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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:

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

with

\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?

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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
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Do you see that $1$ is a remarkable root? –  Sami Ben Romdhane Jun 8 '13 at 21:07

1 Answer 1

up vote 3 down vote accepted

$$36x^3-127x+91=0$$

$$36x^3-36x-91x+91=0$$

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

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

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

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Oh wow, I missed the fact that $36 + 91 = 127$. Thank you. –  timvermeulen Jun 8 '13 at 21:12

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