Complex Roots of a Cubic

One non-real root brings its conjugate along.

A cubic and one of its roots

The cubic z³ − 5z² + 17z − 13 = 0 has real coefficients: 1, −5, 17 and −13. One of its roots is given: z = 2 + 3i. The task is to find the other two.

The given root can be checked by substitution. (2 + 3i)² = 4 + 12i + 9i² = −5 + 12i, and (2 + 3i)³ = (−5 + 12i)(2 + 3i) = −10 − 15i + 24i + 36i² = −46 + 9i. So the cubic at 2 + 3i is (−46 + 9i) − 5(−5 + 12i) + 17(2 + 3i) − 13. The real parts give −46 + 25 + 34 − 13 = 0, and the imaginary parts give 9 − 60 + 51 = 0.

Its conjugate is a root too

When every coefficient is real, the non-real roots come in conjugate pairs, as they did for the quadratic. So 2 − 3i is a root as well.

Here is why. The conjugate of a sum is the sum of the conjugates, and the conjugate of a product is the product of the conjugates. Take the conjugate of every term in the working above: the real coefficients stay as they are, 2 + 3i becomes 2 − 3i, and the total 0 stays 0. So putting 2 − 3i into the cubic also gives 0.

Multiply the pair of factors

Each root gives a factor: z − (2 + 3i) and z − (2 − 3i). Group each as (z − 2) and 3i: their product is ((z − 2) − 3i)((z − 2) + 3i) = (z − 2)² − 9i² = (z − 2)² + 9 = z² − 4z + 13.

The i has gone, as it does whenever a conjugate pair is multiplied. The quick way: the roots add to 4 and multiply to 4 + 9 = 13, so the quadratic is z² − 4z + 13.

Divide out the quadratic

The cubic is z² − 4z + 13 times a linear factor z + c. Multiply that out: (z² − 4z + 13)(z + c) = z³ + (c − 4)z² + (13 − 4c)z + 13c.

The z² terms give c − 4 = −5, so c = −1. The constants check it: 13c = −13. So do the z terms: 13 − 4c = 13 + 4 = 17. The linear factor is z − 1.

Long division gives the same. z³ / z² = z, and z(z² − 4z + 13) = z³ − 4z² + 13z; subtracting that from the cubic leaves −z² + 4z − 13, which is −1 times z² − 4z + 13, with remainder 0.

The three roots

So z³ − 5z² + 17z − 13 = (z² − 4z + 13)(z − 1), and the roots are 2 + 3i, 2 − 3i and 1. Check the real one: 1 − 5 + 17 − 13 = 0.

xy1

The curve y = x³ − 5x² + 17x − 13 crosses the x-axis once, at x = 1, the real root. The roots 2 ± 3i are not real numbers, so they are not crossings.

Or find the real root first

With no root given, the factor theorem finds the real root. Any whole-number root divides the constant term 13, so try ±1 and ±13. At z = 1 the cubic is 1 − 5 + 17 − 13 = 0, so z − 1 is a factor.

Dividing by z − 1 leaves z² − 4z + 13. Its discriminant is 16 − 52 = −36, so z = (4 ± 6i)/2 = 2 ± 3i. The two routes meet at the same three roots.

One real root, or three

A cubic has three roots, counting a repeated root twice. With real coefficients, its non-real roots come two at a time, so it has either none or two of them. That leaves two possibilities: three real roots, or one real root and a conjugate pair.

It cannot have three non-real roots, because the third would need a conjugate of its own, a fourth root. So the graph of a real cubic always crosses the x-axis at least once. x³ − 2x² − 5x + 6 crosses three times, at −2, 1 and 3; z³ − 5z² + 17z − 13 crosses once.

Another cubic

z³ − 3z² + 4z − 2 = 0 has real coefficients and the root 1 + i. So 1 − i is a root too. The pair adds to 2 and multiplies to 1 + 1 = 2, which gives the factor z² − 2z + 2.

Matching coefficients, (z² − 2z + 2)(z − 1) = z³ − 3z² + 4z − 2, so the third root is 1. The roots are 1 + i, 1 − i and 1.

The usual mistakes

Changing both signs. −2 − 3i is −(2 + 3i), not its conjugate. The conjugate changes only the sign of the imaginary part: 2 − 3i.

Swapping the parts. 3 + 2i is a different number; the conjugate of 2 + 3i keeps the real part 2.

Getting a sign wrong in the quadratic. Roots adding to 4 give −4z, not +4z, and the product 13 is +13, because the i² has already turned −9i² into +9.

Reading the root of z − 1 as −1. z − 1 = 0 when z = 1.

Expecting one or three non-real roots. Non-real roots come in pairs, so a real cubic has none or two.

A tank that holds 100 liters

In the application below, a tank with a square base of side x dm and height x − 1 dm must hold 100 liters, which gives x³ − x² − 100 = 0. The real root comes first, by the factor theorem, and the quadratic left over shows that the other two roots are a conjugate pair, so only one tank fits.

Worked example: A Fish Tank That Must Hold 100 Liters: Why Only One Size Works

Question An aquarium maker wants a tank with a square base of side x dm and a height 1 dm less than the side, holding exactly 100 liters, where 1 liter is 1 dm3. (a) Show that x3 − x2 − 100 = 0, and find a real root by the factor theorem. (b) Find the other two roots, and explain why exactly one tank has this shape and this capacity.

  1. 1.The volume is the base area times the height: x2(x − 1) = 100, so x3 − x2 − 100 = 0.

    xx2(x − 1) = 100, x3− x2− 100 = 0
    xx2(x − 1) = 100, x3− x2− 100 = 0
    The base area x2 times the height x − 1 is 100 liters.

  2. 2.Try factors of 100. At x = 4 the cubic is 64 − 16 − 100 = −52, and at x = 5 it is 125 − 25 − 100 = 0. (a) By the factor theorem (x − 5) is a factor, and x = 5 is a root.

    x5x2(x − 1) = 100, x3− x2− 100 = 0x = 5: 125 − 25 − 100 = 0
    x5x2(x − 1) = 100, x3− x2− 100 = 0x = 5: 125 − 25 − 100 = 0
    (a) x = 5 makes the cubic zero, so (x − 5) is a factor.

  3. 3.Divide by (x − 5): x3 − x2 − 100 = (x − 5)(x2 + 4x + 20).

    x5x2(x − 1) = 100, x3− x2− 100 = 0x = 5: 125 − 25 − 100 = 0= (x − 5)(x2+ 4x + 20)
    x5x2(x − 1) = 100, x3− x2− 100 = 0x = 5: 125 − 25 − 100 = 0= (x − 5)(x2+ 4x + 20)
    Dividing by (x − 5) leaves the quadratic x2 + 4x + 20.

  4. 4.Solve the quadratic. The discriminant is 16 − 80 = −64 and √−64 = 8i, so x = −4 ± 8i2 = −2 ± 4i.

    x5x2(x − 1) = 100, x3− x2− 100 = 0x = 5: 125 − 25 − 100 = 0= (x − 5)(x2+ 4x + 20)16 − 80 = −64, x = −2 ± 4i
    x5x2(x − 1) = 100, x3− x2− 100 = 0x = 5: 125 − 25 − 100 = 0= (x − 5)(x2+ 4x + 20)16 − 80 = −64, x = −2 ± 4i
    Its discriminant is −64, and √−64 = 8i, so x = −2 ± 4i.

  5. 5.(b) The other roots are −2 + 4i and −2 − 4i, a conjugate pair. A side must be a real, positive length, so only x = 5 gives a tank: 5 dm by 5 dm by 4 dm, which is 50 cm by 50 cm by 40 cm. Check: 52 × 4 = 100.

    x5one crossingx2(x − 1) = 100, x3− x2− 100 = 0x = 5: 125 − 25 − 100 = 0= (x − 5)(x2+ 4x + 20)16 − 80 = −64, x = −2 ± 4ionly x = 5: 5 dm by 5 dm by 4 dm
    x5one crossingx2(x − 1) = 100, x3− x2− 100 = 0x = 5: 125 − 25 − 100 = 0= (x − 5)(x2+ 4x + 20)16 − 80 = −64, x = −2 ± 4ionly x = 5: 5 dm by 5 dm by 4 dm
    (b) The graph crosses the axis once: x = 5 is the only real root, so there is one tank.

Answer: (a) x = 5; (b) x = −2 ± 4i, which are not lengths, so the only tank is 5 dm by 5 dm by 4 dm (50 cm by 50 cm by 40 cm)

Common mistakes

  • Stopping at x = 5 without showing that no other root is real. A cubic can have three real roots; here the quadratic factor has a negative discriminant, and that is what makes the tank the only one.
  • Writing the quotient as x2 − 4x + 20. Multiplying back, (x − 5)(x2 − 4x + 20) = x3 − 9x2 + 40x − 100, which has −9x2 instead of −x2, so the middle term must be +4x.

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