The Cubic Formula
The quadratic formula tells us the roots of a quadratic polynomial, ap polyb+ b2 4ac
2
2
nomial of
the
form
ax
+
bx
+
c.
The
roots
(if
b
4ac
0)
are
2a
p
b
b2 4ac
and
.
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 on
your point of view) of the complex numbers in the 16th century that allowed
mathematicians to derive the cubic formula, and it was for this reason that
people 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 with
real number coefficients. Specifically,
we’ll look at the two cubic polynomials
p
3
3
x 15x 4 and x 3x + 2, 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 + i2 6 + i2 i
= 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 of
the form ax3 + bx2 + cx + d = 0.
In the chapter “Classification of Conics”, we saw that any quadratic equation in two variables can be modified to one of a few easy equations to understand. In a similar process, mathematicians had known that any cubic
equation 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 solutions
of. If they could find the solutions of equations of the form x3 + ax + b = 0
then 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 sort
a 3
b 2
were equations of the form x3 + ax + b = 0 where
 0. The
3
2
3
a
b 2
more complicated sort were equations x3 + ax + b = 0 where
3
2
was a positive number. We’re going to focus on the more complicated sort
when discussing the cubic formula. Figuring out how to solve these more
complicated equations was the key to figuring out how to solve any cubic
equation, and it was the cubic formula for these equations that lead to the
discovery of complex numbers.
*
*
*
*
*
*
*
388
*
*
*
*
*
*
The Cubic formula
Here are the steps for finding the roots of a cubic polynomial of the form
3
x + ax + b
⇣ a ⌘3
⇣ b ⌘2
r ⇣ ⌘
a 3
3
⇣ b ⌘2
if
3
2
>0
Step 1. Let D be the complex number
b
+i
2
D=
2
Step 2. Find a complex number z 2 C such that z 3 = D.
Step 3. Let R be the real part of z, and let I be the imaginary part of z, so
that R and I are real numbers with z = R + iI.
Step 4. The three roots of x3 + ax + b are the real numbers 2R,
p
and R
3I.
R+
p
3I,
These four steps together are the cubic formula.
It uses complex
numbers
p
p
(D and z) to create real numbers (2R, R + 3I, and R
3I) that are
3
roots of the cubic polynomial x + ax + b.
*
*
*
*
*
*
*
389
*
*
*
*
*
*
We’ll take a look at two examples of cubic polynomials, and we’ll use the
cubic formula to find their roots.
First example
In this example we’ll use the cubic formula to find the roots of the polynomial
x3 15x 4
Notice that this is a cubic polynomial x3 + ax + b where a = 15 and
b = 4. Thus,
⇣ a ⌘3 ⇣ b ⌘2
⇣ 15 ⌘3 ⇣ 4 ⌘2
=
3
2
3
2
3
2
= ( 5)
( 2)
= ( 125)
= 125 4
= 121
(4)
Because 121 is a positive number, we can find the roots of the cubic polynomial 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
r ⇣ ⌘
b
a 3 ⇣ b ⌘2
D=
+i
2
3
2
=
p
4
+ i 121
2
= 2 + i11
Step 2. We need to find a complex number z such that z 3 = 2 + i11. We
saw 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) parts
of z = 2 + i. The real part is 2, and the imaginary part is 1. So
R=2
and
390
I=1
Step 4. The three roots of the cubic polynomial x3
real numbers
2R = 2(2) = 4
p
3I =
(2) +
p
R
p
3I =
(2)
p
3(1) =
3(1) =
2+
p
2
p
4 are the
3
3
+
R+
15x
p
p
We found the real number roots— 4, 2 + 3, and 2
3 —of a polynomial that had real number coefficients— 15 and 4 —using complex
numbers— 2 + i11 and 2 + i. This is why the complex numbers seemed attractive to mathematicians originally. They cared about complex numbers
because they cared about real numbers, and complex numbers were a tool
designed 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 as
important as the real numbers are.
Second example
p
Next we’ll find the roots of the cubic polynomial x3 3x
+
2. It’s a
p
3
polynomial of the form x + ax + b where a = 3 and b = 2. To see if we
can use the cubic formula on page 389, we need to see if the following number
is positive:
391
⇣ a ⌘3
3
⇣ b ⌘2
2
⇣ 3 ⌘3
3
=
=
( 1)3
=
( 1)
=1
=
⇣ p 2 ⌘2
p 22
( 2)
22
2
4
1
2
1
2
oo
Because 12 is positive, we can proceed
with the four steps of the cubic formula
p
3
to find the roots of x
3x + 2.
Step 1. We need to find the number D. Using the equation from Step 1 on
page 389,
r ⇣ ⌘
b
a 3 ⇣ b ⌘2
D=
+i
2
3
2
=
p
2
+i
2
r
1
2
=
p
p
2
1
p p + ip
2 2
2
=
1
1
p + ip
2
2
Step 2. We need to find a complex number z such that z 3 = D. To solve
this problem, notice that D = p12 + i p12 is the number in the unit
circle that is a counterclockwise rotation of 1 by the angle 3⇡
4 . From
what we’ve learned about multiplying complex numbers in the unit
circle, we can see that we can choose z to be the number in the unit
392
k
circle obtained by rotating 1 by an angle of
we’ll choose z to be the number p12 + i p12 .
1 3⇡
3 4
= ⇡4 . That is,
It
It
0
k
It
It
0
To recap, in Step 1 we saw that D =
that if z = p12 + i p12 , then z 3 = D.
Step 3. The real part of z =
equals p12 . That is,
p1
2
+ i p12 is
p1 ,
2
p1
2
+ i p12 . In Step 2 we saw
and it’s imaginary part also
1
R=I=p
2
p
3x + 2 are
p p
⇣ 1 ⌘
2
2 2 p
2R = 2 p = p = p
= 2
2
2
2
p
⇣ 1 ⌘ p ⇣ 1 ⌘
p
1+ 3
p + 3 p =
p
R + 3I =
2
2
2
p
⇣ 1 ⌘ p ⇣ 1 ⌘
p
1
3
p
p
R
3I =
3 p =
2
2
2
Step 4. The three real number roots of x3
393
Exercises
The first example
showed us the roots of x3 15x 4.
p from this chapter
p
They arep4, 2 + 3, p
and 2
3. Another way to put this is to say that
4, 2 + 3, and 2
3 are the solutions of the equation x3 15x p 4 = 0.
Here p
x is a variable, so we could equivalently write that 4, 2 + 3, and
2
3 are the solutions of the equation z 3 15z 4 = 0, for example. Use
this to find the solutions of the following equations. For #1, first solve for
x 2, then solve for x. For #2, first solve for loge (x), and then solve for x.
1.) (x
2)3
2.) loge (x)3
15(x
2)
4=0
15 loge (x)
4=0
All further exercises in this chapter have nothing to do with complex numbers.
394
I
I
Match the functions with their graphs.
3.) f (x) + 1
4.) f (2x)
6.) 12 f (x)
7.) f (x)
9.) f (x
f (x)
1)
10.) f
5.) f (x + 1)
1
8.) 2f (x)
x
2
A.)
B.)
D.)
E.)
a
74
C.)
J.3
a-
G.)
H.)
N
F.)
-a
•1
—a
-2.
395
Match the functions with their graphs.
11.) tan(x)
13.) f (x) =
12.) cot(x)
(
tan(x)
cot(x)
if x  0;
if x > 0.
A.)
14.) g(x) =
B.)
C.)
I
I
/,
7/7’
I
I
I!
II
I
ILI
II
I
ê
D.)
1’
Il
Il
396
(
cot(x)
tan(x)
if x < 0;
if x 0.
To find the solutions of an equation of the form h(x)f (x) = h(x)g(x), you
need 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
16.) (2x
17.) (x
1)(x + 4)
3)(x + 1) = (2x
7) = (x
3)(x + 2)
7)(x + 2)
18.) x2 = x
In the exercises from the previous chapter we reviewed rules for when some
equations have solutions that can be found in one step. Those rules can be
expanded on to give us rules for one part of a larger problem that might
involve several steps. These rules are listed below. Use them to solve the
equations in the remaining exercises.
p
p
• f (x)2 = c implies f (x) = c or f (x) =
c
• af (x)2 + bf (x) + c = 0 implies f (x) =
p
• f (x) = c implies f (x) = c2
19.) (2x
20.)
p
3)2 = 4
p
b2 4ac
2a
or f (x) =
p
22.) ( x)2
3ex + 2 = 0
24.)
397
p
2x
p
b+ b2 4ac
2a
p
5 x+6=0
23.) loge (x)2 = 25
x+4=3
21.) (ex )2
b
3=5