Download MA 3362 Lecture 25 - The Fundamental Theorem of Algebra

MA 3362 Lecture 25 - The Fundamental Theorem of Algebra
Monday, December 1, 2008.
Objectives: Discuss factoring polynomials over the real and complex numbers.
To finish off the semester, I would like to talk about something that all of you should know something about:
The basic fact that all real polynomials can be factored into quadratic and linear factors.
First of all, let’s notice that factoring a polynomial is essentially equivalent to finding zeros.
Consider the polynomial p(x) = x3 − 6x2 + 5x + 6. I claim that x = 2 is a zero. This is easy to check.
p(2) = (2)3 − 6(2)2 + 5(2) + 6 = 8 − 24 + 10 + 6 = 0.
(1)
The fact that x = 2 is a zero corresponds to the fact that (x − 2) is a factor of p(x). We can check this with
long division.
x−2
x3
x3
−
−
(2)
x2
−
4x −
3
6x2
2x2
+
5x +
6
−4x2
−4x2
+
+
5x +
8x
6
−3x +
−3x +
6
6
0
Since we got a remainder of 0, we can conclude that f(x) can be factored as
p(x) = x3 − 6x2 + 5x + 6 = (x − 2)(x2 − 4x − 3).
(3)
Of course, once we know that p(x) factors this way, it is clear that p(2) = 0. The converse is not as clear.
We can kind of see what’s happening by doing the long division with a generic linear polynomial (x − a).
x−a
(4)
x3
x3
−
−
x2
+
(−6 + a)x +
(5 − 6a + a2 )
6x2
ax2
+
5x +
6
(−6 + a)x2
(−6 + a)x2
+
−
5x +
a(−6 + a)x
6
(5 − 6a + a2 )x +
(5 − 6a + a2 )x −
6
a(5 − 6a + a2 )
6 + 5a − 6a2 + a3
The thing to notice here is that the remainder is p(a). This could have been a coincidence, of course, but
with some grunting, you could show that this is always the case. We have the following theorem.
Theorem 1. For a polynomial p(x), if p(a) = 0, then p(x) can be factored p(x) = (x−a)·q(x). The converse
is also true. That is, if p(x) can be factored p(x) = (x − a) · q(x), then p(a) = (a − a) · q(x) = 0.
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MA 3362 Lecture 25 - The Fundamental Theorem of Algebra
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Trying to factor a polynomial
As we try to factor a polynomial, we could look for values for x that make the polynomial equal to zero.
Can we always find such a number?
Question 1. Given a polynomial p(x) is there always a number a such that p(a) = 0?
Consider the polynomial p(x) = x7 − 7x5 + 4x2 − x + 11. Is this polynomial factorable?
What about p(x) = x2 + 1?
What about p(x) = x4 ?
If you look at the graphs of these functions, you can see that the graph tells us if there is a zero, and
therefore, if there is a linear factor. In particular, if the graph crosses the x-axis, then it has a zero and a
linear factor.
For the first polynomial, p(x) = x7 − 7x5 + 4x2 − x + 11 has degree 7, and the x7 -term dominates the
function for large values of x. Therefore, limx→∞ p(x) = ∞ and limx→−∞ p(x) = −∞. All polynomials are
continuous, and a continuous function can’t go from −∞ to +∞ without crossing the x-axis, and so p(x)
has to have a linear factor. The same must be true of any odd-degree polynomial.
1. Complex Numbers
All odd degree polynomials are factorable over the reals. An even degree polynomial might not. For example,
p(x) = x2 + 1 = 0
(5)
has no solutions. The imaginary number i, which is a number whose square is −1, that is i2 = −1, is a
solution, as is its additive inverse −i. Over the complex numbers, x2 + 1 factors as
x2 + 1 = (x + i)(x − i).
(6)
We can even do a long division to see that these factors divide into x2 + 1 evenly.
x−i
x2
x2
x
+
+
−
0x +
ix
1
+
−
1
i2
i
(7)
ix
ix
1 + i2
=1−1=0
Basically, the same algebraic properties hold with the complex numbers, and zeros correspond exactly with
linear factors. What we want to know is the following.
Question 2. Does every polynomial have a complex root?
Homework 25
Given any complex number, z, it can be written in the form z = r ( cos(θ) + i sin(θ) ) = reiθ . This is how
polar coordinates are written in C.
1.
Write 1, i, −1, −i, 1 + i, and 1 − i in polar form.
2.
Let f(z) = z 2 . Compute f(z) for each of the numbers in Problem 1.
3.
Describe geometrically what happens to r and θ under f.