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?

  • 1
    $\begingroup$ You have to write as a product. Your attempt is the difference of two products. $\endgroup$ – Hagen von Eitzen Apr 13 '13 at 20:45
  • $\begingroup$ @HagenvonEitzen Thank you, how would you solve this $a^2-ab-bc+c^2$? Moment, have to think myself... $\endgroup$ – hhh Apr 13 '13 at 20:50

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 $$ a^2-ab-bc+c^2=\left(a+c-\frac{b}{2}+\sqrt{\frac{b^2+8ac}{4}}\right)\left(a+c-\frac{b}{2}-\sqrt{\frac{b^2+8ac}{4}}\right) $$

  • $\begingroup$ What about with arbitrary constants $A,B,C,D$ like $A a^2+B ab+ C bc+D C^2$? Is there any generic way to find the factorization? Anyway +1, very good thinking out of the box -- works even with complex numbers! $\endgroup$ – hhh Apr 14 '13 at 1:52


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.

  • $\begingroup$ The moment you wrote it, I solved it also :D I realized this $a^2-c^2=(a-c)(a+c)$ and $ab+bc=b(a+c)$. Always when I need to speak/write something, things become easier! 9 mins to accept... $\endgroup$ – hhh Apr 13 '13 at 20:47
  • $\begingroup$ @hhh, 9 mins to accept... Accepting an answer should be more a marathon than a sprint. $\endgroup$ – Peter Phipps Apr 13 '13 at 23:29
  • $\begingroup$ It is possible to factorize this if $a,b,c,\alpha,\alpha',\gamma,\gamma' \in \mathbb C$. Choose $\alpha '\gamma=i$ then $1+(\alpha'\gamma)^2=1-1=0$. Thinking now for complex numbers...anyway smart analysis +1! $\endgroup$ – hhh Apr 14 '13 at 1:11

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