MHB Polynomial Challenge: Find $k$ Integral Values

[anemone]

Find all integral values of $k$ such that $q(a)=a^3+2a+k$ divides $p(a)=a^{12}-a^{11}+3a^{10}+11a^3-a^2+23a+30$.

 

[Albert1]

Find all integral values of $k$ such that $q(a)=a^3+2a+k$ divides $p(a)=a^{12}-a^{11}+3a^{10}+11a^3-a^2+23a+30$.

Spoiler

let:$p(a)=a^{12}-a^{11}+3a^{10}+11a^3-a^2+23a+30----(1)$
p(a) has no positive root and p(a) has at most two negative roots
if (1)=0 then the solutions of (1) must be negative
if a<0 is the integer root of (1) then a divides 30
by checking a=-1
$\therefore q(-1)=-1-2+k=0 $
we have :$k=3$

 

[anemone]

Spoiler

let:$p(a)=a^{12}-a^{11}+3a^{10}+11a^3-a^2+23a+30----(1)$
p(a) has no positive root

Spoiler

Thanks for your solution, Albert!:) But, how do you conclude that $p(a)$ has no positive root(s)? I'm sensing perhaps you're using some theorem that I'm not aware of?

 

[Albert1]

Thanks for your solution, Albert!:) But, how do you conclude that $p(a)$ has no positive root(s)? I'm sensing perhaps you're using some theorem that I'm not aware of?

it is easy to conclude that $p(a)$ has no positive root (only by observation)
if $a\geq 1$ then :
$p(a)=(a^{12}-a^{11})+(11a^3-a^2)+3a^{10}+23a+30>0$
if $0<a< 1$ then :
$p(a)=(3a^{10}-a^{11})+(23a-a^2)+a^{12}+11a^3+30>0$

 

[anemone]

it is easy to conclude that $p(a)$ has no positive root (only by observation)
if $a\geq 1$ then :
$p(a)=(a^{12}-a^{11})+(11a^3-a^2)+3a^{10}+23a+30>0$
if $0<a< 1$ then :
$p(a)=(3a^{10}-a^{11})+(23a-a^2)+a^{12}+11a^3+30>0$

Ah I see. Thanks for the clarification reply, Albert!