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I'm having a bit trouble with this excercise:

The problem:
Let there be a polynomial $f(x)=a_1x^{t_1} + a_2x^{t_2} + ... + a_nx^{t_n}$ Where $t_1, t_2, ..., t_n$ are not-negative integers. The polynomial has a root $b$ which occurs $n$ times. Prove that $b = 0$.

What I have so far:
I can presume that $a_1, a_2, ..., a_n \neq 0$
If $n = 1$, then it's obviously true.
If $n = 2$, I tried using this:
$f^{(n-1)}(b) = 0 \\ f^{(n)}(b) = 0$
$f(b) = a_1b^{t_1} + a_2b^{t_2} = 0 \\ f'(b) = t_1*a_1b^{(t_1-1)} + t_2*a_2b^{(t_2-1)} = 0 \\ f''(b) = (t_1-1)t_1*a_1b^{(t_1-2)} + (t_2-1)t_2*a_2b^{(t_2-2)} \neq 0$
Why can't $b \neq 0$ be true?

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So, you presume your polynomial $f(x)$ factors as $(x-b)^n r(x)$, where $r(x)$ is another polynomial. Have you tried comparing coefficients? – J. M. May 4 '12 at 12:41
My first instinct would be to assume (without loss of generality) that $t_1\le t_2\le\cdots\le t_n$ and factor out $x^{t_1}$. – Isaac May 4 '12 at 12:44
Have you counted the number of non-zero coefficients in $(x-b)^n$ when b is not zero and compared it to your number of coefficients? – Chris K. Caldwell May 4 '12 at 12:57
Thanks for your replies. $(x-b)^n$ has $n+1$ number of coefficients (binomial theorem). f(x) has $n$. But how can I prove that $(x-b)^nr(x)$ still has $n+1$ coefficients? – user30620 May 4 '12 at 13:45
@user30620: $(x-b)^n$ has $n+1$ terms if and only if $b\neq0$... – Isaac May 6 '12 at 5:26

You have $b^kf^{(k)}(b)=0$ for $k=0,\dots,n-1$, i.e. $\sum_{j}p_i(t_j)b^{t_j}a_j=0$, where $p_i(x)=x(x-1)\dots(x-i+1)$. If $b^{t_j}a_j\neq 0$ for at least one $j$ then there are some $c_0,\dots,c_{n-1}$, not all zero, such that $\sum_i c_ip_i(t_j)=0$ for all $j$'s (that's where linear algebra is used). But the polynomial $\sum_i c_ip_i(x)$ has degree at most $n-1$, so it can't have $n$ different roots.

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Could you explain "If $b^{t_j}a_j\neq 0$ for at least one j then there are some c0,…,cn−1, not all zero, such that $\sum_i c_ip_i(t_j)=0$ for all j's (that's where linear algebra is used)" a bit further? – user30620 May 4 '12 at 16:58
@user30620: if $M$ is a square matrix and $Mu=0$ for some non-zero vector (column) $u$ then there is a non-zero (row) vector $v$ s.t. $vM=0$ (as row rank = column rank). In our case the entries of $M$ are $p_i(t_j)$. – user8268 May 4 '12 at 20:55
Are $u$ and $v$ taken from the matrix $M$? Also are you implying that $\sum_i c_ip_i(x)$ is equal to $f(x)$? – user30620 May 4 '12 at 22:02

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