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Find the number of monic polynomials $p(x)$ with complex coefficients such that degree of $p(x)$ is at most $7$ and all its roots lie in the set {${1,2,3}$}.

I know the definition of monic polynomial here

I've done :

when degree of monic polynomial is $1$ we have $p(x)= x+a$ , in this case we can get $3$ monic polynomials because roots lie in {${1,2,3}$}.

when degree is $2$ , $p(x)=x^2+ax+b$ , in this case we have ? monic polynomials

But I am unable to think further .

Could anyone please help me ?? Any short method would be highly appreciated .(as Im preparing for an exam where we have to answer within $3$ minutes)

Thanks!

(I'm not sure about the tags .)

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All those polynomials have the form $$p(x)=(x-1)^m(x-2)^n(x-3)^p$$

That's because when working over the complex number you will always have that the polynomial will split into linear factors.

If you impose the condition that $\deg(p)\leq7$, then this translates into $m+n+p\leq7$

Thus you need to solve $m+n+p\leq7$ in the non-negative integers.

Can you do this?

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    $\begingroup$ Which part don't you understand? The part about the polynomials having that form? the part about translating into the inequality? how to solve the inequality? Meet asdf halfway! $\endgroup$ – Gerry Myerson Jun 26 '18 at 7:29
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    $\begingroup$ A consequence of the fundamental theorem of algebra is that every polynomila of degree $n$ over $\mathbb{C}$ has $n$ roots. Hence all of those $n$ roots must be $1,2$ or $3$ counting multiplicities. Also, since your polynomial is monic, you get the representation above. Expanding the brackets gives that the degree is going to be $m+n+p$. But you want this degree to be $\leq 7$ which gives the inequality. The inequality itself can be solved by direct counting $\endgroup$ – asdf Jun 26 '18 at 8:12
  • $\begingroup$ You also need $0 \lt m+n+p$ to exclude the case $\{0,0,0\}$, since $p(x)=1$ is not a solution. $\endgroup$ – gandalf61 Jun 26 '18 at 10:12
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    $\begingroup$ Technically, it is a solution, since all of its roots are in $\{1,2,3\}$ because it doesn't have a root outside of this set $\endgroup$ – asdf Jun 26 '18 at 12:29

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