# Fundamental Theorem of Algebra. Showing polynomial as product of linear polynomials

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?

-
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. – user17794 Jul 31 '12 at 8:21
@Chanyang Ryoo Are you wanting to assume the fact mentioned by Tim, i.e., that every polynomial with complex coefficients has a complex root? – Keenan Kidwell Jul 31 '12 at 12:33
Hint: the inductive step employs the Factor Theorem, a special case of the Division Algorithm. – Bill Dubuque Jul 31 '12 at 14:37

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 $p$.