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To be honest i don't even know where to start. I thought about using diagonalizable matrices and characteristic polynomials, but the class hasn't gotten there yet so there should be a way to solve this with simpler stuff. Any ideas?

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    $\begingroup$ I see an $n$. It must be induction. Perhaps divide $f$ by the product of the first $n-1$ factors of $g$? $\endgroup$ Feb 9, 2020 at 7:44
  • $\begingroup$ I'm doing it by induction, but the other way around. I showed that it works for n=2, and am showing that if you start with the partial fraction decomposition and reassemble it, you'll end up with a n-1 degree polynomial on the numerator. I'm unsure if that's a valid approach but I think it should be? $\endgroup$ Feb 9, 2020 at 22:22
  • $\begingroup$ It sounds to me that you are doing it backwards, you start with a numerator of degree $n-1$ rather than end up with one. $\endgroup$ Feb 10, 2020 at 7:39

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We proceed by strong induction. Given that $n = 1,$ we have that $f(x) = a$ for some real number $a$ so that $$\frac{f(x)}{g(x)} = \frac a {x - c_1},$$ as desired. We will assume inductively that for any polynomial $f(x)$ of degree $\leq n - 1,$ we have that $$\frac{f(x)}{g(x)} = \frac{a_1}{x - c_1} + \cdots + \frac{a_n}{x - c_n}$$ for some real numbers $a_1, \dots, a_n.$ Consider the polynomial $f(x) = b_n x^n + b_{n - 1}x^{n - 1} + \cdots + b_0$ of degree $\leq n.$ We have that $$\frac{f(x)}{g(x)} = \frac{b_n x^n + b_{n - 1}x^{n - 1} + \cdots + b_0}{(x - c_1) \cdots (x - c_n)(x - c_{n + 1})} = \frac 1 {x - c_{n + 1}} \biggl(\frac{b_n x^n}{(x - c_1) \cdots (x - c_n)} + \frac{b_{n - 1} x^{n - 1} + \cdots + b_0}{(x - c_1) \cdots (x - c_n)} \biggr).$$ Given that $b_n = 0,$ the expression in the parentheses is the ratio of a polynomial of degree $\leq n - 1$ and a polynomial of degree $n,$ hence by strong induction, we have that $$\frac{f(x)}{g(x)} = \frac{1}{x - c_{n + 1}} \biggl(\frac{d_1}{x - c_1} + \cdots + \frac{d_n}{x - c_n} \biggr) = \frac{d_1}{(x - c_1)(x - c_{n + 1})} + \cdots \frac{d_n}{(x - c_n)(x - c_{n + 1})}$$ for some real numbers $d_1, \dots, d_n.$ Once again, by strong induction, we can have that $$\frac{f(x)}{g(x)} = \frac{a_1}{x - c_1} + \cdots + \frac{a_{n + 1}}{x - c_{n + 1}}$$ for some real numbers $a_1, \dots, a_{n + 1}.$ (Each of the terms has its own partial fraction decomposition.)

Consider the case that $b_n \neq 0.$ Observe that $(x - c_1) \cdots (x - c_n)$ is a monic polynomial of degree $n$ and $b_n x^n$ is a polynomial of degree $n,$ hence by the Division Algorithm, we have that $$\frac{b_n x^n}{(x - c_1) \cdots (x - c_n)} = b_n + \frac{r(x)}{(x - c_1) \cdots (x - c_n)}$$ for some polynomial $r(x)$ of degree $\leq n - 1.$ Consequently, by strong induction, we have that $$\frac{f(x)}{g(x)} = \frac{1}{x - c_{n + 1}} \biggl(b_n + \frac{d_1}{x - c_1} + \cdots + \frac{d_n}{x - c_n} \biggr)$$ for some real numbers $d_1, \dots, d_n.$ Like before, by strong induction, we have that $$\frac{f(x)}{g(x)} = \frac{a_1}{x - c_1} + \cdots + \frac{a_{n + 1}}{x - c_{n + 1}}$$ for some real numbers $a_1, \dots, a_{n + 1}.$ Our proof is complete by strong induction. QED.

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