Find the numbers at which the polynomial is irreducible over $\mathbb{Q}$

Question

How can I find the integers $a$ at which the polynomial $f(x)$ is irreducible over the field $\mathbb{Q}$? Thank you! $$f(x) = 5x^4 - 6x^3 - ax^2 - 4x + 2$$

Answer 1

By the Rational Roots Theorem, the following rational numbers can be roots of this polynomial: $$\pm \frac 1 5, \pm 1, \pm \frac 2 5, \pm 2$$ Now, plug these eight roots into $f(x)$ to get an expression in terms of $a$ and then solve for $a$. $f(x)$ is reducible for those values of $a$. I will get you started on making the equations: $$x=\frac 1 5 \implies 5\left(\frac 1 5\right)^4 - 6\left(\frac 1 5\right)^3 - a\left(\frac 1 5\right)^2 - 4\left(\frac 1 5\right) + 2=0 \implies f(x) \text{ is reducible for } a=29$$ $$x=-\frac 1 5 \implies 5\left(-\frac 1 5\right)^4 - 6\left(-\frac 1 5\right)^3 - a\left(-\frac 1 5\right)^2 - 4\left(-\frac 1 5\right) + 2=0 \\ \implies f(x) \text{ is reducible for } a=\frac{357}{3}$$ Note that the last equation above does not give us an integer solution, so we ignore it.

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However, we also need to find the values of $a$ for which the polynomial can be factored into two quadratics. The product of the leading coefficients must be $5$ and the product of the constants must be $2$, so assuming positive leading coefficients, there are four cases: $$(5x^2+Ax+1)(x^2+Bx+2)$$ Here, the $x^3$ term is $(A+5B)x^3$ and the $x$ term is $(2A+B)x$, so we have: $$A+5B=-6 \wedge 2A+B=-4 \implies (A,B)=\left(-\frac{14}{9}, -\frac{8}{9}\right)$$ We don't get integer solutions here, so we ignore this. The next case is: $$(5x^2+Ax-1)(x^2+Bx-2)$$ Here, the $x^3$ term is $(A+5B)x^3$ and the $x$ term is $(-2A-B)x$, so we have: $$A+5B=-6 \wedge -2A-B=-4 \implies (A,B)=\left(\frac{26}{9}, -\frac{16}{9}\right)$$ We don't get integer solutions here, so we ignore this. The next case is: $$(5x^2+Ax+2)(x^2+Bx+1)$$ Here, the $x^3$ term is $(A+5B)x^3$ and the $x$ term is $(A+2B)x$, so we have: $$A+5B=-6 \wedge A+2B=-4 \implies (A,B)=\left(-\frac{8}{3}, -\frac{2}{3}\right)$$ We don't get integer solutions here, so we ignore this. The next case is: $$(5x^2+Ax-2)(x^2+Bx-1)$$ Here, the $x^3$ term is $(A+5B)x^3$ and the $x$ term is $(-A-2B)x$, so we have: $$A+5B=-6 \wedge -A-2B=-4 \implies (A,B)=\left(\frac{32}{3}, -\frac{10}{3}\right)$$ We don't get integer solutions here, so we ignore this.

Therefore, factoring into two quadratics yield no solutions for $a$, so we only have to worry about the $a$ we solved for in the first part of this answer.