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 tex2html_wrap_inline115 . 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:

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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

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Now the polynomial tex2html_wrap_inline115 has suddenly become reducible, we can write

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Complex Numbers.

Let's get organized: A number of the form tex2html_wrap_inline145 , where a and b are real numbers, is called a complex number. Here are some examples:

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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:

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In general, multiplication works with the FOIL method:

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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:

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

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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:

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

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and we are done!

You can find more information in our Complex Numbers Section.


Quadratic polynomials with complex roots.

Consider the polynomial

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Using the quadratic formula, the roots compute to

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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

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Its roots are given by

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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

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is called the discriminant.

Consider the discriminant tex2html_wrap_inline181 of the quadratic polynomial tex2html_wrap_inline183 .


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"!


Exercise 1.

Find all (real or complex) roots of the polynomial tex2html_wrap_inline189 .

Answer.

Exercise 2.

Factor the polynomial tex2html_wrap_inline191 completely (a) over the real numbers, (b) over the complex numbers.

Answer.

Exercise 3.

For which values of c does the polynomial tex2html_wrap_inline195 have two complex conjugate roots?

Answer.

Exercise 4.

For which values of a does the polynomial tex2html_wrap_inline199 have two distinct real roots?

Answer.

Exercise 5.

Every quadratic polynomial has either 2 distinct real roots, one real root of multiplicity 2, or 2 complex roots. What cases can occur for a polynomial of degree 3? Give an example for each of these cases.

Answer.

Exercise 6.

(a) Show that every polynomial of degree 3 has at least one x-intercept.

(b) Give an example of a polynomial of degree 4 without any x-intercepts.

Answer.

Exercise 7.

Give an example of a polynomial of degree 5, whose only real roots are x=2 with multiplicity 2, and x=-1 with multiplicity 1.

Answer.

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