First of all, I am so embarrassed by this. I am tutoring this kid in math and this question came up: Find the LCM of $y^2 - 81$ and $9 - y$, which factor into $(y + 9)(y - 9)$ and $-(y - 9)$

My answer: $-(y+9)(y-9)$

Her book's answer: $(y+9)(y-9)$

My question: Why is the negative left off? Isn't it ($-1$) a factor of the second binomial?

  • $\begingroup$ gcds & lcms in polynomial rings are defined only up to unit (invertible) factors. For polynomials (over fields) they usually normalized to be monic, i.e lead coef $= 1.\,$ See here for more on such unit normalization. $\endgroup$ Nov 21, 2018 at 2:02

1 Answer 1


Both answers are valid. Any nonzero constant can be factored out of any polynomial.

Technically speaking, finding the $\operatorname{lcm}$ of two elements is an action done in a ring. If it exists, the $\operatorname{lcm}$ is unique up to multiplication of a unit of the ring, i.e. an invertible element.

In this case, all nonzero constants are units in a polynomial ring over a field (here $\mathbb{R}$), so multiplying by a constant still gives an $\operatorname{lcm}$. If we instead are considering these polynomials over $\mathbb{Z}$, there's still no problem since $-1$ is a unit.

  • $\begingroup$ "all nonzero constants are units". Also better to restrict to domains vs. rings since divisibility theory in general rings is more complex, e.g. associates are no longer the same as unit multiples; furthermore the notions of associate and irreducible bifurcate into a few inequivalent notions in use. $\endgroup$ Nov 21, 2018 at 2:32
  • $\begingroup$ Good point. Could you edit to add the nuance for rings vs domains? My ring theory's a bit rusty. $\endgroup$
    – Sambo
    Nov 21, 2018 at 3:13

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