Solving Cubic Equations

Remark 1. This writeup is a fresh writeup because somehow, and tragically, the original post got permanently deleted.

Our goal is to answer a simple question: how do we solve the cubic equation? Namely, given constants $a,b,c,d$ with $a \neq 0$ , if we know that the real number $x$ satisfies the equation

$ax^3 + bx^2 + cx + d = 0,$

how might we determine the possible values of $x$ ? Firstly, we had better be sure that this equation can be solved, and the power of $3$ is the ticket to why that holds.

Theorem 1. The cubic equation $ax^3 + bx^2 + cx + d = 0$ has at least one real solution.

Proof. Omitted; we can achieve this result by adapting the proof of Theorem 2 in this post.

There is a general “cubic” formula, but we shall work with special cases in which the root in Theorem 1 could be obtained with trial-and-error.

We abbreviate the left-hand side by $f(x) = ax^3 + bx^2 + cx + d$ . Then, we call $\alpha$ a root of the polynomial $f(x)$ if and only if $f(\alpha) = 0$ . In particular,

$a \alpha^3 + b \alpha^2 + c \alpha + d = f(\alpha) = 0.$

Lemma 1. For any real number $\alpha$ ,

$x^3 \pm \alpha^3 = (x \pm \alpha) (x^2 \mp \alpha x + \alpha^2).$

Proof. Firstly, we expand the right-hand side to obtain

$\begin{aligned} (x -\alpha) (x^2 +\alpha x + \alpha^2) &= x^3 + (\alpha - \alpha)x^2 + (\alpha^2 - \alpha^2)x + (-\alpha^3) \\ &= x^3 - \alpha^3. \end{aligned}$

Then we replace $\alpha$ with $-\alpha$ to obtain

$\begin{aligned} x^3 + \alpha^3 &= x^3 - (-\alpha^3) \\ &= x^3 - (-\alpha)^3 \\ &= (x -(-\alpha)) (x^2 +(-\alpha) x + (-\alpha)^2) \\ &= (x+\alpha)(x-\alpha x + \alpha^2). \end{aligned}$

Lemma 2. For any real number $\alpha$ , there exist unique real constants $p,q$ such that

$f(x) - f(\alpha) = a (x-\alpha) (x^2 + kx + m) .$

Proof. Making the substitutions and applying Lemma 1,

$\begin{aligned} f(x) - f(\alpha) &= (a x^3 + b x^2 + c x + d) - (a \alpha^3 + b \alpha^2 + c \alpha + d) \\ &= a(x^3 - \alpha^3) + b(x^2-\alpha^2) + c(x-\alpha) + d(1-1) \\ &= a(x-\alpha)(x^2 + \alpha x + \alpha^2) + b(x-\alpha)(x+\alpha) + c(x-\alpha) \\ &= (x-\alpha)( a (x^2 + \alpha x + \alpha^2) + b (x+\alpha) + c ) \\ &= a(x-\alpha)\left[ (x^2 + \alpha x + \alpha^2) + \frac ba (x+ \alpha) + \frac ca \right] \\ &= a(x-\alpha)\left[ x^2 + \alpha x + \alpha^2 + \frac ba x+ \frac ba \alpha + \frac ca \right] \\ &= a(x-\alpha)\left[ x^2 + \left( \alpha + \frac ba \right) x + \left( \alpha^2 + \frac ba \alpha + \frac ca \right) \right] \\ &= a(x-\alpha) ( x^2 + k x + m ), \end{aligned}$

where we set $k := \alpha+ b/a$ and $m := \alpha^2 + b\alpha/a + c/a$ .

Lemma 2 paves the way for the special case of the remainder and factor theorems—the general case follows a factorisation of $x^n - \alpha^n$ by generalising the argument in Lemma 1.

Theorem 1. Given any real number $\alpha$ , there exists a unique polynomial $q(x)$ and a unique real number $r$ , called the remainder, such that

$f(x) = (x-\alpha)q(x) + r.$

Furthermore, $r = f(\alpha)$ —this result is known as the remainder theorem. Finally, this remainder equals zero if and only if $\alpha$ is a root of $f(x)$ —this result is known as the factor theorem. In this case, we say that $x-\alpha$ is a factor of the polynomial $f(x)$ .

Proof. By Lemma 2,

$f(x) = a(x-\alpha) ( x^2 + k x + m ) + f(\alpha),$

yielding $q(x) = a(x^2 + kx + m)$ and $r = f(\alpha)$ .

For uniqueness, set $x = \alpha$ to obtain $f(\alpha) = r$ . Then

$\begin{aligned} (x-\alpha)q(x) &= f(x) - r \\ &= f(x)-f(\alpha) \\ &= a(x-\alpha) ( x^2 + k x + m ). \end{aligned}$

For $x \neq \alpha$ , dividing by $x-\alpha$ yields $q(x) = x^2 + k x + m$ . For a more complete argument regarding uniqueness, see this post.

While we have discussed the result in an overly abstract manner, perhaps it is worth elucidating the theory with an example.

Example 1. Solve the cubic equation $x^3 - 2x^2 - 3x + 4 = 0$ .

Solution. Define $f(x) := x^3 - 2x^2 - 3x + 4$ . By almost-obvious observation,

$f(1) = 1^3 - 2\cdot 1^2 - 3\cdot 1 + 4 = 1 - 2 - 3 + 4 = 0.$

Therefore, by the factor theorem, $x-1$ is a factor of $f(x)$ . Hence, there exist constants $a,b,c$ such that

$\begin{aligned} f(x) &= (x-1)(ax^2 +bx + c) \\ x^3 - 2x^2 - 3x + 4 &= ax^3 + (b-a)x^2 +(c-b)x - c. \end{aligned}$

Comparing the coefficients of $x^3$ and the constant term respectively, $a = 1$ and $c = -4$ . Comparing the coefficient of $x^2$ ,

$b-a = -2 \quad \Rightarrow \quad b = a-2 = 1-2 = -1.$

Therefore,

$x^3 - 2x^2 - 3x + 4 = (x-1)(x^2 - x - 4).$

Since $x^3 - 2x^2 - 3x + 4 = 0$ , we must have either

$x - 1 = 0\quad \text{or} \quad x^2 - x - 4 = 0.$

In the former, $x = 1$ . In the latter, we solve using the quadratic formula: first compute the discriminant $\Delta = (-1)^2 - 4 \cdot 1 \cdot (-4) = 17$ . Then

$\displaystyle x = \frac{-(-1) \pm \sqrt{17}}{2 \cdot 1} = \frac 12 \pm \frac 12 \sqrt{17}.$

Therefore, $x = 1$ or $x = \frac 12 \pm \frac 12 \sqrt{17}$ .

Remark 2. Solutions can take the form of surds.

Example 2. Solve the cubic equation $4x^3 - 3x^2 - 2x + 1 = 0$ .

Solution. We leave a similar-to-Example 1 solution left as an exercise to the reader. By taking advantage of the similar numbers, we divide the equation by $x^3$ :

$\displaystyle 4 - \frac 3x - \frac 2{x^2} + \frac 1{x^3} = 0.$

Now make the substitution $u = 1/x$ to obtain a rather suspect equation:

$\displaystyle 4 - 3u - 2u^2 + u^3 = 0.$

By re-arranging the terms, we obtain the exact same equation as that in Example 1, just with different letters:

$u^3 - 2u^2 - 3u + 4 = 0.$

Therefore, by Example 1, we have $u = 1$ or $u = \frac 12(1 \pm \sqrt{17})$ . Since $u = 1/x \Rightarrow x = 1/u$ , the former yields $x = 1/1 = 1$ , and the latter takes a bit more work by rationalising the denominator:

$\begin{aligned} x = \frac 1u = \frac 2{1 \pm \sqrt{17}} &= \frac 2{1 \pm \sqrt{17}} \cdot \frac{1 \mp \sqrt{17}}{1 \mp \sqrt{17}} \\ &= \frac 2{1^2 - 17} \cdot (1 \mp \sqrt{17})\\ &= -\frac 1{8} \cdot (1 \mp \sqrt{17}) \\ &= \frac 1{8} \cdot (-1 \pm \sqrt{17}). \end{aligned}$