Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
You have to write as a product. Your attempt is the difference of two products. – Hagen von Eitzen Apr 13 '13 at 20:45
@HagenvonEitzen Thank you, how would you solve this $a^2-ab-bc+c^2$? Moment, have to think myself... – hhh Apr 13 '13 at 20:50
up vote 2 down vote accepted

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

share|cite|improve this answer
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! – 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.

share|cite|improve this answer
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... – hhh Apr 13 '13 at 20:47
@hhh, 9 mins to accept... Accepting an answer should be more a marathon than a sprint. – Peter Phipps Apr 13 '13 at 23:29
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! – hhh Apr 14 '13 at 1:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.