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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:
$f^{(n-1)}(b) = 0 \\ f^{(n)}(b) = 0$
So:
$f(b) = a_1b^{t_1} + a_2b^{t_2} = 0 \\ f'(b) = t_1*a_1b^{(t_1-1)} + t_2*a_2b^{(t_2-1)} = 0 \\ f''(b) = (t_1-1)t_1*a_1b^{(t_1-2)} + (t_2-1)t_2*a_2b^{(t_2-2)} \neq 0$
Why can't $b \neq 0$ be true?

  • 1
    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?2012-05-04
  • 0
    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 $x^{t_1}$.2012-05-04
  • 0
    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?2012-05-04
  • 1
    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?2012-05-04
  • 0
    @user30620: $(x-b)^n$ has $n+1$ terms if and only if $b\neq0$...2012-05-06

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