The problem:

Find the polynomial $r(x)$ with the smallest degree and integer coefficients s.t. it has zeroes -1, 2/3 and -2 w/ resp multiplicities 2, 8 and 4 and $r(1) = -1$.

What I tried:

Without the integer coefficients and $r(1) = -1$ constraints, I think we can have:

$$r(x) = (x+1)^2(x-2/3)^8(x+2)^4$$

With the $r(1) = -1$ constraint, we must find either of the following

  1. a suitable $a$:

$$r(x) = a(x+1)^2(x-2/3)^8(x+2)^4$$

  1. a suitable $b$:

$$r(x) = (x+1)^2(x-2/3)^8(x+2)^4 + b$$

As for the integer coefficient constraint, I kind of have a feeling that that dictates which of the above we should choose. Does it? Where exactly does the integer coefficient constraint come in?

I guess without the $r(1) = -1$ constraint but with the integer coefficient constraint, we could have:

$$r(x) = \frac{1}{3^8}(x+1)^2(3x-2)^8(x+2)^4$$

Wait that doesn't really satisfy the integer coefficient constraint does it?

Okay so I don't know. How do I satisfy both the integer coefficients and $r(1) = -1$ constraints?

  • 1
    $\begingroup$ In case 2. $r(-1) = b$, $-1$ should be a zero. $\endgroup$ – Abstraction Apr 18 '16 at 8:55
  • 2
    $\begingroup$ Adding $b$ will ruin our roots, so this is out of question. As for the integer coefficients constraint, just ignore it until you have the answer. Find that $a$, then see where it gets you. $\endgroup$ – Ivan Neretin Apr 18 '16 at 9:17
  • $\begingroup$ @Abstraction Ah thanks. So it's really just case 1 then? $\endgroup$ – BCLC Apr 18 '16 at 9:18
  • $\begingroup$ @IvanNeretin If roots were the only constraints... obviously, $r(x)=q(x)(x+1)^2(3x-2)^8(x+2)^4$, but $q(1)=-{1 \over 2^2*3^4}$ with integer coefficients of $r$ seems impossible. Can't figure out a strict proof, though. $\endgroup$ – Abstraction Apr 18 '16 at 9:34
  • $\begingroup$ @Abstraction smallest degree so $q(x)$ is constant? $\endgroup$ – BCLC Apr 18 '16 at 9:46

There is no $r(x)$ meeting the required conditions.

Suppose there was. Clearly its coefficients would have gcd 1 (because $r(1)=-1$), so it would be "primitive". By Gauss polynomial lemma, $(3x-2)^8$ must be a factor of $r(x)$. Similarly $(x+1)^2$ and $(x+2)^4$. Hence we must be able to write it as $q(x)(x+1)^2(3x-2)^8(x+2)^4$ where $q(x)$ has integer coefficients. But then $r(1)=-1$ implies $q(1)=-\frac{1}{324}$, which is impossible.

To spell out the first step in more detail: the lemma states that if $r(x)$ is a (non-constant) primitive polynomial with integer coefficients which is reducible over $\mathbb{Q}$, then it is also reducible over $\mathbb{Z}$. We know that $r(x)$ has the factor $(x-\frac{2}{3})^8$ in $\mathbb{Q}[x]$, so we keep taking out factors until we are forced to take out that factor in $\mathbb{Z}[x]$ as $(3x-2)^8$.

  • $\begingroup$ omg. this is a problem in a sample long test in a precalculus class. well thanks almagest :)) it was probably typo in the exam. damn integer coefficient requirement. this isn't a number theory class. :| $\endgroup$ – BCLC Apr 18 '16 at 11:14

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