1
$\begingroup$

I just started learning Algebra 2, and came across the Fundamental Theorem of Algebra.

Here is how my Textbook defines it:

Every Single-Variable Polynomial Function of Degree $n \ge 1$ has at least one zero in the the set of complex numbers.

I cannot really comprehend this rule.

Take, for example, the polynomial $f(x) = x^2 + 3x + 2$ When factored, we get $(x + 2)(x + 1)$

The roots are $ x = -1, -2 $

There are no complex roots in this equation. Doesn't that disprove the Fundamental Theorem of Algebra?

$\endgroup$
6
  • 3
    $\begingroup$ It seems you're thinking that complex numbers are only imaginary; complex numbers can be both real and imaginary (i.e. imaginary numbers are a subset of the complex numbers). $\endgroup$
    – Kman3
    Jun 21, 2021 at 20:54
  • $\begingroup$ Real numbers are complex numbers with zero imaginary part. $\endgroup$
    – herb steinberg
    Jun 21, 2021 at 20:55
  • $\begingroup$ The real numbers are a subset of complex numbers, think like how the integers are a subset of rational numbers. So this is a polynomial of degree 2 with 2 complex roots, which matches the Fundamental Theorem of Algebra. $\endgroup$
    – Stephen Donovan
    Jun 21, 2021 at 20:58
  • 1
    $\begingroup$ The definition is poor. Every n-th degree polynomial has exactly n complex roots (some of which may be duplicates: we say a duplicated root has a multiplicity $> 1$), If the roots are called $c_1, c_2, ... c_n$, then the polynomial (say it is in a variable called $z$) can be rewritten as the product of difference factors $(z - c1)(z - c2) .... (z - cn)$. Here, some of these roots could be the same, e.g. peraps $c3 = c4$. $\endgroup$
    – Kaz
    Jun 22, 2021 at 5:32
  • $\begingroup$ You can see that the formula $(z - c1)(z - c2) .... (z - cn)$ is necessarily zero, whenever $z$ takes on one of the values $c_1, c_2, ... c_n$, because for every such value $z$, one of the difference terms vanishes. Anyway, the big deal is that the complex numbers "cause algebra to be complete". No matter the coefficients in the n-th degree formula, it has that many roots and can be put in the product of differences form. $\endgroup$
    – Kaz
    Jun 22, 2021 at 5:36

2 Answers 2

Reset to default
4
$\begingroup$

Recall that a complex number is a number of the form $a+bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit ($i^2=-1$). Real numbers, as mentioned in the comments, are complex numbers with $b=0$. You have two real roots, and since real numbers are part of the complex numbers, you therefore have two complex roots, and the theorem holds.

$\endgroup$
4
  • $\begingroup$ Thanks everyone. It was just phrased in such a weird format. I think I get it now $\endgroup$
    – Achintya Agrawal
    Jun 21, 2021 at 20:59
  • 5
    $\begingroup$ I see no "weird format" here- just your misunderstanding of "complex number". $\endgroup$
    – user247327
    Jun 21, 2021 at 21:02
  • 1
    $\begingroup$ @AchintyaAgrawal Not a problem. Your textbook does phrase the theorem a bit strangely, at least to me. I would phrase the Fundamental Theorem of Algebra as "Every polynomial of degree $n$ has $n$ complex roots." $\endgroup$
    – Kman3
    Jun 21, 2021 at 21:04
  • 1
    $\begingroup$ @Kman3: Often I have seen the Fundamental Theorem of Algebra been written as "every polynomial has at least $1$ complex root", and from this it can be proven that every polynomial of degree $n$ has $n$ roots by repeatedly using polynomial division. $\endgroup$
    – Joe
    Jun 21, 2021 at 21:08
2
$\begingroup$

In the same way that natural numbers are also integers, integers are also rational numbers, and rational numbers are also real numbers, real numbers are also complex numbers (with imaginary part zero). So $x^2+3x+2$ does have two real roots (indeed, it has two integer roots); but this does not mean that it has doesn't have two complex roots.

If you want a describe a complex number that is not a real number, the correct term is non-real.* However, sometimes people use "complex number" to mean "non-real number", which is what led to your question in the first place.

*If you want to be super-precise, you should say non-real and complex, but usually there is no need to be so explicit.

$\endgroup$

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.