1
$\begingroup$

Polynomial $x^2$ only has the root $(0,0)$, but doesn't that go against the fundamental theorem of algebra?

And if both roots are zero, then does the FToA say we can have roots that are the same number? If so wouldn't it be more appropriate to say any degree polynomial has at most $N$ roots?

$\endgroup$
  • 3
    $\begingroup$ FToA says that a poloynomial, $p(x)$ of degree $n$ has $n$ roots up to multiplicity. The multiplicity of the root of $x^2$ is 2. $\endgroup$ – Eoin Feb 3 '15 at 6:46
  • 2
    $\begingroup$ I much prefer to just say that $p(x)$ has at least one root, and use induction to prove that $p(x)=c(x-a_1)(x-a_2)\cdots(x-a_n)$ for some set of $a_i$. That coincides with the Wikipedia definition of the theorem: "The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root." $\endgroup$ – Thomas Andrews Feb 3 '15 at 6:47
  • $\begingroup$ This has implications for partial fraction decomposition. We can decompose a fraction with distinct factors in the denominator as follows $$\frac{x+2}{x(x-1)}=\frac{3}{x-1}-\frac{2}{x}$$ But if the factors are not distinct then we need to take into consideration the multiplicity of the factors. $$\frac{x+2}{(x-1)^2} = \frac{3}{(x-1)^2}+\frac{1}{x-1}$$ $\endgroup$ – John Joy Feb 3 '15 at 15:26
1
$\begingroup$

Consider $x^2 -3x + 2 = 0$ It factorizes as $(x-1)(x-2) = 0$ and so has two clearly distinct roots $x = 1$ and $x = 2$.

Now consider $x^2 -2x +1 = 0$. It factorizes as $(x-1)(x-1) = 0$ In this case there are two factors which are zero if $x = 1$, so the root $x=1$ is considered to repeat (with "multiplicity" 2).

Now look at $x^2 = 0$. It factorizes as $x.x = 0$, or if it makes it clearer, $(x-0)(x-0) = 0$. Here the root $x = 0$ occurs twice, i.e. with multiplicity 2.

FtoA says that for a polynomial of degree n, there are n (some, possibly complex) roots if you include the multiplicity. This is perhaps more clearly expressed that a polynomial $p_n(x)$ will factorize as $(x-c_1)(x-c_2)....(x-c_n)$ where the $c_i$'s can be repeated and can be zero.

$\endgroup$
  • $\begingroup$ This makes sense, thank you :) $\endgroup$ – Veta Feb 3 '15 at 7:24
0
$\begingroup$

FToA never claimed the roots were distinct.

And if both roots are zero, then does the FToA say we can have roots that are the same number?

Have you considered reading the FToA for yourself? http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

Anyway, the point of the FToA is that every polynomial splits over $\mathbb C$ (into linear factors). This is not true over e.g. $\mathbb R$ where $f=x^2+1$ does not split.

$\endgroup$
  • $\begingroup$ Thanks, I guess that's the disadvantage of reading less technical descriptions :) $\endgroup$ – Veta Feb 3 '15 at 7:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.