How integral values of $k$ exists such that all roots of the polynomial:

$$f(x)=x^3-(k-3)x^2-11x+(4k-8)$$ are also integers.

Could someone please provide me some direction to proceed in this. First I thought first root would be obtained by hit and trial but it is not the case here. I also tried rewriting $x^3-(k-3)x^2-11x+(4k-8)=0$ as $x^3+3x^2-11x-8=k(x^2-4)$ and solve the question graphically but didn't succeed in that. Please give some direction

  • $\begingroup$ So how's the GRE? Lol $\endgroup$
    – BCLC
    Apr 24, 2018 at 20:21

1 Answer 1



Since $$k ={x^3+3x^2-11x-8\over x^2-4}\implies x^2-4 \mid x^3+3x^2-11x-8$$

Since also $$x^2-4\mid (x+3)(x^2-4) = x^3+3x^2 -4x-12$$ we get

$$ x^2-4\mid ( x^3+3x^2-11x-8)-( x^3+3x^2-4x-12)= -7x+4$$

thus $$x-2\mid -7x+4\implies x-2 \mid -7x+4+7(x-2) = -10$$

Finally we have $x-2\in \{-10,-5,-2,-1,1,2,5,10\}$ so $x\in \{-8,-3,0,1,3,4,7,12\}$

Now, of course not all these are good...

  • 4
    $\begingroup$ If you also do $x+2 \mid -7x+4$, you'll get $x+2 \mid 18$. $\endgroup$
    – lhf
    Apr 20, 2018 at 19:10
  • $\begingroup$ Yes of course, but there was not so many candidates so I didn't do that. In the end we have to check which one satisfies $x^2-4\mid -7x+4$. $\endgroup$
    – nonuser
    Apr 20, 2018 at 19:12
  • $\begingroup$ @ChristianF I don't understand last line. "Now, of course not all these are good" $\endgroup$
    – Ananya
    Apr 20, 2018 at 19:21
  • 1
    $\begingroup$ Read my last comment. $\endgroup$
    – nonuser
    Apr 20, 2018 at 19:22

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