The Cubic Formula - Math

The Cubic Formula

The quadratic formula tells us the roots of a quadratic polynomial, appoly-

nomial of the form ax2 + bx + c. The roots (if b2

4ac

0) are

2

b+ b 4ac

p

2a

and b

2

b

4ac .

The cu2baic 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 coe cients. Sppecifically, we'll look at the two cubic polynomials x3 15x 4 and x3 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 + 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 of

the 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 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

were equations of the form x3 + ax + b = 0 where

3 a

b 2 0. The

3

2

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

3 a

2 b

was a positive number. We're going to focus on the more compli3cated s2ort

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

x3 + ax + b if

a

3

b

2

>

0

3

2

Step 1. Let D be the complex number

D=

b

+

r i

a

3

b

2

2

3

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, so that R and I are real numbers with z = R + iI.

p Step 4. The three rpoots of x3 + ax + b are the real numbers 2R, R + 3I,

and R 3I.

These four steps together are the cubic formpula. It uses compplex numbers (D and z) to create real numbers (2R, R + 3I, and R 3I) 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 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

3

b

2

=

15

3

4

2

2

3

2

= ( 5)3 ( 2)2

= ( 125) (4)

= 125 4

= 121

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

D=

b

+

r i

a

3

b

2

2

3

2

=

4 2

+

p i 121

= 2 + i11

Step 2. We need to find a complex number z such that z3 = 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 I = 1

390

Step 4. The three roots of the cubic polynomial x3 15x real numbers 2R = 2(2) = 4

p

p

p

R + 3I = (2) + 3(1) = 2 + 3

p

p

p

R 3I = (2) 3(1) = 2 3

4 are the

+

p

p

We found the real number roots-- 4, 2 + 3, and 2 3 --of a poly-

nomial that had real number coe cients-- 15 and 4 --using complex

numbers-- 2 + i11 and 2 + i. This is why the complex numbers seemed at-

tractive 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 3px + 2. It's a

polynomial of the form x3 + 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

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download