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'm trying to find all solutions for $36x^3-127x+91=0$ with $x \in \mathbb{R}$. So, I tried to factor this polynomial. It can be written in the following way:

$$ (ax^2+bx+c)\cdot(dx+e)\quad (a,b,c,d,e \in \mathbb{Z}) $$


\begin{cases} a \cdot d = 36 = 2^2 \cdot 3^2\\ a \cdot e + b \cdot d = 0\\ b \cdot e + c \cdot d = -127\\ c \cdot e = 91 = 7 \cdot 13 \end{cases}

How do I proceed from here? Should I guess possible values for $e$ ($±1,±7,±13,±91$) and see if it yields integer solutions for $a,b,c,d$? Or is there an easier method?

share|cite|improve this question
Do you know the rational root theorem? It's easier this way. – Ayman Hourieh Jun 8 '13 at 21:02
When you are asked to factor a cubic with integer coefficients, there will almost always be a rational root. Artificial, yes, but the process is too painful otherwise. – André Nicolas Jun 8 '13 at 21:04
Do you see that $1$ is a remarkable root? – user63181 Jun 8 '13 at 21:07
up vote 3 down vote accepted





$$(x-1)(36x(x+1)-91)=0$$ $$(x-1)(36x^2+36x-91)=0$$

share|cite|improve this answer
Oh wow, I missed the fact that $36 + 91 = 127$. Thank you. – Tim Vermeulen Jun 8 '13 at 21:12

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.