Section Week 4 (9/15-9/19)
We will continue to look at polynomials.
Handout Monday 9/15
Summarize what we know about factoring polynomials.
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Polynomials always live in a polynomial ring, \(F[x]\) for some ring or field \(F\) (we usually want a field here, so we have the division algorithm).
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Inside a particular polynomial ring, we can ask whether a polynomial factors or is irreducible. Here irreducible means it cannot be factored into two polynomials of strictly smaller degree; \(6x+3 = 3(2x + 1)\) is irreducible.
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There is a connection between factors and roots: if \(p(x)\) has a root, then it has a factor (\(x - \alpha\text{,}\) where \(\alpha\) is the root). But just because a polynomial has no roots, doesn’t mean it is irreducible (although these are equivalent for polynomials of degree 3 or less).
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When working in \(\Q[x]\) specifically, and when the polynomials only have integer coefficients, we can say more.
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The Rational Roots Theorem tells us what the possible roots can be.
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Eisenstein’s Criterion gives us a test for irreducibility (but only applies to a limited number of polynomials).
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Gauss’s Lemma says that if polynomials factor with rational coefficients, then they factor with integer coefficients.
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What we don’t know anything about yet is what can happen when we expand to larger rings of polynomials, like \(\R[x]\) or \(\C[x]\text{.}\) That’s what we are going to investigate today. So first, a short review of complex numbers.
The complex numbers are the set of numbers \(a+bi\) where \(i^2 = -1\text{.}\) So in some ways, every complex number is determined by two coordinates: the real coordinate and the imaginary coordinate.
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This suggests a way to represent complex numbers graphically. Plot some points on the complex plane.
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We can also describe complex numbers using a polar representation, which uses an angle and distance from the origin. We can convert using trigonometry.
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Instead of referring to a complex numbers as \(z = r\cos(\theta) + i r\sin(\theta)\text{,}\) we will instead use the more compact notation of \(re^{i\theta}\text{.}\) THis might seem like an arbitrary choice, but it is the choice that makes all of complex analysis work nicely together.
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One reason this is so nice is that multiplying complex numbers becomes really easy. Try multiplying some.
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Now suppose we wanted to factor \(x^5 + 1\text{.}\) What we need is a 5th root of -1. How can we write this in polar form?
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How many roots do we expect?
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Do another example: factor \(x^4 - 2\text{.}\)
Handout Wednesday 9/17
Today we explored using roots of unity to factor polynomials in the form \(x^n - a\) over the complex numbers. The textbook has a good number of examples of this in section 2.4.
Handout Friday 9/19
Today we saw that when you multiply the linear factors corresponding to roots that are complex conjugates of each other, that the resulting quadratic has only real coefficients. See the examples in section 2.4 for more details.
