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Is there any general way to find out the coefficients of a polynomial.

Say for e.g.
$(x-a)(x-b)$ the constant term is $ab$, coefficient of $x$ is $-(a+b)$ and coefficient of $x^2$ is $1$.

I have a polynomial $(x-a)(x-b)(x-c)$. What if the number is extended to n terms.?

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On the other hand, if your polynomial is given as a "black box", the question suddenly becomes a bit more interesting... –  J. M. Aug 21 '11 at 10:47

2 Answers 2

up vote 7 down vote accepted

Try opening it to get the feel for yourself; in general, the coefficient of $x^k$ is a sum of all the products on choices of $n-k$ out of the possible $n$ roots, multiplied by $(-1)^k$. So for $(x-a)(x-b)(x-c)(x-d)$ you'll get that the coefficient of $x^2$ is $ab+ac+ad+bc+bd+cd$.

For a precise discussion see

http://en.wikipedia.org/wiki/Vi%C3%A8te%27s_formulas

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This question leads naturally to the elementary symmetric polynomials. If we have a polynomial such as $p(x) = (x - t_1)(x-t_2) \ldots (x-t_n) = \prod_{j=1}^{n}(x-t_i)$, we may write it as $x^{n} + \sum_{j=1}^{n} (-1)^{j}x^{n-j}s_{j}(t_1,\ldots ,t_n).$ Here $t_1,t_2,\ldots ,t_n$ can be elements of any (commutative) ring $R,$ and $x^{0}$ is read as 1.The term $s_{j}(t_1,t_2, \ldots,t_n)$ is a sum of $\left(\begin{array}{clcr} n\\j \end{array} \right)$ terms, and is the sum of all possible products of any $j$ of the $t_i$'s.

Another interesting direction to pursue (though I will not do it here) is that of Newton's identities, which relates the power sum polynomials to the elementary symmetric polynomials. The $k$-th power sum polynomial of the $t_i$ is $\sum_{j=1}^{n} t_{j}^{k}$. The power sum polynomials are expressible in terms of the elementary symmetric polynomials, and if division by every integer can be performed within the ring $R,$ then by Newton's identities, the elementary symmetric polynomials can be expressed (as polynomials) in terms of the power sums. For example, $t_1 t_2 = \frac{1}{2}[(t_1 +t_2)^2 - (t_1^2 + t_2 ^2)].$

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