Cubic Functions and Equations

You can often answer a question about the roots of a cubic without ever finding them. The three relationships below turn "the roots" into three numbers you can read straight off the coefficients.

The shape of a cubic

$$f(x) = ax^3 + bx^2 + cx + d \qquad (a \ne 0)$$

The end behaviour depends only on the sign of $a$:

• $a > 0$: as $x \to -\infty$, $y \to -\infty$; as $x \to +\infty$, $y \to +\infty$. The curve rises from bottom left to top right.
• $a < 0$: the reverse — down from top left to bottom right.

Because the curve runs from $-\infty$ to $+\infty$ and is continuous, it must cross the $x$-axis at least once.

A cubic has at most two turning points — sometimes two, sometimes none (as with $y = x^3$, which merely flattens at the origin).

Roots and coefficients

If $ax^3 + bx^2 + cx + d = 0$ has roots $\alpha$, $\beta$ and $\gamma$, then

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

These come from expanding $a(x-\alpha)(x-\beta)(x-\gamma)$ and comparing coefficients.

For a quadratic $ax^2 + bx + c = 0$ the corresponding pair is

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

For $2x^3 - 5x^2 + 3x - 7 = 0$:

$$\alpha + \beta + \gamma = -\frac{-5}{2} = \frac52$$ $$\alpha\beta + \beta\gamma + \gamma\alpha = \frac{3}{2}$$ $$\alpha\beta\gamma = -\frac{-7}{2} = \frac72$$

Check with a known cubic. $x^3 - 6x^2 + 11x - 6 = 0$ factorises as $(x-1)(x-2)(x-3)$, so the roots are 1, 2 and 3.

• Sum: $1 + 2 + 3 = 6$, and $-\frac{-6}{1} = 6$ ✓
• Pairs: $(1)(2) + (2)(3) + (3)(1) = 2 + 6 + 3 = 11$, and $\frac{11}{1} = 11$ ✓
• Product: $(1)(2)(3) = 6$, and $-\frac{-6}{1} = 6$ ✓

Symmetric functions of the roots

Most questions ask for an expression in $\alpha, \beta, \gamma$ that is symmetric — unchanged if the roots are shuffled. Every such expression can be rebuilt from the three quantities above.

Sum of squares.

$$\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)$$

For $2x^3 - 5x^2 + 3x - 7 = 0$:

$$= \left(\frac52\right)^2 - 2\left(\frac32\right) = \frac{25}{4} - 3 = \frac{13}{4}$$

Sum of reciprocals.

$$\frac1\alpha + \frac1\beta + \frac1\gamma = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma}$$

$$= \frac{3/2}{7/2} = \frac37$$

Forming a new equation

To build the cubic whose roots are $2\alpha$, $2\beta$, $2\gamma$, compute the three symmetric quantities for the new roots.

Starting from $2x^3 - 5x^2 + 3x - 7 = 0$, where the sums are $\frac52$, $\frac32$ and $\frac72$:

$$2\alpha + 2\beta + 2\gamma = 2\left(\frac52\right) = 5$$ $$(2\alpha)(2\beta) + \cdots = 4\left(\frac32\right) = 6$$ $$(2\alpha)(2\beta)(2\gamma) = 8\left(\frac72\right) = 28$$

A monic cubic with roots $p, q, r$ is

$$x^3 - (\text{sum})x^2 + (\text{sum of pairs})x - (\text{product}) = 0$$

so the new equation is

$$x^3 - 5x^2 + 6x - 28 = 0$$

Sketching a cubic

Four things make a sketch complete:

1. End behaviour from the sign of $a$.
2. The $y$-intercept, which is $d$.
3. The roots, from factorising — try the factors of $d$ and use the factor theorem.
4. The turning points, if the question wants them, from calculus.

For $y = x^3 - 6x^2 + 11x - 6$: positive $a$ so bottom-left to top-right; $y$-intercept $(0,-6)$; roots at $x = 1, 2, 3$.

The nature of the roots

A cubic with real coefficients has either:

• three real roots (possibly with repeats), or
• one real root and two complex conjugate roots.

Complex roots of a polynomial with real coefficients always come in conjugate pairs, so they arrive two at a time. With only three roots to distribute, at least one must be real.

Graphically: three distinct real roots means three crossings; one real root means the curve crosses once and its two turning points sit on the same side of the axis (or it has none at all).