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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?

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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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You are going in the wrong direction.

Your initial case $n=1$ is correct.

Now suppose that it is true for $n$, take a polynomial $p$ of degree $n+1$. Then the fundamental theorem shows that there exists some $x_0$ such that $p(x_0) = 0$. Then the division algorithm shows that you can write $p(x) = (x-x_0)q(x) + r(x)$, where the degree of $r$ is less than that of $x \mapsto (x-x_0)$, ie, it is a constant. Since $p(x_0) = 0$, it follows that $r(x) = r(x_0) = 0$, and we have $p(x) = (x-x_0)q(x)$, where the degree of $q$ is $n$. By assumption $q$ can be factored, hence so can $p$.