this problem is from Gelfand & Shen's Algebra book.

Problem 171. The highest coefficient of $P(x)$ is $1$, and we know that $P(1)=0,P(2)=0,P(3)=0,\ldots,P(9)=0,P(10)=0$. What is the minimal possible degree of $P(x)$? Find $P(11)$ for this case.


The minimal degree is $10$ and $P(11)$ is $3628800$ in this case.

I would appreciate if you could give me some hints.


$P(X_0) = 0 \Leftrightarrow (X - X_0) | P$

So you have $P = Q( X- 1)(...)(X-10)$ for some $Q$.

You want to minimize the degree or $P$ so you'll try to minimize the degree of $Q$ because the degree of a product is the sum of the degrees of the operands.

$-\infty$ (meaning $Q=0$) won't work because you would get $P=0$ so the highest coefficient of $P$ wouldn't be $1$.

Then you try $0$ (meaning $Q$ is a constant). The highest coefficient of a product is the product of the highest coefficients of the operands so you want $c(Q)\times1\times...\times1 = 1$ ie $c(Q)=1$ ($c(Q)$ being the highest coefficient of $Q$). Since $Q$ is constant, you get $Q=1$.

So you have $P = ( X- 1)(...)(X-10)$. And you just need to evaluate that at $11$. $P(11) = 10\times 9 \times ... \times 1 = 10! = 3,628,800$

  • $\begingroup$ Thank you for clear explanation. $\endgroup$ – Paul Dirac Nov 13 '12 at 22:02


When looking at a function over the real numbers $f : \mathbb{R} \to \mathbb{R}$, we have $f(k) = 0$ if and only if $(x-k) \mid f(x)$. In other words, $f(k) = 0$ if and only if $f(x) = (x-k) g(x)$, for some function $g(x)$. Note that in the way I worded my answer, it may or may not be true that $g(k) = 0$, but this part doesn't matter here.

  • $\begingroup$ Oh of course, the fact that $P(1)=0,P(2)=0,P(3)=0,\ldots,P(9)=0,P(10)=0$, means that polynomial has at least 10 roots, so it has to be at least degree 10. This is so easy I don't know why I couldn't see it. Thank you very much JavaMan. $\endgroup$ – Paul Dirac Nov 13 '12 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.