Express polynomial from polynomial ring over the integers as a product of irreducible elements

Question

I am asked to solve the following problem.

Express $f(x) = 15x^{4} + 75x^{3} + 10x^{2} + 44x - 30 \in \mathbb{Z}[x]$ as a product of irreducible elements in $\mathbb{Z}[x]$.

So the idea is to translate the problem into a problem of Frac$(\mathbb{Z})[x] = \mathbb{Q}[x]$. So note first that $f$ is primitive since the coefficients are co-prime. Then, if we express $f$ as a product of irreducibles in $\mathbb{Q}[x]$, then it will be irreducible over $\mathbb{Z}$ too.

I tried applying Eisenstein's Criterium, but $f$ is not an Eisenstein polynomial. So next comes to apply the Rational Root Test. And I was able to find a zero of $f$, namely, $-5$. Then, by the Factor Theorem, $x + 5 | f(x)$ and we apply polynomial long division.

The issue that I have with the Rational Root Test in this particular problem is that there were far too many posibilities for zeros of $f$ and computing the possibilities took far too long. Is there a more efficient method to solve this problem?

Answer 1

Suppose ${\large{\frac{p}{q}}}$ is a rational root of $$ f(x) = 15x^{4} + 75x^{3} + 10x^{2} + 44x - 30 $$ where $p,q$ are integers with $p\ne 0,q > 0$ and $\gcd(p,q)=1$.

By the rational root test, we must have $p{\,\mid\,}30$ and $q{\,\mid\,}15$.

From $f\bigl(\frac{p}{q}\bigr)=0$, we get $$ 15p^4+75p^3q+10p^2q^2+44pq^3-30q^4=0\qquad(\text{eq}1) $$ From $(\text{eq}1)$, considering the value of $q$, it follows that

• $q$ can't be a multiple of $3$ else all terms of the $\text{LHS}$ of $(\text{eq}1)$ except the first would be divisible by $3^2$.$\\[4pt]$
• $q$ can't be a multiple of $5$ else all terms of the $\text{LHS}$ of $(\text{eq}1)$ except the first would be divisible by $5^2$.

Hence, since $q{\,\mid\,}15$ and $q > 0$, it follows that $q=1$, so $(\text{eq}1)$ reduces to $$ 15p^4+75p^3+10p^2+44p-30=0\qquad(\text{eq}2) $$ From $(\text{eq}2)$, considering the value of $p$, it follows that

• $p$ can't be positive.
• $p$ can't be a multiple of $2$ else all terms of the $\text{LHS}$ of $(\text{eq}2)$ except the last would be divisible by $2^2$.$\\[4pt]$
• $p$ must be a multiple of $5$ since all terms of the $\text{LHS}$ of $(\text{eq}2)$ except the term $44p$ are divisible by $5$.

So now there only two cases to check, namely $p=-5$ and $p=-15$.

To carry the analysis a little further, consider the reduction of $(\text{eq}2)$, mod $4$ . . .

Since $p$ is odd, we get $p^2\equiv 1\;(\text{mod}\;4)$, so $p^3\equiv p\;(\text{mod}\;4)$ and $p^4\equiv 1\;(\text{mod}\;4)$, hence \begin{align*} & 15p^4+75p^3+10p^2+44p-30=0 \\[4pt] \implies\;& (-1)(1)+(-1)p+2(1)+(0)p-2\equiv 0\;(\text{mod}\;4) \\[4pt] \implies\;& p\equiv -1\;(\text{mod}\;4) \\[4pt] \end{align*} so we can't have $p=-15$.

Hence the only remaining case to check is $p=-5$.