All subjects · Further Mathematics · WFM01 Further Pure Mathematics 1

Quartic equations

Edexcel IAL Further Mathematics · WFM01 Further Pure Mathematics 1

A quartic has four roots. With real coefficients, the number of real roots is 4, 2 or 0. F1 questions usually give you either one complex root or two real roots, and you find the other two.

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Given one complex root

Its conjugate is also a root. Form the real quadratic factor, then find the other quadratic factor by division or by comparing coefficients, and solve it.

Specification example: 2 + i is a root of x4 - x3 - 5x2 + 7x + 10 = 0. The factor is x2 - 4x + 5, and x4 - x3 - 5x2 + 7x + 10 = (x2 - 4x + 5)(x2 + 3x + 2), so the other roots are -1 and -2.

Given two real roots

If x = a and x = b are roots, (x - a)(x - b) is a factor. Divide to get the other quadratic factor and solve it.

Specification example: x = 1 and x = -2 are roots of x4 - x3 + 6x2 + 14x - 20 = 0. The factor is x2 + x - 2, the other factor is x2 - 2x + 10, so the other roots are 1 ± 3i.

Comparing coefficients

For (x2 + px + q)(x2 + sx + t), the x3 coefficient is p + s and the constant is qt. These two usually give the unknown factor quickly; check with the x coefficient.

Worked example

Given that 2 + i is a root of z4 - 2z3 - z2 + 2z + 10 = 0, solve the equation.

  1. 2 - i is also a root, giving the factor z2 - 4z + 5.
  2. Write the quartic as (z2 - 4z + 5)(z2 + sz + t). Constants: 5t = 10, so t = 2. z3 terms: s - 4 = -2, so s = 2.
  3. Check z terms: 5s - 4t = 10 - 8 = 2. Correct.
  4. z2 + 2z + 2 = 0 gives (z + 1)2 = -1, so z = -1 ± i.

Answer: z = 2 ± i, z = -1 ± i

Key idea: Quartic with real coefficients: roots in conjugate pairs, so 0, 2 or 4 are real. Pair roots into real quadratic factors.

Check yourself

  1. 3i is a root of z4 + 2z3 + 14z2 + 18z + 45 = 0. The other two roots are p ± qi with q > 0. Find p.
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    -1. -3i is also a root, giving the factor z2 + 9. The other factor is z2 + 2z + 5 (compare z3 terms and constants), so z = -1 ± 2i.

  2. 3i is a root of z4 + 2z3 + 14z2 + 18z + 45 = 0. The other two roots are p ± qi with q > 0. Find q.
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    2. z2 + 2z + 5 = 0 gives (z + 1)2 = -4, so z = -1 ± 2i and q = 2.

  3. A quartic z4 + az3 + bz2 + cz + d has roots 1 + i and 2 - 3i, and real coefficients. Find d.
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    26. The factors are z2 - 2z + 2 (from 1 ± i) and z2 - 4z + 13 (from 2 ± 3i). d is the product of the constants: 2 × 13 = 26.

  4. The same quartic, with roots 1 ± i and 2 ± 3i. Find a, the coefficient of z3.
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    -6. (z2 - 2z + 2)(z2 - 4z + 13): the z3 terms are -4z3 - 2z3 = -6z3, so a = -6.

  5. Which of these cannot be the number of real roots of a quartic equation with real coefficients?
    • 4
    • 3
    • 2
    • 0
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    3. The 4 roots include non-real roots in conjugate pairs, so the number of real roots is 4, 2 or 0. Three real roots would leave one unpaired non-real root.

  6. x = 1 and x = -2 are roots of x4 - x3 + 6x2 + 14x - 20 = 0. Find the other quadratic factor, in the form x2 + bx + c.
    Show answer

    x2 - 2x + 10. Divide by (x - 1)(x + 2) = x2 + x - 2. Constants: -2c = -20 gives c = 10; x3 terms: b + 1 = -1 gives b = -2.

  7. Solving x2 - 2x + 10 = 0 gives the last two roots of that quartic. Find the positive imaginary part.
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    3. (x - 1)2 + 9 = 0, so x = 1 ± 3i.

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Last updated 8 November 2026. Spotted a mistake? Tell us.