Factoring over the Complex Numbers

Now you'll see mathematicians at work: making easy things harder to make them easier!

The Grandfather of all examples.

Consider the polynomial

. It cannot be factored over the real numbers, since its graph has no x-intercepts. (The graph is just the standard parabola shifted up by one unit!)

How can we tell that the polynomial is irreducible, when we perform square-completion or use the quadratic formula? Let's try square-completion: Not much to complete here, transferring the constant term is all we need to do to see what the trouble is:

We can't take square roots now, since the square of every real number is non-negative!

Here is where the mathematician steps in: She (or he) imagines that there are roots of -1 (not real numbers though) and calls them i and -i. So the defining property of this imagined number i is that

Now the polynomial has suddenly become reducible, we can write

Complex Numbers.

Let's get organized: A number of the form

, where a and b are real numbers, is called a complex number. Here are some examples:

The number a is called the real part of a+bi, the number b is called the imaginary part of a+bi.

Luckily, algebra with complex numbers works very predictably, here are some examples:

In general, multiplication works with the FOIL method:

eqnarray24

Two complex numbers a+bi and a-bi are called a complex conjugate pair. The nice property of a complex conjugate pair is that their product is always a non-negative real number:

Using this property we can see how to divide two complex numbers. Let's look at the example

The magic trick is to multiply numerator and denominator by the complex conjugate companion of the denominator, in our example we multiply by 1+i:

Since (1+i)(1-i)=2 and (2+3i)(1+i)=-1+5i, we get

and we are done!

You can find more information in our Complex Numbers Section.

Quadratic polynomials with complex roots.

Consider the polynomial

Using the quadratic formula, the roots compute to

It is not hard to see from the form of the quadratic formula, that if a quadratic polynomial has complex roots, they will always be a complex conjugate pair!

Here is another example. Consider the polynomial

Its roots are given by

The discriminant.

We can see from the graph of a polynomial, whether it has real roots or is irreducible over the real numbers. How can we tell algebraically, whether a quadratic polynomial has real or complex roots? The symbol i enters the picture, exactly when the term under the square root in the quadratic formula is negative. This term

is called the discriminant.

Consider the discriminant of the quadratic polynomial .

If the discriminant is positive, the polynomial has 2 distinct real roots.
If the discriminant is negative, the polynomial has 2 complex roots, which form a complex conjugate pair.
If the discriminant is zero, the polynomial has one real root of multiplicity 2.

The Fundamental Theorem of Algebra, Take Two.

We already know that every polynomial can be factored over the real numbers into a product of linear factors and irreducible quadratic polynomials. But now we have also observed that every quadratic polynomial can be factored into 2 linear factors, if we allow complex numbers. Consequently, the complex version of the The Fundamental Theorem of Algebra is as follows:

Over the complex numbers, every polynomial (with real-valued coefficients) can be factored into a product of linear factors.

We can state this also in root language:

Over the complex numbers, every polynomial of degree n (with real-valued coefficients) has n roots, counted according to their multiplicity.

The usage of complex numbers makes the statements easier and more "beautiful"!