Factorize $a^2-ab-bc\pm c^2$

Question

I got this question in a test but it did not specify the variable with respect to which I was supposed to factorize

$$a^2-ab-bc\pm c^2$$

where it could be just $a(a-b)-c(b\pm c)$ but no common factor over all terms. I feel I may be missing something. The $\pm$ is there because I cannot remember whether the last sign was minus or plus.

Is there some trick to factorize this or is this question vacuous? What does it mean to factorize this?

Answer 1

We observe that

\begin{eqnarray} a^2-ab-bc+c^2&=&a^2+c^2-b(a+c)\\ &=&(a+c)^2-b(a+c)-2ac\\ &=&(a+c)^2-b(a+c)+\frac{b^2}{4}-\frac{b^2+8ac}{4}\\ &=&\left(a+c-\frac{b}{2}\right)^2-\frac{b^2+8ac}{4}. \end{eqnarray} Hence, if $b^2+8ac\geq 0$ then $$ a^2-ab-bc+c^2=\left(a+c-\frac{b}{2}+\sqrt{\frac{b^2+8ac}{4}}\right)\left(a+c-\frac{b}{2}-\sqrt{\frac{b^2+8ac}{4}}\right) $$

Answer 2

$a^2-ab-bc-c^2=a^2-c^2-ab-bc=(a-c)(a+c)-b(a+c)=(a-c-b)(a+c)$.

If $a^2-ab-bc+c^2$ is factorizable, it is equal to $(\alpha a + \beta b + \gamma c)(\alpha' a + \beta' b + \gamma' c)$ where $\alpha,\alpha',\beta, \beta',\gamma, \gamma' \in \mathbb{R}$. So, $\alpha \alpha'=1$, $\gamma \gamma'=1$ and $\alpha \gamma' + \alpha' \gamma=0$.

$0=\alpha' \gamma( \alpha \gamma' + \alpha' \gamma)=(\alpha \alpha')(\gamma \gamma')+ (\alpha' \gamma)^2=1 +( \alpha' \gamma)^2$. It's impossible.

$a^2-ab-bc+c^2$ is not factorizable.