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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 \(x^4 - 2\text{.}\)
