Problem Of The Week #430 August 18th, 2020

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anemone

MHB POTW Director
Staff member
Here is this week's POTW:

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Find all integers $n$ such that all roots of the following polynomial are also integers:

$P(x)=x^3-(n-3)x^2-11x+4n-8$

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anemone

MHB POTW Director
Staff member
No one answered last week's POTW, but you can find the suggested solution below:

Let $a,\,b$ and $c$ be the integral roots of the function, $P(x)=x^3-(n-3)x^2-11x+4n-8$.

Therefore we have

$P(x)=(x-a)(x-b)(x-c)$

$P(2)=(2)^3-(n-3)(2)^2-11(2)+4n-8=8-4n+12-22+4n-8=-10=(2-a)(2-b)(2-c)$

This implies $(2-a)(2-b)(2-c)=-10$ and $ab+bc+ca=-11$.

Solving for integers values we find $(a,\,b,\,c)$ can be any permutation set of $(1,\,4,\,-3)$.

$abc=8-4n\\-12=8-4n\\ \therefore n=5\,\text{is the only solution.}$

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