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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.
