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Create polynomial coefficients from its roots

I am reading the first chapter titled Numerical Solutions Of Equations And Interpolation by K.A. Stroud (Advanced Engineering Math) page 4. This is how it goes

Let $\alpha$,$\beta$,$\gamma$ be the roots of $x^3 + px^2 + qx + r = 0$. Then,writing the expression $x^3 + px^2 + qx + r$ in terms of $\alpha$, $\beta$ and $\gamma$ gives $(x - \alpha)(x - \beta)(x - \gamma)$.

$\therefore$ $x^3 + px^2 + qx + r = (x - \alpha)(x - \beta)(x - \gamma)$.

$= (x^2 - [\alpha + \beta]x + \alpha\beta)(x - \gamma)$

$= x^3 - (\alpha + \beta)x^2 + \alpha\beta x - \gamma x^2 + (\alpha + \beta)\gamma x - \alpha\beta\gamma$

$= x^3 - (\alpha + \beta + \gamma) x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha) x - \alpha\beta\gamma$

$\therefore$ equating coefficients

(a) $\alpha + \beta + \gamma = -p$.

(b) $\alpha\beta + \beta\gamma + \gamma\alpha = q$.

(c) $\alpha\beta\gamma = -r$.

This, of course, applies to a cubic equation. Let us extend this to a more general equation

In general, if $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3},\ldots,\alpha_{n}$ are the roots of the equation $p_{0}x^n + p_{1} x^{n-1} + p_{2}x^{n-2} + \cdots + p_{n-1}x + p_{n} = 0$ where ($p_{0}\neq0$) then

sum of the roots $= \frac{-p_{1}}{p_{0}}$

sum of products of the roots, two at a time $= \frac{p_{2}}{p_{0}}$

sum of products if the roots, three at a time $= \frac{-p_{3}}{p_{0}}$

sum of products of the roots, $n$ at a time = $(-1)^n,\frac{p_{n}}{p}_{0}$

I was able to understand the cubic equation's part but I am completely lost with the general part (i.e an $n$th degree polynomial). I am looking for a simpler explanation of what that means. However I understand that it can be used as formulas for finding the roots but i need to know how did he obtain the above formulas.

Thanks in advance.


marked as duplicate by J. M. is a poor mathematician, Arturo Magidin, Asaf Karagila, Sasha, Mike Spivey Dec 7 '11 at 4:20

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The Factor Theorem says that if a polynomial $p(x)$ has root $r$, then $x-r$ divides $p(x)$.

Since polynomial with complex coefficients always have exactly the same number of roots as its degree (counting multiplicity), and they have unique factorization, that means that if $p(x)$ is the polynomial $$x^n + p_1x^{n-1} + \cdots + p_{n-1}x + p_0$$ (note that I am doing the monic case first; I'll deal with the general case later) and $r_1,\ldots,r_n$ are the roots of $p(x)$, then we can factor $p(x)$ as $$p(x) = x^n + p_1x^{n-1}+\cdots + p_{n-1}x + p_0 = (x-r_1)(x-r_2)\cdots(x-r_n).$$

When you multiply out $(x-r_1)\cdots(x-r_n)$, and then put together the powers of $x$, you get a polynomial expression in which the coefficients are written in terms of the roots. For example, to get the coefficient of $x$, note that you will get an $x$ when you expand and multiply the $x$ in the first binomial by each of the constants in the rest, giving you $(-1)^{n-1}(r_2\cdots r_n)x$. Another when you multiply the $x$ in the second binomial by each of the constants in the rest; another when you multiply the $x$ in the third binomial by the constants in the rest; etc.

If you work this out, you'll find that when you expand $(x-r_1)\cdots(x-r_n)$ and then group together the powers of $x$, you will have:

  • The coefficient of $x^n$ is $1$.
  • The coefficient of $x^{n-1}$ is $-(r_1+\cdots + r_n)$.
  • The coefficient of $x^{n-2}$ is $$r_1r_2 + r_1r_3 + \cdots +r_1r_n + r_2r_3+\cdots + r_{n-1}r_n.$$
  • The coefficient of $x^{n-3}$ is $$-\bigl(r_1r_2r_3 + r_1r_2r_4+\cdots + r_1r_2r_n + r_1r_3r_4 + \cdots + r_{n-2}r_{n-1}r_n\bigr).$$
  • $\vdots$
  • The coefficient of $x$ is $$(-1)^{n-1}\bigl(r_1r_2\cdots r_{n-1} + r_1r_2\cdots r_{n-2}r_n +\cdots + r_2\cdots r_n\bigr).$$
  • The constant coefficient is $$(-1)^n\bigl(r_1\cdots r_n\bigr).$$

But for two polynomials to be equal they have to be equal coefficient by coefficient. So that means that: $$\begin{align*} p_1 &= (-1)^1\bigl( r_1+\cdots + r_n\bigr)\\ p_2 &= (-1)^2\bigl( r_1r_2+r_1r_3+\cdots + r_1r_n + r_2r_3+\cdots + r_{n-1}r_n\bigr)\\ p_3 &= (-1)^3\bigl( r_1r_2r_3 + r_1r_2r_4+\cdots + r_1r_2r_n + r_1r_3r_4+\cdots + r_{n-2}r_{n-1}r_n\bigr)\\ &\vdots\\ p_n &= (-1)^n\bigl(r_1\cdots r_n\bigr). \end{align*}$$ That is:

  • The sum of the roots is $-p_1$;
  • The sum of all products of two roots is $(-1)^2p_2$;
  • The sum of all products of three roots is $(-1)^3p_3$;
  • The sum of all products of four roots is $(-1)^4p_4$;
  • $\vdots$
  • The sum of all products of $n-1$ roots is $(-1)^{n-1}p_{n-1}$;
  • The product of all roots is $(-1)^np_n$.

Which is what the text you cite says.

How about a polynomial that is not monic, $$p(x) = p_0x^n + p_1x^{n-1}+\cdots + p_{n-1}x + p_n,\qquad p_0\neq 0\ ?$$

Note that $r$ is a root of $p(x)$ if and only if it is a root of $$P(x) = \frac{1}{p_0}p(x) = x^n + \frac{p_1}{p_0}x^{n-1} + \cdots + \frac{p_{n-1}}{p_0}x + \frac{p_n}{p_0}.$$ So the argument above applies to $P(x)$.

  • $\begingroup$ for some reason i'am unable to understand the part where it says "When you multiply out (x−$r_{1}$ )⋯(x−$r_{n}$ )" till the end.I've been reading this over and over again but i'am sorry i don't get it. $\endgroup$ – alok Dec 7 '11 at 7:44
  • $\begingroup$ Don't just read it; try it out with some examples (say, with $n=4$; with $n-5$; etc). When you multiply out $(x-r_1)\cdots(x-r_n)$, how do you get a term that has $x^{n-1}$ in it? You get one when you take the product of the $x$s in the first $n-1$ binomials and the $-r_n$ in the last monomial; you get another one when you take the product of the $x$s in the first $n-2$ binomials, the $x$ in the last binomial, and the $-r_{n-1}$ in the penultimate binomial. Another one when you multiply the $x$s in the first $n-3$ and last two binomials, times $-r_{n-2}$ in the antepenultimate one (cont) $\endgroup$ – Arturo Magidin Dec 7 '11 at 19:55
  • $\begingroup$ @alok: And so on. These are the only terms with $x^{n-1}$ in it. In the end, you get one for each constant terms, so you will have $$-r_nx^{n-1}-r_{n-1}x^{n-1}-r_{n-2}x^{n-1}-\cdots - r_1x^{n-1} = -(r_1+\cdots+r_n)x^{n-1}.$$ How do you get terms with $x^{n-2}$ in them? When you multiply out with $n-2$ $x$s and two constants from the binomials. So you get one term for each pair of constants you can pick. In the end, you end up with $$r_1r_2x^{n-2}+r_1r_3x^{n-2}+\cdots+r_{n-1}r_{n}x^{n-2} = (r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)x^{n-2}.$$Same with the other powers of $x$. $\endgroup$ – Arturo Magidin Dec 7 '11 at 19:58
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    $\begingroup$ @alok: P.S. If you don't understand, don't accept the answer! $\endgroup$ – Arturo Magidin Dec 7 '11 at 19:58

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