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I am aware of the fundamental theorem of algebra, i.e., the degree of a polynomial is the number of roots of the polynomial. For example, $x^2 - 9 = 0$ would have two solutions: $x=3$ and $x=-3$. However, sometimes I come across quadratic polynomials that only have one root, e.g.,

$$t^2 - 2 t + 1 = (t-1)(t-1) = 0$$

which only has the solution $1$, or so I think. Is there some underlying concept that I am overlooking?

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  • $\begingroup$ Welcome to Math Stack Exchange. A degree n polynomial with complex coefficients has n complex roots when they are counted with multiplicity. In your example, root $1$ has multiplicity $2$ $\endgroup$ – J. W. Tanner Apr 7 at 4:40
  • $\begingroup$ Can you go further in depth , I still don't quite understand $\endgroup$ – RonRon Scores Apr 7 at 4:49
  • $\begingroup$ It appears someone else did ;-) $\endgroup$ – J. W. Tanner Apr 7 at 4:55
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    $\begingroup$ You can see it as "1 appears twice, so we double count it." This is usually written "the solution 1 has multiplicity 2". A particular solution showing up more than once has an effect, so we count each time it shows up as one solution. Think of the graphs of (x-1)(x-1)=0 and x-1=0. These are different as one is a parabola and one is a straight line. Hope this helps. $\endgroup$ – Edcookie274 Apr 7 at 5:38
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This concept is called multiplicity. The Fundamental Theorem of Algebra tells us that if we have a polynomial $p$ of degree $n$ with complex coefficients, then we can express $p$ as $p(x)=(x-r_1)(x-r_2)(x-r_3)\cdots(x-r_{n-1})(x-r_{n})$, where $r_1,r_2,r_3,...,r_{n-1},r_n$ are complex numbers (and the roots).

The Fundamental Theorem of Algebra is not violated in your example because $t^2-2t+1=(t-1)(t-1),$ and our root $r=1$ is definitely a complex number (you can also think of it think of it like $r_1=r_2=1)$. Therefore, we say that the multiplicity of our root $r=1$ is $2.$

Also, by the quadratic formula, if $b^2-4ac=0,$ then $\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt0}{2a}=\frac{-b\pm0}{2a}=\frac{-b}{2a}$, which implies that we will only have one root $r=\frac{-b}{2a}$ with multiplicity $2$.

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Two main things overlooked as far as I can tell:

  1. Multiplicities, the number of times a root occurs.

  2. Complex roots, $x^2+1=0$ has no real roots, but two complex roots i, and -i.

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