Section Week 3 (9/8-9/12)
Let’s do some algebra with polynomials!
Handout Monday 9/8: Division in Polynomials
Solve for \(x\text{:}\)
\begin{equation*}
x^3 + x^2 - 5x - 2 = 0
\end{equation*}
Note that we can factor this as \((x-2)(x^2 + 3x + 1) = 0\text{,}\) so \(x=2\) is a root, and the other roots are the roots of \(x^2 + 3x + 1\text{,}\) which we can find using the quadratic formula. But how did we know to try \(x-2\) as a factor?
When \(F\) is a field, we can consider the polynomial ring \(F[x]\text{.}\) This is NOT a field, but we can use the cancellation property (it is an integral domain). Since it is not a field, we cannot divide by polynomials. Or more precisely, there are polynomials that do not have multiplicative inverses.
However, we can sometimes “divide by” polynomials, in the sense that some polynomials factor, and then we can cancel factors (using the cancellation law). Just like in the integers.
In \(\Z\text{,}\) we can even do more than factor and cancel. We can “divide with remainder”. Let’s see what this looks like and see how it carries over to polynomials.
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In \(\Z\text{,}\) we have that \(4 | 12\text{,}\) which means that \(12 = 4\cdot k\) for some \(k \in \Z\) (namely \(k=3\)).
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However, \(5 \nmid 17\text{.}\) We might say that \(17/5 = 3.4\text{,}\) or better, use fractions: \(12/5 = 3\frac{2}{5}\text{.}\) That means \(5\) goes into 17 three times with a remainder of 2.
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We can make this precise using the division algorithm (which is really a theorem): For any integers \(a\) and \(b \gt 0\) there exist unique integers \(q\) and \(r\) such that\begin{equation*} a = qb + r \end{equation*}where \(0 \le r \lt b\text{.}\)
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Why is this true? Keep adding \(b\) to itself until we are “close” to \(a\text{.}\) That is, close enough that if we add \(b\) again, we would go over. That means that \(r = a - qb\) will be less than \(b\text{.}\)
How does the division algorithm translate to polynomial rings \(F[x]\) for \(F\) a field?
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Let \(a(x)\) and \(b(x) \ne 0\) be two polynomials in \(F[x]\) for some field \(F\text{.}\) Then there are polynomials \(q(x)\) and \(r(x)\) in \(F[x]\) such that\begin{equation*} a(x) = q(x)b(x) + r(x) \end{equation*}where \(r(x) = 0\) or \(\deg r(x) \lt \deg b(x)\text{.}\)
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Our argument for the division algorithm for integers doesn’t really work here, since you can’t just add \(b(x)\) to itself enough times. That doesn’t increase the degree.
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Instead, we just do long division. Do an example, and then try it with \(a(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\) and \(b(x) = b_mx^m + b_{m-1}x^{m-1} + \cdots + b_1x + b_0\text{.}\) Note that after the first step, we are left with a smaller degree dividend, so we can apply induction.
Handout Wednesday 9/10
Last time we uncovered a connection between roots of polynomials and factors of polynomials. But what numbers can be roots? Let’s explore this a bit by working on the Rational Roots activity.
The conclusion? If \(p(x)\) is a polynomial with integer coefficients and \(\frac{s}{t}\) is a root, then \(s\) must divide the constant term and \(t\) must divide the leading coefficient.
Handout Friday 9/12
Can you factor \(x^4 + x^3 + 3x^2 + x + 6\) over \(\Q\text{,}\) or is it irreducible? Does the polynomial have any rational roots?
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Careful. Just because the polynomial has no rational roots does not mean it is irreducible. It could factor into two quadratics with no linear factors. Let’s explore that.
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Let’s try to factor \(x^4 + x^3 + 3x^2 + x + 6\) as \((x^2 + ax + b)(x^2 + cx + d)\) for some rational numbers \(a, b, c, d \in \Q\text{.}\) Multiplying out, we get \(x^4 + (a+c)x^3 + (ac+b+d)x^2 + (ad+bc)x + bd\text{.}\) Setting coefficients equal, we get the system of equations:\begin{equation*} \begin{aligned} a + c \amp= 1 \\ ac + b + d \amp= 3 \\ ad + bc \amp= 1 \\ bd \amp= 6 \end{aligned} \end{equation*}Can we solve this system? If so, what are the values of \(a, b, c, d\text{?}\)
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Let’s try this again with a different polynomial. How about \(a(x) = 8x^4 + 6x^3 + 3x^2 - 15x + 12\text{.}\)
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Let’s suppose that this factors as \(b(x)c(x)\) for some polynomials \(a(x), b(x) \in \Q[x]\text{.}\) In general, we have\begin{equation*} b(x) = b_0 + b_1x + b_2x^2 \end{equation*}and\begin{equation*} c(x) = c_0 + c_1x + c_2x^2.\text{.} \end{equation*}(We assume that there are no linear factors).
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Now look at the different terms when we multiply out \(b(x)c(x)\text{.}\) What will the constant 12 term be? \(b_0c_0 = 12\text{.}\) What will the leading term be? \(b_2c_2 = 8\text{.}\) Notice that \(b_0c_0\) is a multiple of 3 but not of \(3^2 = 9\text{.}\) That means that one of \(b_0\) or \(c_0\) is a multiple of 3, but not both. Let’s assume \(b_0 \nmid 3\text{.}\)
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Now look at \(-15 = b_1c_0 + b_0c_1\text{.}\) Since \(b_0 \nmid 3\text{,}\) we must have \(c_1 \div 3\text{.}\)
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Okay, what about \(3 = b_2c_0 + b_1c_1 + b_0c_2\text{?}\) The LHS is a multiple of 3, and \(c_0\) and \(c_1\) are too, but \(b_0\) is not. So \(c_2\) is a multiple of 3.
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Keep going... wait, what is \(c_2b_2\text{?}\) That’s 8, but this is impossible if \(c_2\) is a multiple of 3. Unless \(b_2\) is not an integer.
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However, it turns out that if a polynomial factors as two polynomials with rational coefficients, then it must also factor as two polynomials with integer coefficients. So we have proven that this polynomial doesn’t factor.
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Let’s generalize this. If there is some prime number \(p\) that divides all the coefficients of some polynomial \(a(x) \in \Z[x]\text{,}\) except the leading coefficient, and \(p^2\) does not divide the constant coefficient, then the polynomial is irreducible over \(\Z\text{,}\) and therefore over \(\Q\text{.}\) This is called Eisenstein’s Criterion.
