The Cubic Formula

The quadratic formula tells us the roots of a quadratic polynomial, a poly- 2 2 b+pb2 4ac nomial of the form ax + bx + c. The roots (if b 4ac 0) are 2a b pb2 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 coecients. Speciﬁcally, we’ll look at the two cubic polynomials x3 15x 4andx3 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 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 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 3 a 3 b 2 were equations of the form x + ax + b =0where 3 2 0. The 3 2 more complicated sort were equations x3 + ax + b =0where a b 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.

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388 The Cubic formula Here are the steps for ﬁnding the roots of a cubic polynomial of the form

a 3 b 2 x3 + ax + b if > 0 3 2 ⇣ ⌘ ⇣ ⌘

Step 1. Let D be the complex number

b a 3 b 2 D = + i 2 3 2 r ⇣ ⌘ ⇣ ⌘

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

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, R + p3I, and R p3I.

These four steps together are the cubic formula. It uses complex numbers (D and z)tocreaterealnumbers(2R, R + p3I,and R p3I)thatare roots of the cubic polynomial x3 + ax +b.

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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 ⇣ ⌘ ⇣ ⌘ = (⇣ 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 4usingthe4stepsoutlinedonthepreviouspage. Step 1. We need to find the complex number D. It’s given by b a 3 b 2 D = + i 2 3 2 r ⇣ ⌘ ⇣ ⌘ 4 = + ip121 2

=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)andimaginary(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 4arethe real 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—ofapoly- nomial that had real number coecients— 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 Next we’ll find the roots of the cubic polynomial x3 3x + p2. It’s a polynomial of the form x3 + ax + b where a = 3andb = p2. 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 b 2 3 3 p2 2 = 3 2 3 2 ⇣ ⌘ ⇣ ⌘ ⇣ ⌘ (p⇣2)2 ⌘ = ( 1)3 22 2 = ( 1) 4 1 =1 2 1 = 2 1 Because 2 is positive, we can proceed with the four stepso of the cubic formula to find the roots of x3 3x + p2. Step 1. We need to find the number D. Using the equation from Step 1 on page 389,

b a 3 b 2 D = + i 2 3 2 r ⇣ ⌘ ⇣ ⌘ p2 1 = + i 2 2 r o

p2 p1 = + i p2p2 p2

1 1 = + i p2 p2

Step 2. We need to ﬁnd a complex number z such that z3 = D. To solve this problem, notice that D = 1 + i 1 is the number in the unit p2 p2 3⇡ circle that is a counterclockwise rotation of 1 by the angle 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 a a k 1 3⇡ ⇡ circle obtained by rotating 1 by an angle of 3 4 = 4 . That is, we’ll choose z to be the number 1 + i 1 . p2 p2 0 It It k

0 1 1 It It To recap, in Step 1 we saw that D = + i . In Step 2 we saw p2 p2 that if z = 1 + i 1 , then z3 = D. p2 p2 Step 3. The real part of z = 1 + i 1 is 1 , and it’s imaginary part also p2 p2 p2 equals 1 . That is, p2 1 R = I = p2

Step 4. The three real number roots of x3 3x + p2are 1 2 p2p2 2R =2 = = = p2 p2 p2 p2 ⇣ ⌘ 1 1 1+p3 R + p3I = + p3 = p2 p2 p2 ⇣ ⌘ ⇣ ⌘ 1 1 1 p3 R p3I = p3 = p2 p2 p2 ⇣ ⌘ ⇣ ⌘

393 Exercises The 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 p3arethesolutionsoftheequationz3 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 log (x), and then solve for x. e 1.) (x 2)3 15(x 2) 4=0 2.) log (x)3 15 log (x) 4=0 e e

All further exercises in this chapter have nothing to do with complex numbers.

394 I I I

Match the functions with their graphs. 3.) f(x)+1 4.) f(2x)5.)f(x +1)

6.) 1f(x)7.)f(x) 18.)2f(x) 2 9.) f(x 1) 10.) f x 2 f(x) A.) B.)

a- I J.3 C.) D.) E.)

74 a

F.) G.) H.) N

-a •1

—a -2.

395 Match the functions with their graphs. r 11.) tan(x)12.)cot(x)

tan(x)ifx 0; cot(x)ifx<0; 13.) f(x)=  14.) g(x)= cot(x)ifx>0. tan(x)ifx 0. ( (

A.) B.)

C.) D.)

I I 1’

I Il I I! I Il II II ê 7/7’I ILI /,

396 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 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 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.

f(x)2 = c implies f(x)=pc or f(x)= pc • 2 b pb2 4ac b+pb2 4ac af(x) + bf(x)+c =0impliesf(x)= or f(x)= • 2a 2a f(x)=c implies f(x)=c2 • p 19.) (2x 3)2 =4 22.)(px)2 5px +6=0 2 20.) px +4=3 23.)loge(x) =25

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

397