# Integral values of $k$ for which polynomial has integral roots.

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

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

Hint:

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...

• If you also do $x+2 \mid -7x+4$, you'll get $x+2 \mid 18$.
– lhf
Apr 20, 2018 at 19:10
• 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 $x^2-4\mid -7x+4$. Apr 20, 2018 at 19:12
• @ChristianF I don't understand last line. "Now, of course not all these are good" Apr 20, 2018 at 19:21
• Read my last comment. Apr 20, 2018 at 19:22