has no real roots. Step 1 - find a root by the factor theorem. Any integer root must divide the constant term 16, so test x = ±1, ±2, ±4, ±8, ±16. f(1) = 3 + 2 + 16 = 21, not zero. f(−1) = −3 + 2 + 16 = 15, not zero. f(−2) = 3(−2)³ + 2(−2)² + 16 = 3(−8) + 2(4) + 16 = −24 + 8 + 16 = 0 Since f(−2) = 0, the factor theorem says (x + 2) is a factor. Step 2 - divide out. Write the cubic with every power present: 3x³ + 2x² + 0x + 16, and divide by (x + 2): 3x³ ÷ x = 3x²; 3x²(x + 2) = 3x³ + 6x²; subtracting leaves −4x² + 0x + 16 −4x² ÷ x = −4x; −4x(x + 2) = −4x² − 8x; subtracting leaves 8x + 16 8x ÷ x = 8; 8(x + 2) = 8x + 16; subtracting leaves 0 So f(x) = (x + 2)(3x² − 4x + 8). Step 3 - test the quadratic factor with the discriminant b² − 4ac, using a = 3, b = −4, c = 8: b² − 4ac = (−4)² − 4 × 3 × 8 = 16 − 96 = −80 Because −80 < 0, the equation 3x² − 4x + 8 = 0 has no real solutions. Step 4 - conclude. As f(x) = (x + 2)(3x² − 4x + 8) and the quadratic factor is never zero for real x, the only way to make f(x) = 0 is x + 2 = 0. So x = −2 is the only real root.