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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.

Answer.

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

I would appreciate if you could give me some hints.

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

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 $11$. $P(11) = 10\times 9 \times ... \times 1 = 10! = 3,628,800$

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

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.

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  • $\begingroup$ 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

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