The geometric motivation for this question comes from the study of some familiar curves in the plane, in particular lines and parabolas.

The linear case: degree 1 in x

The equation of a line in the plane with slope m is given by y = mx + b. The quantity b is the y-intercept. The solution to the polynomial equation mx+b = 0 answers the question: for what value of x do we have y=0?, in other words, what is the x-intercept of this line? It was well-known even at that time, and even though the equation was expressed in words rather than with symbols, that the answer is given by isolating x on one side of the equation, which is done in two steps, by:

*first isolating the constant on one side, so that the equation becomes mx = -b,

*then dividing by the coefficient m of x, so that the equation becomes x = -b/m.

The quadratic case: degree 2 in x

The equation of a parabola in the plane has the form y = ax² + bx + c, where c corresponds to the y-intercept of the parabola. The solution to the polynomial equation ax² + bx + c = 0 answers the question: where, if anywhere, does the parabola y = ax² + bx + c cross the x-axis?

Here, also, although the question was phrased in words without the aid of formulas, and the solution was given as an algorithm, or procedure, rather than a formula, the answer was known thousands of years before the Renaissance; it predates the Greeks, reaching back to ancient Babylonian civilization. It is obtained in six simple steps, as follows: by "completing the square", isolating the constant on one side, dividing by the coefficient, extracting a square root, isolating x on one side, and pulling out the factor of 1/2a:

* Completing the square:  ax² + bx + c = a(x+b/2a)² - b²/4a +c = 0

* Isolating the constant:   a(x+b/2a)² = b²/4a - c

* Dividing by the coefficient:   (x+b/2a)² = b²/4a2 - c/a

* Extracting a square root:  x + b/2a = ± √ (b²/4a²-c/a)

* Isolating x:   x = -b/2a ± √ (b²/4a²-c/a)

* Take the factor 1/2a outside the square root:   x = -b/2a ± √ (b²-4ac))/2a

* Pull out the factor of 1/2a:   x = (-b ± √ (b²-4ac))/2a

Example. A problem on a cuneiform tablet from Old Babylonian times states that the area of a rectangle is 10 and its length exceeds its width by 7; one must compute the length and width of the rectangle. Before even starting, we must notice that Babylonians wrote number in base 60!! So their number 10 is equal to 1x60 + 0 = 60 in base 10. For us, the area equation would be written lw=60, so plugging in the given information l=w+7, this gives (w+7)w=w²+7w=60. Using the quadratic formula we find

Please read this complete and amusing historical account of the development of the solution to the cubic equation, and the Cardano-Tartaglia duel.

In order to solve the cubic equation, we will use a slightly simpler method developed some decades after Cardano published Tartaglia's solution, by the British mathematician Thomas Harriot (1560-1621).

Suppose we want to solve a cubic equation Ax³ + Bx² + CX + D = 0. We start by dividing everything by A in order to bring it to the simpler form x³+ax²+bx+c=0. Notice that this makes no difference at all to the solutions: solutions of one of these equations will also be solutions of the other.

In the quadratic case, we added a term to x in order to obtain an expression without a linear term. Here, we perform an analogous operation to obtain an expression without a quadratic term (called a depressed cubic): we make the substitution x = t - a/3 in our polynomial. We obtain

Now Thomas Harriot proposes to make the following substitution in this polynomial: t = s - p/3s. After wading through the algebra, this simple idea magically reduces the cubic to a quadratic!

We are looking for the solutions of
0 = t³ + pt + q
= (s - p/3s)³ + p(s-p/3s) + q
= s³ - 3(p/3s)s² + 3(p²/9s ²)s - (p³/27s³) + ps - (p²/3s) + q
= s³ - ps + (p²/3s) - (p³/27s³) + ps - (p²/3s) + q
= s³ + q - (p³/27s³) = 0.

Multiplying both sides of this equation by s³ yields the new form s⁶ + qs³ = p³/27, having the same solutions s as the version just above. Now, if we make the substitution r = s³, the equation simply becomes the quadratic
r² + qr = p³/27, whose solutions we know! Nothing is left but to retranslate the solutions from terms of r back to terms of s, then t, and finally x.

Example: Find a real root of the cubic equation x³ - 24x - 72.

Solution: The first step in our algorithm above, the substitution t = x-a/3, has the goal of eliminating the quadratic term. However, there is no quadratic term here, so this step is not needed. We can just take t = x,
p = -24 and q = -72, and proceed directly to the next substitution x = t = s - p/3s = s + 24/3s = s + 8/s. Plugging this into the polynomial yields

Like the cubic, the quartic equation can be solved by radicals. Ludovico Ferrari proposed a rather complicated method, but we give the simplest possible one here. If a₄x⁴+ a₃x³+a₂x²+a₁x+a₀= 0 is our quartic, we start by dividing both sides by a₄, and making the substitution t=x-a₃/4a₄ to obtain a depressed quartic t⁴+ct²+dt+e=0. Now, we firmly assume that this quartic is a product of two quadratic polynomials, and write

Now compute

Set P=p². This equation then becomes

Multiplying out and multiplying both sides by P gives

Example. Factor the quadratic polynomial x⁴-6x²-16x-15.

Solution. We have c=-6, d=-16, e=-15, and we have to solve the cubic P³+2cP²+(c²-4e)P-d² = P³-12P²+96P-256=0.

We first make this into a depressed cubic by substituting P=Q+4 into the cubic, obtaining Q³+48Q. Thus, we see at once that Q=0 is a root, so P=4 is a root of the cubic in P. Now we can take p=√P=√4=2 and we obtain r=-2, s=-5 and q=3, so our quartic factors into two quadratics

The roots of the quartic are, finally, two real and two complex roots:
(1/2)(-2 ± √-8) = (1/2)(-2 ± 2√-2) = -1± √-2 and
(1/2)(2 ± √24) = (1/2)(2 ± 2√6) = 1 ± √6,

The next step in the theory of the solution of polynomial equations took another two hundred years. In 1824, Niels Henrik Abel finally published a proof that some polynomials of degree 5 or higher cannot be solved by radicals, although others (such as xⁿ-a=0, for instance) can be. The proof works by the absurd, assuming that there is such a solution and then arriving at a contradiction. In work done around 1830, at the age of 18, Evariste Galois went farther than Abel, giving a specific and verifiable criterion for whether a given polynomial is or is not solvable by radicals. His work had incredibly deep and wide-ranging applications and developments, but there is a particularly simple statement of his theorem:

Galois' theorem.  A polynomial of degree n is solvable by radicals if and only if all of its roots are expressible as rational functions in two of them.

Lecture 14: February 18, 2008

The complex numbers

Please read the MacTutor biography of Rafael Bombelli (1526-1572).

The final great contribution of the Italian Renaissance to mathematics was the acceptance of the fact that negative and complex numbers could be manipulated by the same simple arithmetic rules as the more familiar positive real numbers. Note that in fact the structure of the set of real numbers was not at all understood at the time either, but people were familiar with the idea of a row or line of increasing numbers containing many rather mysterious quantities that were not fractions.

This acceptance was not something that emerged slowly little by little. The question was raised openly by Cardan, who came on square roots of negative numbers in his work. He wrote these down; for instance, he explicitly multiplied out (5+√ -15)(5-√ -15) and obtained 40. But the result puzzled him, and he left it aside, qualifying it as a result as subtle as it is useless

Rafael Bombelli was 19 at the time that Cardan's great work Ars Magna, containing the solutions of the cubic and quartic, was published in 1545. He was an engineer, but mathematics was his great love. He published three books on algebra in 1572, the year that he unfortunately died. In these books, he approaches the question of negative numbers and complex numbers completely explicitly, giving the set of rules almost in the form of a poem:

Plus times plus makes plus
Minus times minus makes plus Minus times plus makes minus
Plus 8 times plus 8 makes plus 64
Minus 5 times minus 6 makes plus 30
Minus 4 times plus 5 makes minus 20
Plus 5 times minus 4 makes minus 20.

More than this, he attempted to give a real geometric explanation of why this is so. We all know these rules yet this why still seems extremely puzzling today.

From the acceptance of negative numbers, Bombelli took the step to accepting also square roots of negative numbers, and doing arithmetic with these quantities, adding, subtracting, multiplying and dividing them according to the usual rules of arithmetic (commutativity, associativity, distributivity).

He justified the usefulness of accepting to work with these numbers with examples such as the following.

Example of usefulness of working with complex numbers. Consider the cubic x³-15x-4. Let's go through our procedure for solving this cubic. It is already depressed, so we have p=-15, q=-4 and we set x=s+5/s. Expanding the cubic in terms of s, we get

In order to investigate this possibility, he tried to see if the cube root of the number 2+√-121=2+i√121=2+11i could itself be a number of the form a+ib with a and b real. So he set (a+ib)³=(a³-3ab²) +(3a²-b²)+2+11i, and found that a=2, b=1 gives a solution to this. In other words, Bombelli discovered that formally, (2+i)³=2+11i, so (2+i) is the cube root of 2+11i, so s=2+i. Similarly, he showed that 2-i is the cube root of 2-11i, so 5/s=2-i, and the solution to the cubic x³-15x-4 is given by x=s+5/s=2+i+2-i=4. This computation was Bombelli's justification for using and calculating with square roots of negative numbers.

This work on complex numbers led to the natural question about finding roots of polynomials: can we always find a complex root? The answer is always yes, whereas of course it is no if one asks whether one can always find a real root (for example x²+1=0 has two complex roots i and -i but no real roots). This result, which was first proved much later, by Gauss, is known as the:

Fundamental theorem of algebra.   Every polynomial with complex coefficients has a complex root.

Exercises. (1) Find the complex number a+ib which is the square root of the complex number 3+4i.

(2) What can you say about the product of two complex numbers (a+ib)(a-ib) (foil it)?

(3) The above exercise shows that every complex number a+ib has a kind of partner a-ib called the complex conjugate. Prove that if a quadratic equation has no real roots (i.e. if b²-4ac is negative), then it has two complex conjugate roots.

(4) Prove that if a complex number a+ib is a root of a polynomial p(x) with real coefficients, then a-ib is also a root of p(x). Thus, when looking for the roots of a polynomial p(x), we can always be sure that the non-real roots occur in pairs.