Now you'll see mathematicians at work: making easy things harder to make them easier!
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
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:
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.
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
is called the discriminant.
Consider the discriminant of the quadratic polynomial .
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"!
(b) Give an example of a polynomial of degree 4 without any x-intercepts.
Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.