O-Level Polynomials and partial fractions: worked solution

Question

Hence solve 4x³ + ax + b = 0, writing the non-integer roots as (c + √d)/2.

The values found in part (a) are a = -9 and b = 5, so the equation to solve is 4x³ - 9x + 5 = 0. The word 'Hence' means: use the factor found in part (a) rather than starting again. Step 1 - confirm the factor with the factor theorem. The factor theorem says (x - k) is a factor of f(x) exactly when f(k) = 0. Try the simplest candidate, k = 1: f(1) = 4(1)³ - 9(1) + 5 = 4 - 9 + 5 = 0. Since f(1) = 0, (x - 1) is a factor and x = 1 is one root. Step 2 - divide out that factor. Because the cubic starts with 4x³ and there is no x² term, write 4x³ + 0x² - 9x + 5 = (x - 1)(4x² + px + q) and compare coefficients: x³ terms: 4 = 4, which fixes the leading 4x² in the bracket. constant terms: -q = 5, so q = -5. x² terms: on the right the x² terms are px² from x × px and -4x² from -1 × 4x², so p - 4 = 0, giving p = 4. Therefore 4x³ - 9x + 5 = (x - 1)(4x² + 4x - 5). Step 3 - solve the quadratic factor. Setting 4x² + 4x - 5 = 0, there is no pair of whole numbers that works, so use the formula with a = 4, b = 4, c = -5: x = (-b ± √(b² - 4ac)) / (2a) x = (-4 ± √(4² - 4 × 4 × (-5))) / (2 × 4) x = (-4 ± √(16 + 80)) / 8 x = (-4 ± √96) / 8. Simplify the surd: √96 = √(16 × 6) = √16 × √6 = 4√6. x = (-4 ± 4√6)/8. Every term has a factor 4, so divide top and bottom by 4: x = (-1 ± √6)/2. So the roots are x = 1, x = (-1 + √6)/2 and x = (-1 - √6)/2, the two non-integer roots being of the required form (c + √d)/2 with c = -1 and d = 6.