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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.
