The Cubic Formula

The quadratic formula tells us the roots of a quadratic polynomial, a poly-nomial of the form ax2 + bx + c. The roots (if b2 � 4ac � 0) are �b+

pb

2�4ac

2a

and �b�pb

2�4ac

2a

.The cubic formula tells us the roots of a cubic polynomial, a polynomial of

the form ax3+ bx2+ cx+ d. It was the invention (or discovery, depending onyour point of view) of the complex numbers in the 16th century that allowedmathematicians to derive the cubic formula, and it was for this reason thatpeople became interested in complex numbers.In this chapter we’ll see what the cubic formula is. We’ll see how it uses

complex numbers to tell us the real number roots of cubic polynomials withreal number coe�cients. Specifically, we’ll look at the two cubic polynomialsx3� 15x� 4 and x3� 3x+

p2, and we’ll use complex numbers and the cubic

formula to determine what their real number roots are.

A Computation we’ll need later

Before we begin, we should make a quick digression. We’ll want to find(2+ i)3. We have two options. We can multiply it out as (2+ i)(2+ i)(2+ i)using the distributive law many times. Or, we can use the binomial theorem.You may have seen the binomial theorem in Math 1050. It says that

(2 + i)3 = 23 + 3(22)i+ 3(2)i2 + i3

Simplifying, we have

(2 + i)3 = 8 + 12i+ 6i2 + i3

= 8 + i12 + i26 + i2i

= 8 + i12� 6� i

= 2 + i11

We’ll need to know that (2 + i)3 = 2 + i11 later in this chapter.

387

A Simplification for cubics

The cubic formula tells us the roots of polynomials of the form ax3+ bx2+cx + d. Equivalently, the cubic formula tells us the solutions of equations ofthe form ax3 + bx2 + cx+ d = 0.In the chapter “Classification of Conics”, we saw that any quadratic equa-

tion in two variables can be modified to one of a few easy equations to un-derstand. In a similar process, mathematicians had known that any cubicequation in one variable—an equation of the form ax3 + bx2 + cx + d = 0—could be modified to look like a cubic equation of the form x3+ax+ b = 0,so it was these simpler cubic equations that they were looking for solutionsof. If they could find the solutions of equations of the form x3 + ax + b = 0then they would be able to find the solutions of any cubic equation.Of the simpler cubic equations that they were trying to solve, there was an

easier sort of equation to solve, and a more complicated sort. The easier sortwere equations of the form x3 + ax + b = 0 where �

�

a

3

�

3 ��

b

2

�

2 0. The

more complicated sort were equations x3 + ax + b = 0 where ��

a

3

�

3 ��

b

2

�

2

was a positive number. We’re going to focus on the more complicated sortwhen discussing the cubic formula. Figuring out how to solve these morecomplicated equations was the key to figuring out how to solve any cubicequation, and it was the cubic formula for these equations that lead to thediscovery of complex numbers.

* * * * * * * * * * * * *

388

The Cubic formula

Here are the steps for finding the roots of a cubic polynomial of the form

x3 + ax+ b if �⇣a

3

⌘

3

�⇣b

2

⌘

2

> 0

Step 1. Let D be the complex number

D = �b

2+ i

r

�⇣a

3

⌘

3

�⇣b

2

⌘

2

Step 2. Find a complex number z 2 C such that z3 = D.

Step 3. Let R be the real part of z, and let I be the imaginary part of z, sothat R and I are real numbers with z = R + iI.

Step 4. The three roots of x3 + ax+ b are the real numbers 2R, �R+p3I,

and �R�p3I.

These four steps together are the cubic formula. It uses complex numbers(D and z) to create real numbers (2R, �R +

p3I, and �R�

p3I) that are

roots of the cubic polynomial x3 + ax+ b.

* * * * * * * * * * * * *

389

We’ll take a look at two examples of cubic polynomials, and we’ll use thecubic formula to find their roots.

First exampleIn this example we’ll use the cubic formula to find the roots of the polyno-

mialx3 � 15x� 4

Notice that this is a cubic polynomial x3 + ax + b where a = �15 andb = �4. Thus,

�⇣a

3

⌘

3

�⇣b

2

⌘

2

= �⇣�15

3

⌘

3

�⇣�4

2

⌘

2

= �(�5)3 � (�2)2

= �(�125)� (4)

= 125� 4

= 121

Because 121 is a positive number, we can find the roots of the cubic polyno-mial x3 � 15x� 4 using the 4 steps outlined on the previous page.

Step 1. We need to find the complex number D. It’s given by

D = �b

2+ i

r

�⇣a

3

⌘

3

�⇣b

2

⌘

2

= ��4

2+ i

p121

= 2 + i11

Step 2. We need to find a complex number z such that z3 = 2 + i11. Wesaw earlier in this chapter that (2 + i)3 = 2 + i11, so we can choose

z = 2 + i

Step 3. In this step, we write down the real (R) and imaginary (I) partsof z = 2 + i. The real part is 2, and the imaginary part is 1. So

R = 2 and I = 1390

Step 4. The three roots of the cubic polynomial x3 � 15x� 4 are thereal numbers

2R = 2(2) = 4

�R +p3I = �(2) +

p3(1) = �2 +

p3

�R�p3I = �(2)�

p3(1) = �2�

p3

We found the real number roots— 4, �2 +p3, and �2�

p3 —of a poly-

nomial that had real number coe�cients— �15 and �4 —using complexnumbers— 2 + i11 and 2 + i. This is why the complex numbers seemed at-tractive to mathematicians originally. They cared about complex numbersbecause they cared about real numbers, and complex numbers were a tooldesigned to give them information about real numbers.Over the last few centuries, complex numbers have proved their usefulness

in mathematics in many other ways. They are now viewed as being just asimportant as the real numbers are.

Second exampleNext we’ll find the roots of the cubic polynomial x3 � 3x +

p2. It’s a

polynomial of the form x3 + ax + b where a = �3 and b =p2. To see if we

can use the cubic formula on page 389, we need to see if the following numberis positive:

391

+

�⇣a

3

⌘

3

�⇣b

2

⌘

2

= �⇣�3

3

⌘

3

�⇣

p2

2

⌘

2

= �(�1)3 � (p2)2

22

= �(�1)� 2

4

= 1� 1

2

=1

2

Because 1

2

is positive, we can proceed with the four steps of the cubic formulato find the roots of x3 � 3x+

p2.

Step 1. We need to find the number D. Using the equation from Step 1 onpage 389,

D = �b

2+ i

r

�⇣a

3

⌘

3

�⇣b

2

⌘

2

= �p2

2+ i

r

1

2

= �p2p

2p2+ i

p1p2

= � 1p2+ i

1p2

Step 2. We need to find a complex number z such that z3 = D. To solvethis problem, notice that D = � 1p

2

+ i 1p2

is the number in the unit

circle that is a counterclockwise rotation of 1 by the angle 3⇡

4

. Fromwhat we’ve learned about multiplying complex numbers in the unitcircle, we can see that we can choose z to be the number in the unit

392

oo

circle obtained by rotating 1 by an angle of 1

3

�

3⇡

4

�

= ⇡

4

. That is,we’ll choose z to be the number 1p

2

+ i 1p2

.

To recap, in Step 1 we saw that D = � 1p2

+ i 1p2

. In Step 2 we saw

that if z = 1p2

+ i 1p2

, then z3 = D.

Step 3. The real part of z = 1p2

+ i 1p2

is 1p2

, and it’s imaginary part also

equals 1p2

. That is,

R = I =1p2

Step 4. The three real number roots of x3 � 3x+p2 are

2R = 2⇣ 1p

2

⌘

=2p2=

p2p2p

2=

p2

�R +p3I = �

⇣ 1p2

⌘

+p3⇣ 1p

2

⌘

=�1 +

p3p

2

�R�p3I = �

⇣ 1p2

⌘

�p3⇣ 1p

2

⌘

=�1�

p3p

2

393

It0It

a

k It0It

a

k

ExercisesThe first example from this chapter showed us the roots of x3 � 15x � 4.

They are 4, �2 +p3, and �2�

p3. Another way to put this is to say that

4, �2+p3, and �2�

p3 are the solutions of the equation x3� 15x� 4 = 0.

Here x is a variable, so we could equivalently write that 4, �2 +p3, and

�2�p3 are the solutions of the equation z3� 15z� 4 = 0, for example. Use

this to find the solutions of the following equations. For #1, first solve forx� 2, then solve for x. For #2, first solve for log

e

(x), and then solve for x.

1.) (x� 2)3 � 15(x� 2)� 4 = 0

2.) loge

(x)3 � 15 loge

(x)� 4 = 0

All further exercises in this chapter have nothing to do with complex num-bers.

394

Match the functions with their graphs.

3.) f(x) + 1 4.) f(2x) 5.) f(x+ 1)

6.) 1

2

f(x) 7.) f(x)� 1 8.) 2f(x)

9.) f(x� 1) 10.) f�

x

2

�

f(x) A.) B.)

C.) D.) E.)

F.) G.) H.)

395

I

J.3

I

74

a-

-a•1

—aI

N

a

-2.

I

Match the functions with their graphs.

11.) tan(x) 12.) cot(x)

13.) f(x) =

(

tan(x) if x 0;

cot(x) if x > 0.14.) g(x) =

(

cot(x) if x < 0;

tan(x) if x � 0.

A.) B.)

C.) D.)

396

7/7’/,I 1’II Il

I I! I IlIIII ê

I ILI

r

To find the solutions of an equation of the form h(x)f(x) = h(x)g(x), youneed to find the solutions of the following two equations:

h(x) = 0 and f(x) = g(x)

Find the solutions of the following equations.

15.) (x� 1)(x+ 3) = (x� 1)(x+ 4)

16.) (2x� 3)(x+ 1) = (2x� 3)(x+ 2)

17.) (x� 7) = (x� 7)(x+ 2)

18.) x2 = x

In the exercises from the previous chapter we reviewed rules for when someequations have solutions that can be found in one step. Those rules can beexpanded on to give us rules for one part of a larger problem that mightinvolve several steps. These rules are listed below. Use them to solve theequations in the remaining exercises.

• f(x)2 = c implies f(x) =pc or f(x) = �

pc

• af(x)2 + bf(x) + c = 0 implies f(x) = �b�pb

2�4ac

2a

or f(x) = �b+

pb

2�4ac

2a

•p

f(x) = c implies f(x) = c2

19.) (2x� 3)2 = 4 22.) (px)2 � 5

px+ 6 = 0

20.)px+ 4 = 3 23.) log

e

(x)2 = 25

21.) (ex)2 � 3ex + 2 = 0 24.)p2x� 3 = 5

397