Cubic equations and the factor theorem

What a cubic gives you

• A cubic $az^3 + bz^2 + cz + d = 0$ has exactly three roots, counted with repetition.
• If the coefficients are real, complex roots come in conjugate pairs. Since three is odd, a real cubic has:
  • three real roots, or
  • one real root and one conjugate pair of complex roots
• A real cubic can never have exactly two complex roots and one other complex root — the pairing forbids it. That fact alone answers several exam questions.

The remainder theorem

• When a polynomial $P(z)$ is divided by $(z - k)$, the remainder is $P(k)$.
• So you can find a remainder by substituting, without doing any division at all.
• If $P(z) = z^3 - 2z + 5$, the remainder on dividing by $(z - 2)$ is $P(2) = 8 - 4 + 5 = 9$.

The factor theorem

• The special case where the remainder is zero:

$$(z - k) \text{ is a factor of } P(z) \quad \Longleftrightarrow \quad P(k) = 0$$

• Substituting a value and getting zero proves you have found a root and therefore a factor.
• This is the standard route into a cubic: find one root by trial, then divide out and solve the quadratic that remains.

Finding the first root

• For a cubic with integer coefficients, any rational root has a numerator dividing the constant term and a denominator dividing the leading coefficient.
• In practice: try the factors of the constant term — $\pm 1, \pm 2, \pm 3, \ldots$ — until one gives $P(k) = 0$.
• For $z^3 - 3z^2 + 4z - 12 = 0$, the constant is $-12$, so try $\pm1, \pm2, \pm3, \pm4, \pm6, \pm12$. $P(3) = 27 - 27 + 12 - 12 = 0$, so $(z - 3)$ is a factor.

Dividing out the factor

• Once you have a factor, divide to reduce the cubic to a quadratic.
• Comparing coefficients is usually quicker than long division. Write:

$$z^3 - 3z^2 + 4z - 12 = (z - 3)(z^2 + pz + q)$$

• Expand the right-hand side and match coefficients term by term:
  • $z^2$ terms: $p - 3 = -3$, so $p = 0$
  • constant terms: $-3q = -12$, so $q = 4$
• Giving $(z - 3)(z^2 + 4) = 0$, and $z^2 = -4$ so $z = \pm 2i$.
• The three roots are $3$, $2i$ and $-2i$ — one real, one conjugate pair, as predicted.

When you are given a complex root

• This is the common exam form: "one root of $z^3 + az^2 + bz + c = 0$ is $2 - i$; find $a$, $b$ and $c$".

• The method:

  • Write down the conjugate. $2 + i$ is also a root, because the coefficients are real.
  • Build the quadratic factor from that pair, using sum and product:
    • Sum $= 4$, product $= 4 + 1 = 5$, so the factor is $z^2 - 4z + 5$.
  • Divide the cubic by that quadratic, or write the cubic as $(z^2 - 4z + 5)(z - k)$ and expand.
  • Compare coefficients to find $k$ and the unknowns.

• Note the shortcut: forming the real quadratic factor from the conjugate pair avoids ever doing arithmetic with $i$ in the division step.

Sum and product of roots for a cubic

• For $az^3 + bz^2 + cz + d = 0$ with roots $\alpha$, $\beta$, $\gamma$:

$$\alpha + \beta + \gamma = -\frac{b}{a} \qquad \alpha\beta\gamma = -\frac{d}{a}$$

• These give quick checks, and sometimes a quick route to a third root when you already know two.