AN EASY LOOK AT THE CUBIC FORMULA

Introduction

All students learn the quadratic formula for finding the roots of a quadratic
equation. The cubic formula for solving cubic polynomials is seldom used, even though it
has been known since the 1545 when Girolamo Cardano published his Ars Magna [2].
This cubic formula, like the quadratic formula, gives the exact answer in closed form.
Fifty years ago, when this author was a schoolboy, algebra text books frequently included
a detailed discussion of the cubic formula. Precalculus texts of today rarely consider the
subject. Why? Because the cubic formula, unlike the quadratic formula, frequently
involves awkward cube roots of complex numbers. Besides, excellent numerical methods
are available, such as Newton’s iterative method, which converge very rapidly to
approximations with many accurate digits . However, there are cases where the exact
closed form answer is appealing, and where the effort involved in using the cubic formula
is not overwhelming.

It is the purpose of this brief note to show how the cubic formula can be presented
easily at the precalculus level. While n one of this material is new, the selection of items
and their presentation is designed to avoid the difficulties mentioned above. We give a
nice canonical form for the cubic formula that is relatively easy to remember. We show
how to verify that the formula is correct, and we identify when it is profitable to use it.

The cubic formula in simplest form

To solve the cubic equation

we must first remove the quadratic term. This is always achieved with the substitution

Substituting (2) into (1) we get

where

(Equation (3) is known as the reduced cubic .) Now we can write our cubic formula for
the real root of (3).

Theorem:

Let

Then a real root of x³ −3cx − 2a = 0 is

(Cubic formula)

We interpret all the roots in (6) as real numbers. (Actually, with proper
interpretation of the radicals involved , formula (6) can give all three roots of (3)
regardless of the values of the coefficients c and a. This does get messy when b < 0 , and
we will not consider that case here.)

Proof:
First notice that

Cubing both sides of (6) we get

Using (6) and (7) we can rewrite this last expression as
x³ = 3cx + 2a .
Thus we have verified that (6) is a root of (3) and the theorem is proved.

Examples

Example 1: Find a real root of x³ + 3x² + 6y + 2 = 0 .
Solution: Comparing our problem with (1), we see that p = 3 so we begin by making the
substitution (2) y = x − p / 3 = x −1. This converts the original problem to x³ + 3x − 2 = 0 ,
in which the quadratic term does not appear. Comparing this with (3) we see that c = −1
and a =1. From (5) we get b = a² − c³ = 2 . Since b > 0 our theorem says that (6) gives us

a real root . Since the second cube root is negative, it is best

written as . Finally, a root of our original cubic is given by

Example 2: Find a real root of y³ − 7y² +14y − 20 = 0 .

Solution: We compare our problem with (1) and see that p = −7 . We start with the

substitution (2) Our original equation is now reduced to

in which the quadratic term has been removed.

Comparing this last equation with (3) we see that Calculating b from

(5) we get . Since b is positive , a real root of our cubic is given by

(6) as Finally, a root of our original cubic is
given by Since this root is an integer, it is easy to find the other

two roots by dividing our original cubic y³ − 7y² +14y − 20 by y − 5 . This gives us the
quadratic y² − 2y + 4 = 0 which has roots

Example 3: Find a real root of x³ − x −1 = 0

Solution: This problem comes from the interesting article [6] in which the “plastic
number” is defined as the root of our equation. Our equation has no quadratic term, so
there is no need to use the linear substitution . Comparing our equation with (3) we see
that Using (4) we get Since b is positive,
we can use (6) to get the root

Example 4: It is clear that x = 1 is a root of the cubic x³ + 3x − 4 = 0 . Use the cubic
formula to obtain a surprising expression for this root.
Solution: Comparing our cubic with (3) we see at once that c = −1 and a = 2 .
Calculating b we get b = a² − c³ = 5 . Our cubic formula now gives us

This does not look like x = 1, but a quick
check with a calculator helps to convince us that it probably is 1. The reader might try to
simplify this difference of two cube roots into the number 1, but all attempts to do this
simply lead back to the original cubic x³ + 3x − 4 = 0 . The paper [4] discusses how to
recognize when radical expressions of the form for n = 2, 3, 4, …,

reduce to simple numbers like integers or rational values.

Final remarks

There is much more that could be said about the cubic formula. How do you find
the two complex roots when b > 0 , and how do we find any roots when b < 0 ? To
answer these questions requires a quantum leap in the difficulty of our presentation. This
is not our purpose. We hope that we have shown that there is partial information about
the cubic formula that is both interesting and useful. The reader can find complete
presentations of this subject in many algebra text books dating from before 1960 such as
[5] and nice summaries in handbooks such as [3].