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Section A.1 Motivation

We begin with a few examples.

Example A.1.1.
  • The polynomial equation \(x^4 - 5x^2 + 4 = 0\) has four solutions: \(x = 1\text{,}\) \(x = -1\text{,}\) \(x = 2\text{,}\) and \(x = -2\text{.}\)
  • The polynomial equation \(x^4 - 1 = 0\) has two solutions: \(x = 1\) and \(x = -1\text{.}\)
  • The polynomial equation \(x^4 + 1 = 0\) has no solutions.

Solutions of a polynomial equation are called the roots of the equation, and are also often referred to as roots of the polynomial involved. From the examples above, different polynomials of the same degree can have different numbers of roots.

Question A.1.2.

What can we say about the possibilities for the number of roots a polynomial can have?

A proper proof would use the method of proof by induction, but we will be a little less formal. If the polynomial equation \(p(x) = 0\) has no solutions at all, then clearly the number of solutions is not more than the degree \(n\text{.}\) On the other hand, if this equation has distinct solutions \(x_1,x_2,\dotsc,x_\ell\text{,}\) then it is a fact of algebra that we can factor \(p(x)\) as

\begin{equation*} p(x) = (x-x_1)^{m_1} (x-x_1)^{m_2} \dotsm (x-x_\ell)^{m_\ell} q(x), \end{equation*}

where each exponent \(m_i\) is equal to the number of times solution \(x=x_i\) is repeated as a root of the equation (called the multiplicity of the root), and \(q(x)\) is another polynomial with no roots.

Since the degree of the factorization of \(p(x)\) above must be the same as the degree of \(p(x)\text{,}\) the sum \(m_1 + m_2 + \dotsb + m_\ell\) cannot be greater than \(n\text{,}\) and this sum of multiplicities is exactly the number of roots of the equation \(p(x)=0\) (including repetition).

In some sense, the optimal situation is when a polynomial has a full slate of roots (including repetition), so that the polynomial can be fully factored into linear factors. In order to “fix” the problem of some polynomials having fewer than a full slate of roots, we will just “invent” solutions for these root-deficient polynomials.

The quintessential example of a polynomial equation with no solution is \(x^2 + 1 = 0 \text{.}\) Let's start by inventing a solution to this equation.

Definition A.1.4. The imaginary number \(\ci\) .

An imaginary “number” that satisfies \(\ci^2 + 1 = 0\text{.}\)

Note A.1.5.

Notice in particular that \(\ci^2 = -1\text{,}\) which will be the main way that we use the above definition of \(\ci\) in algebraic computations.

Remark A.1.6.

It is common to think of \(\ci\) as being equal to \(\sqrt{-1}\text{,}\) because

\begin{equation*} (\sqrt{-1})^2 + 1 = -1 + 1 = 0 \end{equation*}

suggests that if \(-1\) had a square root then it would be a solution to \(x^2+1=0\text{,}\) which is what's desired. However, the notation \(\sqrt{-1}\) implies that this is a positive square root, i.e. greater than zero. Where we're going, concepts of “greater than” and “less than” will no longer be meaningful.

If we want to treat this new imaginary number just like our ordinary real numbers that we're used to, we would like to be able to perform operations invovling it and other numbers, such as addition, subtraction, multiplication, division, and taking negatives. To do this, we will treat \(\ci\) algebraically as if it were a variable.

More on that in the next section. For now, let's just start with taking negatives. Since \((-x)^2 = x^2\) for every real number \(x\text{,}\) we will require the same of our new imaginary number: \((-\ci)^2 = \ci^2\text{.}\) But then if we stipulate that \(-\ci\) is not the same imaginary number as \(\ci\) (a reasonable requirement, since no nonzero real number is equal to its own negative), then with the invention of the imaginary number \(\ci\text{,}\) the polynomial equation \(x^2 + 1 = 0\) now has just as many solutions as the degree of the polynomial involved!

In summary, in the example of the polynomial \(x^2 + 1\text{,}\) we can go from having no solutions to the full amount allowed by The Fundamental Theorem of Algebra (Real Version) by inventing one solution.