# Show that a Polynomial has certain factorization

$P(x)$ is a polynomial in $x$ of degree $\leq n-1$.

Show that $P(x)$ has $n-1$ distinct roots and thus has the factorization $$k\Pi_{i=2}^n(x-a_i)$$, where the constant $k$ is the coefficient of $x^{n-1}$.

Why does $i$ start from 2? Why is the $k$ there?

Edit: I've seen the "correct" answer. It says "A polynomial with degree (n-1) has n-1 distinct roots. Then it must take the form $k\Pi_{i=2}^n(x-a_i)$".

This seems that the answer just restates the question. So I'm not any better off understanding this question.

• What context, algebra-precalculus, analysis, or abstract algebra? Apr 7, 2014 at 2:04
• linear algebra. In the context of the Vandermonde matrix Apr 7, 2014 at 2:11
• In that case you need to state the entire problem. Apr 7, 2014 at 2:23
• It's not true that every polynomial of degree $n-1$ in the world has $n-1$ distinct roots (e.g., $x^{n-1}$). You'll have to use some information about your specific polynomial. I'm guessing that you're supposed to see directly that each $a_i$ is a root (probably because the determinant defining $P$ obviously vanishes when $x=a_i$). Apr 7, 2014 at 2:51

Echoing Greg Martin: In case that you are exploring the Vandermonde matrix $V_n(x,a_2,a_3,...,a_n)$, the usual task is to show

1. first that it is a polynomial that can not have a degree higher than $n-1$,
2. then that $x=a_i$, $k=i,...,n$ are roots of this polynomial
3. and that since $n-1$ roots have been located, the factorization $$V_n(x,a_2,...,a_n)=k\prod_{i=2}^n(x-a_i)$$ can only have $k$ as a constant;
4. in a last step, one then concludes that $k=(-1)^{n-1}\,V_{n-1}(a_2,...,a_n)$.

That the factorization has this form is a consequence of the (Horner-)Ruffini theorem/rule.

This then gives the recursion $$V_n(a_1,...,a_n)=(-1)^{n-1}\prod_{i=2}^n(a_1-a_i)\,V_{n-1}(a_2,...,a_n)=\prod_{i=2}^n(a_i-a_1)\,V_{n-1}(a_2,...,a_n)$$ which leads to $$V_n(a_1,...,a_n)=\prod_{i<j}(a_j-a_i)$$