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Using the division algorithm repeatedly, show

$$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = a_n (x-k) (x-j) \cdots (x-b)$$ for $n$ greater than or equal to $1$.

My attempt: (Proof by induction) Consider the case $n=1$. Then, we can write $$ax + b=a\left(x-\left(-\frac{b}{a}\right)\right).$$ Now assume it is true that $$a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = a_n (x-k) (x-j)\cdots(x-b)$$ for some constants $k,j,\ldots,b$.

We will show that $$a_{n+1} x^{n+1} + a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 = a_{n+1}(x-k')(x-j')\cdots(x-b')$$ for some constants $k',j',\ldots,b'$.

I am stuck here, do I use the division algorithm here?

  • 2
    The main difficulty associated with your approach is that the only nontrivial aspect of proving the Fundamental Theorem of Algebra is showing the existence of one complex root. This would allow you to apply your inductive hypothesis. There are a number of ways to show this. Arguably the most accessible proofs use elementary complex analysis (see Spivak's *Calculus* or any basic complex variables book), though there also exist proofs using algebraic and topological techniques.2012-07-31
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    @Chanyang Ryoo Are you wanting to assume the fact mentioned by Tim, i.e., that every polynomial with complex coefficients has a complex root?2012-07-31
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    Hint: the inductive step employs the [Factor Theorem,](http://en.wikipedia.org/wiki/Factor_theorem) a special case of the Division Algorithm.2012-07-31

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